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Efficiency of 200-ohm hairpin matching
I have recently mounted a 6-meter long Yagi. The driven element is a
(non-folded) dipole whose impedance is brought up to 200 ohm by significantly shortening it (so introducing a high series capacitive reactance) and using a (rather small) hairpin to resonate the residual reactance. The system is fed by a 4-to-1 balun (200-to-50 ohm) made of a half-wavelength cable. The SWR at resonance is almost perfect, but the SWR response vs. frequency is very sharp, this meaning that the system has a high Q. I am wondering whether such a 200-ohm feed technique could cause non negligible losses caused by the higher current circulating in the hairpin (due to its low reactance as well as to the high voltage on the high-impedance feed point, that is 200 ohm + reactance of the shortened driven element). 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On 6 abr, 23:43, "Antonio Vernucci" wrote:
I have recently mounted a 6-meter long Yagi. The driven element is a (non-folded) dipole whose impedance is brought up to 200 ohm by significantly shortening it (so introducing a high series capacitive reactance) and using a (rather small) hairpin to resonate the residual reactance. The system is fed by a 4-to-1 balun (200-to-50 ohm) made of a half-wavelength cable. The SWR at resonance is almost perfect, but the SWR response vs. frequency is very sharp, this meaning that the system has a high Q. I am wondering whether such a 200-ohm feed technique could cause non negligible losses caused by the higher current circulating in the hairpin (due to its low reactance as well as to the high voltage on the high-impedance feed point, that is 200 ohm + reactance of the shortened driven element). 73 Tony I0JX Hello Antonio, When I understand you correctly, the series component of the dipole input impedance is 200 Ohm? I would expect value lower then 70 Ohms. When 200 Ohms real component of Zi, you used a series hairpin to cancel the capacitive component. You can guess the loss by calculating the AC resistance of the pin. Assuming homogeneous current distribution around the circumference, you can calculate the AC resistance with (assuming length of hairpin 0.15 lambda): Rac = 80*10^-9*sqrt(f)*le/d , Rac in Ohm, f in Hz, le (wire length) in m, d (diameter of wire) in m, valid for copper. For other materials (non magnetic), when conductance is factor 4 less (so 4 times higher spec. resistance), Rac doubles. When the result for your hairpin 200 Ohm, the loss is negligible. In case of a parallel pin, you can use same technique, but in that case you have to calculate the inductance of the pin. With the inductance you know the reactance and the current that runs through it (for example with 100Vrms across it). With Rac you can calculate the power loss. Hope this helps a bit. Best regards, Wim PA3DJS www.tetech.nl remove abc from the mail address. |
Efficiency of 200-ohm hairpin matching
When I understand you correctly, the series component of the dipole
input impedance is 200 Ohm? I would expect value lower then 70 Ohms. the series component is probably just 25 ohm or so (due to the effect of the parasitic elements), but it is brought up to 200 ohm by shortening the dipole length and putting an hairpin in parallel (very well known technique to increase impedance). As a matter fact it matches very well with a 4-to-1 balun When 200 Ohms real component of Zi, you used a series hairpin to cancel the capacitive component. No, the hairpin is in parallel to the dipole You can guess the loss by calculating the AC resistance of the pin. Assuming homogeneous current distribution around the circumference, you can calculate the AC resistance with (assuming length of hairpin 0.15 lambda): Rac = 80*10^-9*sqrt(f)*le/d , Rac in Ohm, f in Hz, le (wire length) in m, d (diameter of wire) in m, valid for copper. For other materials (non magnetic), when conductance is factor 4 less (so 4 times higher spec. resistance), Rac doubles. When the result for your hairpin 200 Ohm, the loss is negligible. In case of a parallel pin, you can use same technique, but in that case you have to calculate the inductance of the pin. With the inductance you know the reactance and the current that runs through it (for example with 100Vrms across it). With Rac you can calculate the power loss. Having all data available, it is clearly possible to calculate losses using the Ohm law. Mine was a question of more general nature. Does anyone have direct experience on whether the extra loss of a 200-ohm hairpin match (with respect to the more conventional 50-ohm hairpin match) is in pratice noticeable? Thanks & 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
"Antonio Vernucci" wrote in message ... I have recently mounted a 6-meter long Yagi. The driven element is a (non-folded) dipole whose impedance is brought up to 200 ohm by significantly shortening it (so introducing a high series capacitive reactance) and using a (rather small) hairpin to resonate the residual reactance. The system is fed by a 4-to-1 balun (200-to-50 ohm) made of a half-wavelength cable. The SWR at resonance is almost perfect, but the SWR response vs. frequency is very sharp, this meaning that the system has a high Q. I am wondering whether such a 200-ohm feed technique could cause non negligible losses caused by the higher current circulating in the hairpin (due to its low reactance as well as to the high voltage on the high-impedance feed point, that is 200 ohm + reactance of the shortened driven element). 73 Tony I0JX I wonder what you consider narrow bandwidth. 500 KHz seems to be about par for a decent 6 m beam. Tam/WB2TT |
Efficiency of 200-ohm hairpin matching
On 7 abr, 00:33, "Antonio Vernucci" wrote:
When I understand you correctly, the series component of the dipole input impedance is 200 Ohm? I would expect value lower then 70 Ohms. the series component is probably just 25 ohm or so (due to the effect of the parasitic elements), but it is brought up to 200 ohm by shortening the dipole length and putting an hairpin in parallel (very well known technique to increase impedance). As a matter fact it matches very well with a 4-to-1 balun When 200 Ohms real component of Zi, you used a series hairpin to cancel the capacitive component. No, the hairpin is in parallel to the dipole You can guess the loss by calculating the AC resistance of the pin. Assuming homogeneous current distribution around the circumference, you can calculate the AC resistance with (assuming length of hairpin 0.15 lambda): Rac = 80*10^-9*sqrt(f)*le/d , Rac in Ohm, f in Hz, le (wire length) in m, d (diameter of wire) in m, valid for copper. For other materials (non magnetic), when conductance is factor 4 less (so 4 times higher spec. resistance), Rac doubles. When the result for your hairpin 200 Ohm, the loss is negligible. In case of a parallel pin, you can use same technique, but in that case you have to calculate the inductance of the pin. With the inductance you know the reactance and the current that runs through it (for example with 100Vrms across it). With Rac you can calculate the power loss. Having all data available, it is clearly possible to calculate losses using the Ohm law. Mine was a question of more general nature. Does anyone have direct experience on whether the extra loss of a 200-ohm hairpin match (with respect to the more conventional 50-ohm hairpin match) is in pratice noticeable? Thanks & 73 Tony I0JX Hello Tony, Given your data and some assumptions about hairpin construction and use, loss is negligible. Best regards, Wim PA3DJS www.tetech.nl please remove abc from the mail address. |
Efficiency of 200-ohm hairpin matching
I wonder what you consider narrow bandwidth. 500 KHz seems to be about par
for a decent 6 m beam. This antenna shows an SWR of 1.7 at 150 kHz below the resonant frequency, where the SWR is just 1, so I consider it narrow. Another antenna I have, also using an hairpin match but at 50 ohm instead of 200 ohm, is by far broader. On both antennas I have about 100 feet of low-loss 1/4" Andrew hardline, so the SWR is only slightly influenced by cable loss. Talking with the manfacturer, he told me that he preferred rasing the antenna impedance up to 200 ohm, instead than to 50 ohm, because a 4:1 cable balun can me more easily realized than a 1:1 balun. But doing so the feed system Q factor increases quite a lot, causing a significant bandwidth reduction. Moreover ohmic losses increase, due to the higher circulating currents (for a given RF power), but I am not able to predict whether such extra losses are significant or can be disregarded for all practical purposes. 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
"Antonio Vernucci" wrote in message ... I wonder what you consider narrow bandwidth. 500 KHz seems to be about par for a decent 6 m beam. This antenna shows an SWR of 1.7 at 150 kHz below the resonant frequency, where the SWR is just 1, so I consider it narrow. Another antenna I have, also using an hairpin match but at 50 ohm instead of 200 ohm, is by far broader. On both antennas I have about 100 feet of low-loss 1/4" Andrew hardline, so the SWR is only slightly influenced by cable loss. Talking with the manfacturer, he told me that he preferred rasing the antenna impedance up to 200 ohm, instead than to 50 ohm, because a 4:1 cable balun can me more easily realized than a 1:1 balun. But doing so the feed system Q factor increases quite a lot, causing a significant bandwidth reduction. Moreover ohmic losses increase, due to the higher circulating currents (for a given RF power), but I am not able to predict whether such extra losses are significant or can be disregarded for all practical purposes. 73 Tony I0JX I don't think it is the balun. I measured a 4:1, 1/2 wave 6m balun made of LMR240 and got a 2:1 SWR bandwidth from 40 to 60 MHz. This was with an MFJ-269 and about a foot of 50 Ohm coax. Load was a 1/2 W 180 Ohm resistor (measured high). Tam/WB2TT |
Efficiency of 200-ohm hairpin matching
On Apr 7, 8:36 am, "Antonio Vernucci" wrote:
I wonder what you consider narrow bandwidth. 500 KHz seems to be about par for a decent 6 m beam. This antenna shows an SWR of 1.7 at 150 kHz below the resonant frequency, where the SWR is just 1, so I consider it narrow. Another antenna I have, also using an hairpin match but at 50 ohm instead of 200 ohm, is by far broader. On both antennas I have about 100 feet of low-loss 1/4" Andrew hardline, so the SWR is only slightly influenced by cable loss. Talking with the manfacturer, he told me that he preferred rasing the antenna impedance up to 200 ohm, instead than to 50 ohm, because a 4:1 cable balun can me more easily realized than a 1:1 balun. But doing so the feed system Q factor increases quite a lot, causing a significant bandwidth reduction. Moreover ohmic losses increase, due to the higher circulating currents (for a given RF power), but I am not able to predict whether such extra losses are significant or can be disregarded for all practical purposes. 73 Tony I0JX I'm trying to find a good way to wrap my mind around this "general" problem. I'm not at a point I'm really comfortable with yet, but I offer the following ideas as "food for thought." First, the matching is being done essentially with an "L" network (or rather the balanced version of an "L" network), where there is a load resistance (the resistive part of the feedpoint impedance, which includes radiation resistance and element loss resistance reflected to the feedpoint), the series capacitive reactance of the feedpoing element, and a shunt inductive reactance, provided by the hairpin. Because shortening the driven element causes a decrease in resistance and an increase in capacitive reactance, it's possible to find a length that allows matching to any of a wide range of resistances. But the higher the resistance to which you match, the shorter you need to make the element and the lower the feedpoint resistance. The ratio of feedpoint resistance to matched resistance determines the loaded Q of the matching network; as you make the matched resistance higher, the loaded Q goes up rather quickly. If you know the loaded Q and the unloaded Q of the hairpin, you have a good handle on the amount lost to heat in the hairpin: if the hairpin Q is two times the loaded Q, half the power is dissipated in the hairpin, for example. However, unless you build an antenna with a very low feedpoint resistance at resonance, there almost certainly won't be an efficiency problem: the reactance changes quickly enough with changes in driven element length that the resistance won't drop much by the time you reach enough reactance to get a match to 200 ohms. It appears that the loaded Q of the match to 200 ohms for your case will be less than 4. I would think unless you really messed up badly, the hairpin unloaded Q should be well in excess of 100, and if that's the case, the power lost in the hairpin would correspond to well under 0.1dB signal level change. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
Tony I0JX
I don't think it is the balun. I measured a 4:1, 1/2 wave 6m balun made of LMR240 and got a 2:1 SWR bandwidth from 40 to 60 MHz. This was with an MFJ-269 and about a foot of 50 Ohm coax. Load was a 1/2 W 180 Ohm resistor (measured high). You are right, it is not a balun problem. It is a problem due to the reasons I had explained in my post, which had nothing to do with the balun. 73 Tony I0JX. |
Efficiency of 200-ohm hairpin matching
Hello Tom,
First, the matching is being done essentially with an "L" network (or rather the balanced version of an "L" network), where there is a load resistance (the resistive part of the feedpoint impedance, which includes radiation resistance and element loss resistance reflected to the feedpoint), the series capacitive reactance of the feedpoing element, and a shunt inductive reactance, provided by the hairpin. Because shortening the driven element causes a decrease in resistance and an increase in capacitive reactance, it's possible to find a length that allows matching to any of a wide range of resistances. But the higher the resistance to which you match, the shorter you need to make the element and the lower the feedpoint resistance. The ratio of feedpoint resistance to matched resistance determines the loaded Q of the matching network; as you make the matched resistance higher, the loaded Q goes up rather quickly. If you know the loaded Q and the unloaded Q of the hairpin, you have a good handle on the amount lost to heat in the hairpin: if the hairpin Q is two times the loaded Q, half the power is dissipated in the hairpin, for example. However, unless you build an antenna with a very low feedpoint resistance at resonance, there almost certainly won't be an efficiency problem: the reactance changes quickly enough with changes in driven element length that the resistance won't drop much by the time you reach enough reactance to get a match to 200 ohms. It appears that the loaded Q of the match to 200 ohms for your case will be less than 4. I would think unless you really messed up badly, the hairpin unloaded Q should be well in excess of 100, and if that's the case, the power lost in the hairpin would correspond to well under 0.1dB signal level change. All OK. I however reckon that, due to the parasitic elements effect, the radiation resistance of the driven element (before shortening it) would be in the order of 20 ohm. So, impedance gets brought up by a factor of 10 or so. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. His argument is that, for high-power operation (say 1500W), it is more convenient for him to provide a quarter-wavelength 4:1 balun made of RG-142 teflon cable than a sealed box with a 1:1 coil on an ironpowder toroidal core. I mentioned him that, just using some extra length of RG-142 cable, he can easily build a 1:1 balun, and hence design the antenna matching system for 50 ohm instead of 200 ohm. I hope he will listen to me, because that antenna is really narrowband! 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
"K7ITM" wrote in message ... On Apr 7, 8:36 am, "Antonio Vernucci" wrote: .................................................. .................. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. Cheers, Tom Also, I think the 4:1 , 1/2 wave would be a voltage balun. I am considering adding a ferrite 1:1 on the feedline side of the the coax balun. There is some feedline radiation right now. Tam/WB2TT |
Efficiency of 200-ohm hairpin matching
On 7 abr, 23:15, "Antonio Vernucci" wrote:
Hello Tom, First, the matching is being done essentially with an "L" network (or rather the balanced version of an "L" network), where there is a load resistance (the resistive part of the feedpoint impedance, which includes radiation resistance and element loss resistance reflected to the feedpoint), the series capacitive reactance of the feedpoing element, and a shunt inductive reactance, provided by the hairpin. Because shortening the driven element causes a decrease in resistance and an increase in capacitive reactance, it's possible to find a length that allows matching to any of a wide range of resistances. But the higher the resistance to which you match, the shorter you need to make the element and the lower the feedpoint resistance. The ratio of feedpoint resistance to matched resistance determines the loaded Q of the matching network; as you make the matched resistance higher, the loaded Q goes up rather quickly. If you know the loaded Q and the unloaded Q of the hairpin, you have a good handle on the amount lost to heat in the hairpin: if the hairpin Q is two times the loaded Q, half the power is dissipated in the hairpin, for example. However, unless you build an antenna with a very low feedpoint resistance at resonance, there almost certainly won't be an efficiency problem: the reactance changes quickly enough with changes in driven element length that the resistance won't drop much by the time you reach enough reactance to get a match to 200 ohms. It appears that the loaded Q of the match to 200 ohms for your case will be less than 4. I would think unless you really messed up badly, the hairpin unloaded Q should be well in excess of 100, and if that's the case, the power lost in the hairpin would correspond to well under 0.1dB signal level change. All OK. I however reckon that, due to the parasitic elements effect, the radiation resistance of the driven element (before shortening it) would be in the order of 20 ohm. So, impedance gets brought up by a factor of 10 or so. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. His argument is that, for high-power operation (say 1500W), it is more convenient for him to provide a quarter-wavelength 4:1 balun made of RG-142 teflon cable than a sealed box with a 1:1 coil on an ironpowder toroidal core. I mentioned him that, just using some extra length of RG-142 cable, he can easily build a 1:1 balun, and hence design the antenna matching system for 50 ohm instead of 200 ohm. I hope he will listen to me, because that antenna is really narrowband! 73 Tony I0JX Hello, Imagine 10 Ohms radiation resistance + capacitive component after shortening the open dipole radiator. When you convert that to 200 Ohms via a parallel inductance (hairpin), the Q factor of such a network is about 4.4. At 50 MHz that will result in a bandwidth (VSWR=2) of 8 MHz. A 200 Ohms to 50 Ohms coaxial balun will have a lower Q factor. So the combination of balun and L-network will certainly have a useful bandwidth 340 kHz. Maybe the radiation resistance is less (you can derive that from the hairpin inductance) and/or the antenna is by nature (very) narrow band. Best regards, Wim PA3DJS www.tetech.nl remove abc from the address. |
Efficiency of 200-ohm hairpin matching
On Apr 7, 4:02 pm, Wimpie wrote:
On 7 abr, 23:15, "Antonio Vernucci" wrote: Hello Tom, First, the matching is being done essentially with an "L" network (or rather the balanced version of an "L" network), where there is a load resistance (the resistive part of the feedpoint impedance, which includes radiation resistance and element loss resistance reflected to the feedpoint), the series capacitive reactance of the feedpoing element, and a shunt inductive reactance, provided by the hairpin. Because shortening the driven element causes a decrease in resistance and an increase in capacitive reactance, it's possible to find a length that allows matching to any of a wide range of resistances. But the higher the resistance to which you match, the shorter you need to make the element and the lower the feedpoint resistance. The ratio of feedpoint resistance to matched resistance determines the loaded Q of the matching network; as you make the matched resistance higher, the loaded Q goes up rather quickly. If you know the loaded Q and the unloaded Q of the hairpin, you have a good handle on the amount lost to heat in the hairpin: if the hairpin Q is two times the loaded Q, half the power is dissipated in the hairpin, for example. However, unless you build an antenna with a very low feedpoint resistance at resonance, there almost certainly won't be an efficiency problem: the reactance changes quickly enough with changes in driven element length that the resistance won't drop much by the time you reach enough reactance to get a match to 200 ohms. It appears that the loaded Q of the match to 200 ohms for your case will be less than 4. I would think unless you really messed up badly, the hairpin unloaded Q should be well in excess of 100, and if that's the case, the power lost in the hairpin would correspond to well under 0.1dB signal level change. All OK. I however reckon that, due to the parasitic elements effect, the radiation resistance of the driven element (before shortening it) would be in the order of 20 ohm. So, impedance gets brought up by a factor of 10 or so. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. His argument is that, for high-power operation (say 1500W), it is more convenient for him to provide a quarter-wavelength 4:1 balun made of RG-142 teflon cable than a sealed box with a 1:1 coil on an ironpowder toroidal core. I mentioned him that, just using some extra length of RG-142 cable, he can easily build a 1:1 balun, and hence design the antenna matching system for 50 ohm instead of 200 ohm. I hope he will listen to me, because that antenna is really narrowband! 73 Tony I0JX Hello, Imagine 10 Ohms radiation resistance + capacitive component after shortening the open dipole radiator. When you convert that to 200 Ohms via a parallel inductance (hairpin), the Q factor of such a network is about 4.4. At 50 MHz that will result in a bandwidth (VSWR=2) of 8 MHz. A 200 Ohms to 50 Ohms coaxial balun will have a lower Q factor. So the combination of balun and L-network will certainly have a useful bandwidth 340 kHz. Maybe the radiation resistance is less (you can derive that from the hairpin inductance) and/or the antenna is by nature (very) narrow band. Best regards, Wim PA3DJSwww.tetech.nl remove abc from the address. Yes, I had similar thoughts, but a bit different. First, I think it's safe to say that if, at resonance, the driven element presents about 20 ohms at the feedpoint, shortening the D.E. only a little will give enough capacitive reactance to allow the hairpin match to 200 ohms. Even if the D.E. looks like 5 ohms, the "L" network match still gives a 3dB bandwidth of 8MHz at a 50MHz center, or 3MHz 1.5:1 SWR bandwidth. I think we need to look somewhere else for the answer to the narrow bandwidth. My working hypothesis at the moment is that it's in the antenna, or perhaps rather in the combination of antenna and matching network. Note that the calc for the L match assumed a constant capacitance, but the antenna will not, in general, look anything like a constant C even over a fairly narrow frequency range. I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. Over that range, the equivalent series capacitance changes from 59pF at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. Is it possible to lengthen the D.E., causing it to present an inductive reactance at the feedpoint, and match that (to 200 ohms, or to 50 ohms) with a shunt capacitance? That may work better, giving a broader SWR bandwidth. The resistive component should be higher, further lowering the Q, and I suspect the reactance won't change so quickly with frequency. I don't have time at the moment to compare the two, but may this evening. It may also be possible to raise the resistive part to 50 ohms and match to 50 with a series capacitance. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
K7ITM wrote:
Is it possible to lengthen the D.E., causing it to present an inductive reactance at the feedpoint, and match that (to 200 ohms, or to 50 ohms) with a shunt capacitance? That may work better, giving a broader SWR bandwidth.... Good, constructive suggestion, Tom. Kudos for putting in a bit of design/analysis effort on this rather than just shooting from the hip. This is definitely a weird way of driving a yagi. It makes me yearn for the old TV-antenna schemes that used folded dipoles for the driven element, suitably split-up between upper and lower wires so as to give a 300-ohm terminal impedance even with the resistance-lowering effects of the reflector and directors. Jim, K7JEB |
Efficiency of 200-ohm hairpin matching
Over that range, the equivalent series capacitance changes from 59pF
at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. That is exactly the point! It would not be correct to calculate bandwidth on the basis of the Q factor at resonance and assuming that the capacitive antenna reactance is equivalent to that of a fixed capacitor. Today I have discovered another shortcoming of that antenna. After raining cats and dogs, the antenna resonant frequency gets lowered by about 130 kHz due to the influence of the wet terrain. That is really a lot if you consider that, after making very accurate measurements with a Bird wattmeter, the antenna bandwidth is only 100 kHz at 1.4 SWR! I am considering to re-build the driven element for 50-ohm match, by using a longer driven element and a 1:1 balun. However it will not be easy to find the optimum situation because there are two variables to be adjusted, that is the driven element length and the hairpin length. Also, I am not too sure on to which extent using a longer driven element would influence the antenna radiation pattern. Any comment? 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
Over that range, the equivalent series capacitance changes from 59pF
at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. That is exactly the point! It would not be correct to calculate bandwidth on the basis of the Q factor at resonance and assuming that the capacitive antenna reactance is equivalent to that of a fixed capacitor. Today I have discovered another shortcoming of that antenna. After raining cats and dogs, the antenna resonant frequency gets lowered by about 130 kHz due to the influence of the wet terrain. That is really a lot if you consider that, after making very accurate measurements with a Bird wattmeter, the antenna bandwidth is only 100 kHz at 1.4 SWR! I am considering to re-build the driven element for 50-ohm match, by using a longer driven element and a 1:1 balun. However it will not be easy to find the optimum situation because there are two variables to be adjusted, that is the driven element length and the hairpin length. Also, I am not too sure on to which extent using a longer driven element would influence the antenna radiation pattern. Any comment? 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
Over that range, the equivalent series capacitance changes from 59pF
at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. That is exactly the point! It would not be correct to calculate bandwidth on the basis of the Q factor at resonance and assuming that the capacitive antenna reactance is equivalent to that of a fixed capacitor. Today I have discovered another shortcoming of that antenna. After raining cats and dogs, the antenna resonant frequency gets lowered by about 130 kHz due to the influence of the wet terrain. That is really a lot if you consider that, after making very accurate measurements with a Bird wattmeter, the antenna bandwidth is only 100 kHz at 1.4 SWR! I am considering to re-build the driven element for 50-ohm match, by using a longer driven element and a 1:1 balun. However it will not be easy to find the optimum situation because there are two variables to be adjusted, that is the driven element length and the hairpin length. Also, I am not too sure on to which extent using a longer driven element would influence the antenna radiation pattern. Any comment? 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On Apr 8, 12:31 pm, "Antonio Vernucci" wrote:
Over that range, the equivalent series capacitance changes from 59pF at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. That is exactly the point! It would not be correct to calculate bandwidth on the basis of the Q factor at resonance and assuming that the capacitive antenna reactance is equivalent to that of a fixed capacitor. Today I have discovered another shortcoming of that antenna. After raining cats and dogs, the antenna resonant frequency gets lowered by about 130 kHz due to the influence of the wet terrain. That is really a lot if you consider that, after making very accurate measurements with a Bird wattmeter, the antenna bandwidth is only 100 kHz at 1.4 SWR! I am considering to re-build the driven element for 50-ohm match, by using a longer driven element and a 1:1 balun. However it will not be easy to find the optimum situation because there are two variables to be adjusted, that is the driven element length and the hairpin length. Also, I am not too sure on to which extent using a longer driven element would influence the antenna radiation pattern. Any comment? 73 Tony I0JX Though the Q calculation doesn't give the right SWR bandwidth for the antenna/matching system, it does tell you that (with such a low loaded Q), it should not be difficult to make a hairpin or even standard helical coil inductor that has low enough loss that you can ignore the effect. I believe that the physical length of the driven element in a Yagi is much less important than the tuning and spacing of the parasitic elements. The question becomes something like this: what is the relative amplitude and phase of the current in each parasitic element, for some excitation of the driven element? A Yagi is a system of coupled resonators, like a system of coupled pendulums. If one of the pendulums is driven at a particular amplitude and frequency, even if it's not that pendulum's natural frequency, the rest of the pendulums will follow along pretty much the same as if the driven pendulum was tuned to have that natural frequency. In the antenna, the difference will only be in the coupling from the driven element to the others, and I believe that changes only slightly as the length of the driven element changes. But I may be wrong about that, and await my re-education. ;-) But I just ran EZNec on the example "NBS" 3-element 50.1MHz Yagi, varying the nominal 110 inch long D.E. by +/- 10 inches, and saw the expected fairly large variation in impedance, but only 0.02dB change in gain over that whole range, with similarly small variation in F/B ratio and beam width. The longest D.E. I ran was also the highest gain (by that tiny amount), and provided enough inductive reactance that the feedpoint could be tuned to resonance and present 200 ohms by shunting with about 55pF capacitance. Next to try: compare the SWR bandwidths of the hairpin (inductive) shunt of a shortened D.E. and the capacitive shunt of a lengthened D.E.. Unless someone offers a better test case, I'll use the NBS 3 element Yagi... Cheers, Tom |
Efficiency of 200-ohm hairpin matching
I believe that the physical length of the driven element in a Yagi is much less important than the tuning and spacing of the parasitic elements. The question becomes something like this: what is the relative amplitude and phase of the current in each parasitic element, for some excitation of the driven element? A Yagi is a system of coupled resonators, like a system of coupled pendulums. If one of the pendulums is driven at a particular amplitude and frequency, even if it's not that pendulum's natural frequency, the rest of the pendulums will follow along pretty much the same as if the driven pendulum was tuned to have that natural frequency. In the antenna, the difference will only be in the coupling from the driven element to the others, and I believe that changes only slightly as the length of the driven element changes. But I may be wrong about that, and await my re-education. ;-) But I just ran EZNec on the example "NBS" 3-element 50.1MHz Yagi, varying the nominal 110 inch long D.E. by +/- 10 inches, and saw the expected fairly large variation in impedance, but only 0.02dB change in gain over that whole range, with similarly small variation in F/B ratio and beam width. The longest D.E. I ran was also the highest gain (by that tiny amount), and provided enough inductive reactance that the feedpoint could be tuned to resonance and present 200 ohms by shunting with about 55pF capacitance. Next to try: compare the SWR bandwidths of the hairpin (inductive) shunt of a shortened D.E. and the capacitive shunt of a lengthened D.E.. Unless someone offers a better test case, I'll use the NBS 3 element Yagi... Cheers, Tom Hi Tom, the results you got on EZNEC are encouraging. Nevertheless I would not like to try using a lengthened element in conjunction with a capacitor, as the difference between that configuration and the original configuration would be the maximum (although it would be much easier to adjust a capacitor than the inductance of an hairpin). What puzzles me is that the antenna manufacturer reported me having sold several hundreds of those antennas, and no one has reported him the bandwidth being too narrow or the exagerated wet terrain influence. I am not sure on what I am going to do, also because I am not 100% sure on whether the bandwidth problem is only due to the matching system, or it is also due to the particular antenna design. My original intention was to compare this 50-MHz long Yagi antenna (32-foot boom) against a smaller antenna (11-foot boom) I have on another tower, so as to determine how much a bigger antenna really helps during multiple-hop sporadic openings to US and Japan. Probably for the forecoming sporadic-E season (May-August) I will leave things as they are, and just try to assess the practical advantages of the bigger antenna. After that I will see what I shall do. Thanks very much for the useful discussion. 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On 8 abr, 18:52, K7ITM wrote:
On Apr 7, 4:02 pm, Wimpie wrote: On 7 abr, 23:15, "Antonio Vernucci" wrote: Hello Tom, First, the matching is being done essentially with an "L" network (or rather the balanced version of an "L" network), where there is a load resistance (the resistive part of the feedpoint impedance, which includes radiation resistance and element loss resistance reflected to the feedpoint), the series capacitive reactance of the feedpoing element, and a shunt inductive reactance, provided by the hairpin. Because shortening the driven element causes a decrease in resistance and an increase in capacitive reactance, it's possible to find a length that allows matching to any of a wide range of resistances. But the higher the resistance to which you match, the shorter you need to make the element and the lower the feedpoint resistance. The ratio of feedpoint resistance to matched resistance determines the loaded Q of the matching network; as you make the matched resistance higher, the loaded Q goes up rather quickly. If you know the loaded Q and the unloaded Q of the hairpin, you have a good handle on the amount lost to heat in the hairpin: if the hairpin Q is two times the loaded Q, half the power is dissipated in the hairpin, for example. However, unless you build an antenna with a very low feedpoint resistance at resonance, there almost certainly won't be an efficiency problem: the reactance changes quickly enough with changes in driven element length that the resistance won't drop much by the time you reach enough reactance to get a match to 200 ohms. It appears that the loaded Q of the match to 200 ohms for your case will be less than 4. I would think unless you really messed up badly, the hairpin unloaded Q should be well in excess of 100, and if that's the case, the power lost in the hairpin would correspond to well under 0.1dB signal level change. All OK. I however reckon that, due to the parasitic elements effect, the radiation resistance of the driven element (before shortening it) would be in the order of 20 ohm. So, impedance gets brought up by a factor of 10 or so. On the other hand, I'm surprised by the comment from the manufacturer about difficulties making a 1:1 balun. I have had good luck using ferrites and/or self-resonant coils of feedline and/or coils of feedline specifically resonated with additional capacitance. A 4:1 balun from a "hairpin" of 1/2 wave of coax, arranged symmetrically, is easy enough to make, but I would not rule out using a 1:1, if it has advantages for you. His argument is that, for high-power operation (say 1500W), it is more convenient for him to provide a quarter-wavelength 4:1 balun made of RG-142 teflon cable than a sealed box with a 1:1 coil on an ironpowder toroidal core. I mentioned him that, just using some extra length of RG-142 cable, he can easily build a 1:1 balun, and hence design the antenna matching system for 50 ohm instead of 200 ohm. I hope he will listen to me, because that antenna is really narrowband! 73 Tony I0JX Hello, Imagine 10 Ohms radiation resistance + capacitive component after shortening the open dipole radiator. When you convert that to 200 Ohms via a parallel inductance (hairpin), the Q factor of such a network is about 4.4. At 50 MHz that will result in a bandwidth (VSWR=2) of 8 MHz. A 200 Ohms to 50 Ohms coaxial balun will have a lower Q factor. So the combination of balun and L-network will certainly have a useful bandwidth 340 kHz. Maybe the radiation resistance is less (you can derive that from the hairpin inductance) and/or the antenna is by nature (very) narrow band. Best regards, Wim PA3DJSwww.tetech.nl remove abc from the address. Yes, I had similar thoughts, but a bit different. First, I think it's safe to say that if, at resonance, the driven element presents about 20 ohms at the feedpoint, shortening the D.E. only a little will give enough capacitive reactance to allow the hairpin match to 200 ohms. Even if the D.E. looks like 5 ohms, the "L" network match still gives a 3dB bandwidth of 8MHz at a 50MHz center, or 3MHz 1.5:1 SWR bandwidth. Hello Tom, Fully agree with you. I gave the values for a 20 to 200 Ohms match to show that the problem is not in the matching, but in the antenna. Even matching from 20 to 50 Ohms will not give sufficient bandwidth, because actual BW is far below the BW of the matching network (with 20 Ohms termination). I think we need to look somewhere else for the answer to the narrow bandwidth. My working hypothesis at the moment is that it's in the antenna, or perhaps rather in the combination of antenna and matching network. Note that the calc for the L match assumed a constant capacitance, but the antenna will not, in general, look anything like a constant C even over a fairly narrow frequency range. That rapid Im(Zant) change is the reason for the narrow BW. I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. Did you also scale the thickness of the elements? I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. Over that range, the equivalent series capacitance changes from 59pF at the low end to 138pF at the high end, and at least by NEC2's prediction, the impedance changes especially quickly around 51MHz-- both reactive and resistive parts. 50.75MHz: 10.3-j31.76; 51MHz: 3.91- j22.56, quite a large percentage change in 250kHz. Having the effective series capacitance change that quickly will cause the matching network to behave very differently than it would with a capacitance element that is fixed. Is it possible to lengthen the D.E., causing it to present an inductive reactance at the feedpoint, and match that (to 200 ohms, or to 50 ohms) with a shunt capacitance? That may work better, giving a broader SWR bandwidth. The resistive component should be higher, further lowering the Q, and I suspect the reactance won't change so quickly with frequency. I don't have time at the moment to compare the two, but may this evening. It may also be possible to raise the resistive part to 50 ohms and match to 50 with a series capacitance. I did use length extension for very thick mesh dipoles for VHF air band. Their resonant impedance is less then 50 Ohms (because they become short). Some additional length gives some inductance to use a parallel capacitor match and (most important) bandwidth increase. I don't know whether this will give sufficient BW improvement for the Yagi as the Q is also determined by the reflector en directors. I am looking forward to your simulation results. Cheers, Tom Wim PA3DJS www.tetech.nl Please remove abc from the address. |
Efficiency of 200-ohm hairpin matching
"Antonio Vernucci" wrote in
: .... the results you got on EZNEC are encouraging. Nevertheless I would not like to try using a lengthened element in conjunction with a capacitor, as the difference between that configuration and the original configuration would be the maximum (although it would be much easier to adjust a capacitor than the inductance of an hairpin). Tony, Some thoughts. You are suggesting that it is easier to make a low loss capacitor that is located at the feedpoint in a hostile environment, than it is to make a low loss inductor (the hairpin). Just as the hairpin is a s/c stub for inductive reactance, you could use an o/c stub... but remember that transmission line elements are a path to low Q reactors, use thick conductors for the transmision line (which for an o/c stub will need to be much longer than for the s/c stub). What puzzles me is that the antenna manufacturer reported me having sold several hundreds of those antennas, and no one has reported him the bandwidth being too narrow or the exagerated wet terrain influence. Only hundreds? Hy-Gain have used this feed system on 2m antennas for a very long time. Yes, their gain figures seem a bit generous, but the hairpin is a viable commercial option. I am not sure on what I am going to do, also because I am not 100% sure on whether the bandwidth problem is only due to the matching system, or it is also due to the particular antenna design. My gut feeling is that optimised long Yagis have narrow bandwidth because of the large number of elements with role that is frequency critical. A short Yagi has wider bandwidth with the same feed system. It is a long time since I read your first post, but narrow bandwidth can be an advantage. It reduces out of band signal reach your first amplifier where it will mix and produce IMD products that may be in-band. Narrow band antennas help to provide the selectivity that is lacking in many / most modern radios. Owen |
Efficiency of 200-ohm hairpin matching
Hi Wim (and lurkers),
On Apr 8, 1:56 pm, Wimpie wrote: On 8 abr, 18:52, K7ITM wrote: .... I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. Did you also scale the thickness of the elements? Yes, EZNEC lets you scale everything in the same proportion in one quick operation. I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. .... I don't know whether this will give sufficient BW improvement for the Yagi as the Q is also determined by the reflector en directors. I am looking forward to your simulation results. Yes, of course. If the antenna itself is not wide-band, there is rather little you can do about it. It may be possible with a more-or- less complicated network to make the SWR bandwidth somewhat greater (assuming still low loss--if you have a lossy network, you can make the SWR bandwidth very large -- ;-) But making the SWR bandwidth large does not make the antenna's gain bandwidth large. That is, for a given set of directors and reflectors, the pattern will change with frequency without much regard for what you do to get power into the antenna. Thanks for the encouragement about running some simulations. I'll report results as I have a chance, probably this evening (US Pacific coast time). Cheers, Tom |
Efficiency of 200-ohm hairpin matching
On 9 abr, 01:26, K7ITM wrote:
Hi Wim (and lurkers), On Apr 8, 1:56 pm, Wimpie wrote: On 8 abr, 18:52, K7ITM wrote: ... I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. Did you also scale the thickness of the elements? Yes, EZNEC lets you scale everything in the same proportion in one quick operation. I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. ... I don't know whether this will give sufficient BW improvement for the Yagi as the Q is also determined by the reflector en directors. I am looking forward to your simulation results. Yes, of course. If the antenna itself is not wide-band, there is rather little you can do about it. It may be possible with a more-or- less complicated network to make the SWR bandwidth somewhat greater (assuming still low loss--if you have a lossy network, you can make the SWR bandwidth very large -- ;-) But making the SWR bandwidth large does not make the antenna's gain bandwidth large. That is, for a given set of directors and reflectors, the pattern will change with frequency without much regard for what you do to get power into the antenna. Thanks for the encouragement about running some simulations. I'll report results as I have a chance, probably this evening (US Pacific coast time). Cheers, Tom Hello Tom, I do not have experience with EZNEC, so I didn't know that it also scales element diameter. Fully agree with regards to VSWR BW, gain BW and wide band dummy loads.... Here it is UTC+2, so it is time for me te visit my bed for some hours. Tommorow VAT administration is waiting. Best regards, Wim |
Efficiency of 200-ohm hairpin matching
On Apr 8, 4:40 pm, Wimpie wrote:
On 9 abr, 01:26, K7ITM wrote: Hi Wim (and lurkers), On Apr 8, 1:56 pm, Wimpie wrote: On 8 abr, 18:52, K7ITM wrote: ... I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. Did you also scale the thickness of the elements? Yes, EZNEC lets you scale everything in the same proportion in one quick operation. I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. ... I don't know whether this will give sufficient BW improvement for the Yagi as the Q is also determined by the reflector en directors. I am looking forward to your simulation results. Yes, of course. If the antenna itself is not wide-band, there is rather little you can do about it. It may be possible with a more-or- less complicated network to make the SWR bandwidth somewhat greater (assuming still low loss--if you have a lossy network, you can make the SWR bandwidth very large -- ;-) But making the SWR bandwidth large does not make the antenna's gain bandwidth large. That is, for a given set of directors and reflectors, the pattern will change with frequency without much regard for what you do to get power into the antenna. Thanks for the encouragement about running some simulations. I'll report results as I have a chance, probably this evening (US Pacific coast time). Cheers, Tom Hello Tom, I do not have experience with EZNEC, so I didn't know that it also scales element diameter. Fully agree with regards to VSWR BW, gain BW and wide band dummy loads.... Here it is UTC+2, so it is time for me te visit my bed for some hours. Tommorow VAT administration is waiting. Best regards, Wim Hi Wim and lurkers, Yes, EZNEC has some nice features. I exercised a couple of them tonight, doing frequency sweeps with inductive and capacitive matching. I did cheat: since I do not know the dimensions of the hairpin match, I elected to just use a pure lumped inductance. I suppose the error compared with a transmission line stub (hairpin) won't be great. I also used for the first time ASCII file import for the wire description, making use of Excel to generate the wires. (Why did it take me so long?? ;-) Somewhat to my surprise, I found that the SWR bandwidth of the shortened D.E. (driven element) loaded with a shunt inductance and that of the lengthened D.E. loaded with a shunt capacitance was not so different. The antenna I picked to model was one in the Orr and Cowan book, "The Beam Antenna Book." It's a 6-element 1.20 wave long design, with 0.2 waves D.E. to reflector, and 0.25 waves D.E. to first director and also between each adjacent pair of directors. For 6 meters it is a little shorter than Tony's antenna, but not very much shorter. But the elements (per Orr and Cowan's numbers) are 2 inches (about 50mm) diameter. The large diameter almost certainly contributes to the fairly broad bandwidth. All models below are done in freespace. For the shortened D.E.: D.E. length = 95.4 inches. Feedpoint impedance at 50.1MHz is 16.32-j54.63 ohms. Gain varies from 11.78dBi at 49.5MHz to 11.98dBi at 50.1 to 11.68 at 50.5. f/b varies uniformly from 19.43dB at 49.5 to 8.21dB at 50.5. Matched at 50.1MHz to 200 ohms with 189.4nH shunt at the feedpoint yields a 1.4:1 SWR bandwidth of about 49.89MHz to 50.28MHz. For the lengthened D.E.: length = 115.8 inches; impedance at 50.1 = 29.28+j70.7 ohms. Gain: , 12.05max, . f/b shows same uniform degradation with increased freq; to . 38.35pF shunt across feedpoint yields 1.4:1 SWR BW of about 49.89 to 50.34MHz, just slightly greater than the shortened one. (Neither of these includes frequency effects of the 4:1 balun to get back to 50 ohms...) I also tried a much-lengthened D.E., to get it up to 50+j204.1 ohms. Then a series 15.57pF gives a 50 ohm match. Gain is similar to above, 11.83dBi -- 12.11dBi -- 11.93dBi, and f/b is also similar, 22.82 -- 8.83. I notice both the gain and the f/b improve slightly as the D.E. is lengthened, though it's small enough that you'd never notice it--it would be swamped out by other effects. The 1.4:1 SWR BW for this configuration is about 49.85MHz to 50.32MHz; this of course against a 50 ohm reference which is different from the first two examples above which are against a 200 ohm reference. OK--I think I got all the critical stuff in there... Cheers, Tom |
Efficiency of 200-ohm hairpin matching
K7ITM wrote in
: .... Yes, EZNEC has some nice features. I exercised a couple of them tonight, doing frequency sweeps with inductive and capacitive matching. I did cheat: since I do not know the dimensions of the hairpin match, I elected to just use a pure lumped inductance. I suppose the error compared with a transmission line stub (hairpin) won't be great. I also used for the first time ASCII file import for It probably isn't. However, you could get an idea from my Two Wire Line Loss Calculator (http://www.vk1od.net/tl/twllc.htm). It looks like a hairpin made of 4mm dia aluminium 50mm spacing and 150mm in length give an impedance of 0.02 +j61... so the Q is probably mainly determined by the end connections. a o/c stub made of 4mm dia aluminium 50mm spacing and 1300mm in length give an impedance of 0.07-j80... so, it is not quite as good electrically, it is unweildly and end connection resistance will probably still be significant. This is no doubt why people use a hairpin in preference to an o/c stub! Owen |
Efficiency of 200-ohm hairpin matching
On 9 abr, 08:40, K7ITM wrote:
On Apr 8, 4:40 pm, Wimpie wrote: On 9 abr, 01:26, K7ITM wrote: Hi Wim (and lurkers), On Apr 8, 1:56 pm, Wimpie wrote: On 8 abr, 18:52, K7ITM wrote: ... I just ran EZNEC on a frequency-scaled version of the 14MHz 5 element Yagi included in the sample files, with the D.E. slightly shortened to allow a decent hairpin match to 200 ohms at the design center frequency. Did you also scale the thickness of the elements? Yes, EZNEC lets you scale everything in the same proportion in one quick operation. I did a frequency sweep, 49 to 51 MHz, in 0.25MHz steps. ... I don't know whether this will give sufficient BW improvement for the Yagi as the Q is also determined by the reflector en directors. I am looking forward to your simulation results. Yes, of course. If the antenna itself is not wide-band, there is rather little you can do about it. It may be possible with a more-or- less complicated network to make the SWR bandwidth somewhat greater (assuming still low loss--if you have a lossy network, you can make the SWR bandwidth very large -- ;-) But making the SWR bandwidth large does not make the antenna's gain bandwidth large. That is, for a given set of directors and reflectors, the pattern will change with frequency without much regard for what you do to get power into the antenna. Thanks for the encouragement about running some simulations. I'll report results as I have a chance, probably this evening (US Pacific coast time). Cheers, Tom Hello Tom, I do not have experience with EZNEC, so I didn't know that it also scales element diameter. Fully agree with regards to VSWR BW, gain BW and wide band dummy loads.... Here it is UTC+2, so it is time for me te visit my bed for some hours. Tommorow VAT administration is waiting. Best regards, Wim Hi Wim and lurkers, Yes, EZNEC has some nice features. I exercised a couple of them tonight, doing frequency sweeps with inductive and capacitive matching. I did cheat: since I do not know the dimensions of the hairpin match, I elected to just use a pure lumped inductance. I suppose the error compared with a transmission line stub (hairpin) won't be great. I also used for the first time ASCII file import for the wire description, making use of Excel to generate the wires. (Why did it take me so long?? ;-) Somewhat to my surprise, I found that the SWR bandwidth of the shortened D.E. (driven element) loaded with a shunt inductance and that of the lengthened D.E. loaded with a shunt capacitance was not so different. The antenna I picked to model was one in the Orr and Cowan book, "The Beam Antenna Book." It's a 6-element 1.20 wave long design, with 0.2 waves D.E. to reflector, and 0.25 waves D.E. to first director and also between each adjacent pair of directors. For 6 meters it is a little shorter than Tony's antenna, but not very much shorter. But the elements (per Orr and Cowan's numbers) are 2 inches (about 50mm) diameter. The large diameter almost certainly contributes to the fairly broad bandwidth. All models below are done in freespace. For the shortened D.E.: D.E. length = 95.4 inches. Feedpoint impedance at 50.1MHz is 16.32-j54.63 ohms. Gain varies from 11.78dBi at 49.5MHz to 11.98dBi at 50.1 to 11.68 at 50.5. f/b varies uniformly from 19.43dB at 49.5 to 8.21dB at 50.5. Matched at 50.1MHz to 200 ohms with 189.4nH shunt at the feedpoint yields a 1.4:1 SWR bandwidth of about 49.89MHz to 50.28MHz. For the lengthened D.E.: length = 115.8 inches; impedance at 50.1 = 29.28+j70.7 ohms. Gain: , 12.05max, . f/b shows same uniform degradation with increased freq; to . 38.35pF shunt across feedpoint yields 1.4:1 SWR BW of about 49.89 to 50.34MHz, just slightly greater than the shortened one. (Neither of these includes frequency effects of the 4:1 balun to get back to 50 ohms...) I also tried a much-lengthened D.E., to get it up to 50+j204.1 ohms. Then a series 15.57pF gives a 50 ohm match. Gain is similar to above, 11.83dBi -- 12.11dBi -- 11.93dBi, and f/b is also similar, 22.82 -- 8.83. I notice both the gain and the f/b improve slightly as the D.E. is lengthened, though it's small enough that you'd never notice it--it would be swamped out by other effects. The 1.4:1 SWR BW for this configuration is about 49.85MHz to 50.32MHz; this of course against a 50 ohm reference which is different from the first two examples above which are against a 200 ohm reference. OK--I think I got all the critical stuff in there... Cheers, Tom Hello Tom, Regarding the "cheating", no problem with that. When length of hairpin is far below 1/8 lambda, difference with lumped component is negligible (for this situation). Owen mentioned that a 150mm hairpin has the required reactance. The 0.02 Ohms loss resistance in the hairpin does not dissipate significant power (not measurable in practice within the antenna setup). This also shows that Tony does not have to worry about hairpin loss, even in his 200 Ohm intermediate impedance. Your simulation made fully clear that Tony's problem is in the antenna and not in the matching. Thanks for sharing your results, Wim PA3DJS www.tetech.nl please remove abc from the address. |
Efficiency of 200-ohm hairpin matching
Tom and Wim,
I agree with your concluion that the bandwidth narrowness is due to the antenna design and not to the matching system. I have simulated the antenna on EZNEC and the narrow band property of the antenna is very evident. I would like to also put the hairpin in the EZNEC model, but I am not very good at EZNEC and I would not now know to do it. If you are interested I can send you the EZNEC file via e-mail. 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On Apr 9, 2:08 am, Owen Duffy wrote:
K7ITM wrote : ... Yes, EZNEC has some nice features. I exercised a couple of them tonight, doing frequency sweeps with inductive and capacitive matching. I did cheat: since I do not know the dimensions of the hairpin match, I elected to just use a pure lumped inductance. I suppose the error compared with a transmission line stub (hairpin) won't be great. I also used for the first time ASCII file import for It probably isn't. However, you could get an idea from my Two Wire Line Loss Calculator (http://www.vk1od.net/tl/twllc.htm). It looks like a hairpin made of 4mm dia aluminium 50mm spacing and 150mm in length give an impedance of 0.02 +j61... so the Q is probably mainly determined by the end connections. a o/c stub made of 4mm dia aluminium 50mm spacing and 1300mm in length give an impedance of 0.07-j80... so, it is not quite as good electrically, it is unweildly and end connection resistance will probably still be significant. This is no doubt why people use a hairpin in preference to an o/c stub! Owen Thanks, Owen. Actually, the error I was thinking of was not the Q, since both a good hairpin and a good helical coil will have Qu's very much higher than the loaded Q of the matching network -- but rather of the slope of reactance versus frequency, since the hairpin is a transmission line and will presumably show a sharp resonance at a different frequency from the coil's self resonance. But that error should also be small--negligible as Wim notes. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
On Apr 9, 9:02 am, "Antonio Vernucci" wrote:
Tom and Wim, I agree with your concluion that the bandwidth narrowness is due to the antenna design and not to the matching system. I have simulated the antenna on EZNEC and the narrow band property of the antenna is very evident. I would like to also put the hairpin in the EZNEC model, but I am not very good at EZNEC and I would not now know to do it. If you are interested I can send you the EZNEC file via e-mail. 73 Tony I0JX Hi Tony, I would be happy to have a look at the file and add the hairpin if you wish. But I can also guide you through doing it yourself. It should be quite easy. Simply click on the " Trans Lines" button, the second one below Sources in the main window. Specify the first end the same as the source--perhaps 50% along wire 2, assuming the D.E. is wire 2. That end will then be in parallel with the source. You can type "S" in for the other end's wire #, or if you click in the End 2 Wire # box, you should see a list of "open" and "short" and you can select "short" there. Then enter the length, Z0, velocity factor (1) and loss. Reverse or normal doesn't matter since only one end is connected. In as much as the hairpin actually IS in the field of the antenna, and couples to it to some degree, there may be a small error in using the EZNEC transmission line, instead of entering it as wires, but for a short stub I would not expect that to be a significant issue. I do know that the 1/4 wave stub used to separate elements of a collinear does couple to the antenna fields and that coupling is a significant factor in the performance of the antenna--or at least that is what R.W.P. King tells us. And...thanks much for the interesting discussion topic. I learned a few things from it. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
You are suggesting that it is easier to make a low loss capacitor that is
located at the feedpoint in a hostile environment, than it is to make a low loss inductor (the hairpin). What I was maning to say that is easier to determine the correct capacitance (just using a variable capacitor) rather than the correct inductance (using a sliding short on the hairpin). Once one determined the correct value, the variable element should anyway be replaced with an equiavalent fixed element Only hundreds? Hy-Gain have used this feed system on 2m antennas for a very long time. Yes, their gain figures seem a bit generous, but the hairpin is a viable commercial option. Well, this is an antenna just for 6-meter enthusiasts fabricated in Italy, and there are not too many of them around here. It is a long time since I read your first post, but narrow bandwidth can be an advantage. It reduces out of band signal reach your first amplifier where it will mix and produce IMD products that may be in-band. Narrow band antennas help to provide the selectivity that is lacking in many / most modern radios. yes, but the bandwidth is so tight that it barely fits my needs. Also the SWR response shift when it rains is a problem to which I have no solution 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
Hi Tony,
I would be happy to have a look at the file and add the hairpin if you wish. But I can also guide you through doing it yourself. It should be quite easy. Simply click on the " Trans Lines" button, the second one below Sources in the main window. Specify the first end the same as the source--perhaps 50% along wire 2, assuming the D.E. is wire 2. That end will then be in parallel with the source. You can type "S" in for the other end's wire #, or if you click in the End 2 Wire # box, you should see a list of "open" and "short" and you can select "short" there. Then enter the length, Z0, velocity factor (1) and loss. Reverse or normal doesn't matter since only one end is connected. Thanks on you guidance. My hairpin is a U of aluminum tube connected directly to the antenna feedpoints. The U center is connected to the boom. The hairpin tube diameter is about 0.8 cm. The three sides of the U are approximately 35 cm, 10 cm and 35 cm. I presume that I shall enter 35 cm as the transmission line length, but how to calculate Z0? Moreover what about the velocity factor, perhaps 0.98 for a tube in open air? Would you have a formula at hand? Thanks and 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On Apr 9, 11:36 am, "Antonio Vernucci" wrote:
Hi Tony, I would be happy to have a look at the file and add the hairpin if you wish. But I can also guide you through doing it yourself. It should be quite easy. Simply click on the " Trans Lines" button, the second one below Sources in the main window. Specify the first end the same as the source--perhaps 50% along wire 2, assuming the D.E. is wire 2. That end will then be in parallel with the source. You can type "S" in for the other end's wire #, or if you click in the End 2 Wire # box, you should see a list of "open" and "short" and you can select "short" there. Then enter the length, Z0, velocity factor (1) and loss. Reverse or normal doesn't matter since only one end is connected. Thanks on you guidance. My hairpin is a U of aluminum tube connected directly to the antenna feedpoints. The U center is connected to the boom. The hairpin tube diameter is about 0.8 cm. The three sides of the U are approximately 35 cm, 10 cm and 35 cm. I presume that I shall enter 35 cm as the transmission line length, but how to calculate Z0? Moreover what about the velocity factor, perhaps 0.98 for a tube in open air? Would you have a formula at hand? Thanks and 73 Tony I0JX Hi Tony, Since the spacing is so large (10cm), it may be better to enter the hairpin as wires in the model. It might be interesting to compare the results between wires and a transmission line. First the wires: I assume that the D.E. is one wire in the model you have now. I suppose it has an odd number of segments, so that the source can go just in the middle. I'm assuming the hairpin parallel tube segments are center-to-center 10cm apart. I will assume the D.E. is currently 250cm long, all at some x value, and y extending from -1.25m to +1.25m, as one "wire" of some diameter. To be able to connect the hairpin at the right places, we need wire ends at -.05m and +.05m, so we can change the D.E. wire to be one segment between y=-.05m and +. 05m, and then add 2 wires the same diameter as the D.E., one from y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about half as many segments as the original D.E., all at the original x value. (Or exchange x and y if I have guessed wrong.) So if the original D.E. had 11 segments, put 5 segments on each of the two new wires. Now add three new wires to represent the hairpin U. Assuming the hairpin is oriented perpendicular to the plane of the antenna, it will be all at the same x value as the D.E. I'll assume the antenna is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-. 05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm or .008m. For the transmission line model: the diameter (8mm) and spacing (100mm center to center) tells me the line impedance is about 380 ohms. It should be OK to assume VF=1.00, if there is no insulation around the tube. One problem is that the short across the end is not really a short--it is more an inductance. We could go to the trouble of calculating the inductance and putting that as a load on the end of the transmission line instead of just a short, but it should be a good approximation to just lengthen the line by half the length of the "short"; I would enter the line length just as 40cm instead of 35. (I won't be surprised if Owen pops in here with either a confirmation or a better suggestion!) Cheers, Tom |
Efficiency of 200-ohm hairpin matching
I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go just in the middle. I'm assuming the hairpin parallel tube segments are center-to-center 10cm apart. I will assume the D.E. is currently 250cm long, all at some x value, and y extending from -1.25m to +1.25m, as one "wire" of some diameter. To be able to connect the hairpin at the right places, we need wire ends at -.05m and +.05m, so we can change the D.E. wire to be one segment between y=-.05m and +. 05m, and then add 2 wires the same diameter as the D.E., one from y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about half as many segments as the original D.E., all at the original x value. (Or exchange x and y if I have guessed wrong.) So if the original D.E. had 11 segments, put 5 segments on each of the two new wires. Now add three new wires to represent the hairpin U. Assuming the hairpin is oriented perpendicular to the plane of the antenna, it will be all at the same x value as the D.E. I'll assume the antenna is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-. 05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm or .008m. Thanks Tom, I did the EZNEC model exactly as you suggested, but I now have another doubt (I must re-read the EZNEC help one of these days...) I would say that I shall still feed the driven element at its center, and not separately feed the two junctions between the hairpin and the driven element Is that right? 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
K7ITM wrote in
: On Apr 9, 2:08 am, Owen Duffy wrote: K7ITM wrote om: ... Yes, EZNEC has some nice features. I exercised a couple of them tonight, doing frequency sweeps with inductive and capacitive matching. I did cheat: since I do not know the dimensions of the hairpin match, I elected to just use a pure lumped inductance. I suppose the error compared with a transmission line stub (hairpin) won't be great. I also used for the first time ASCII file import for It probably isn't. However, you could get an idea from my Two Wire Line Loss Calculator (http://www.vk1od.net/tl/twllc.htm). It looks like a hairpin made of 4mm dia aluminium 50mm spacing and 150mm in length give an impedance of 0.02 +j61... so the Q is probably mainly determined by the end connections. a o/c stub made of 4mm dia aluminium 50mm spacing and 1300mm in length give an impedance of 0.07-j80... so, it is not quite as good electrically, it is unweildly and end connection resistance will probably still be significant. This is no doubt why people use a hairpin in preference to an o/c stub! Owen Thanks, Owen. Actually, the error I was thinking of was not the Q, since both a good hairpin and a good helical coil will have Qu's very much higher than the loaded Q of the matching network -- but rather of the slope of reactance versus frequency, since the hairpin is a transmission line and will presumably show a sharp resonance at a different frequency from the coil's self resonance. But that error should also be small--negligible as Wim notes. Hi Tom, Warning bells sound to me when applications call for reactors fabricated from TL sections. Not to say that are always bad, but they aren't always good, and they bear examination. I think that may have been in Tony's mind over the matching network. In the case of the hairpin, fabricated from substantial material, it looks good in this application. Because the line section is so short, inductive reactance is almost linearly proportional to length, and self resonance of the line section itself is nearly a decade higher. My experience with one of the HyGain (now MFJ) 2m Yagis was that VSWR was stable with weather and over time. I had taken some care to exclude air + water from key connections, in fact I replaced all the fastners used for electrical connection with SS when the beam was less than a year old because of corrosion. Tweny years later the remaining fastners used only for mechanical retention required replacement due to corrosion. But, after the first rework, the matching system seemed robust (bird proof) and reliable. I have used them since on home made antennas with good success. Owen |
Efficiency of 200-ohm hairpin matching
On Apr 9, 2:59 pm, Owen Duffy wrote:
... Hi Tom, Warning bells sound to me when applications call for reactors fabricated from TL sections. Not to say that are always bad, but they aren't always good, and they bear examination. .... :-) Yes, indeed. I just this afternoon fabricated a 500MHz LPF that I wanted to give about 40dB attenuation up to several GHz. My design was a 5th order elliptical, so I had two parallel tanks for the series paths and three shunt capacitors. Because the coils (10nH and 15nH) are so small, I though about using a shorted stub instead, but then realized that my shorted stub would look like a short when it was 1/2 wave long, not a good thing to have as a series element at a frequency you want to block. Probably would have worked OK since the two stubs were different and wouldn't have the same resonance frequency; and by the time you reached that freq., the shunt caps would be pretty effective on their own. Anyway, it works fine with tiny coils. Thank goodness for good microscopes. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
On Apr 9, 1:22 pm, "Antonio Vernucci" wrote:
I assume that the D.E. is one wire in the model you have now. I suppose it has an odd number of segments, so that the source can go just in the middle. I'm assuming the hairpin parallel tube segments are center-to-center 10cm apart. I will assume the D.E. is currently 250cm long, all at some x value, and y extending from -1.25m to +1.25m, as one "wire" of some diameter. To be able to connect the hairpin at the right places, we need wire ends at -.05m and +.05m, so we can change the D.E. wire to be one segment between y=-.05m and +. 05m, and then add 2 wires the same diameter as the D.E., one from y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about half as many segments as the original D.E., all at the original x value. (Or exchange x and y if I have guessed wrong.) So if the original D.E. had 11 segments, put 5 segments on each of the two new wires. Now add three new wires to represent the hairpin U. Assuming the hairpin is oriented perpendicular to the plane of the antenna, it will be all at the same x value as the D.E. I'll assume the antenna is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-. 05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm or .008m. Thanks Tom, I did the EZNEC model exactly as you suggested, but I now have another doubt (I must re-read the EZNEC help one of these days...) I would say that I shall still feed the driven element at its center, and not separately feed the two junctions between the hairpin and the driven element Is that right? 73 Tony I0JX Hi Tony, I can understand your confusion and doubt. I hope this says it in clear terms: In the revised model with the hairpin done as three wires, the original D.E. single wire was replaced by three wires: a 10cm long center section which should have the source in its center, and a "left" and a "right" outer section. Those three sections are three separate wires. The reason to break the original single wire into three wires is so that there will be wire "ends" to connect the hairpin wires to. You can only connect wires at their ends, even though they have multiple segments. So yes, in the new model the source is just in the middle of the wire which is the middle section of the D.E. Cheers, Tom |
Efficiency of 200-ohm hairpin matching
I can understand your confusion and doubt. I hope this says it in
clear terms: In the revised model with the hairpin done as three wires, the original D.E. single wire was replaced by three wires: a 10cm long center section which should have the source in its center, and a "left" and a "right" outer section. Those three sections are three separate wires. The reason to break the original single wire into three wires is so that there will be wire "ends" to connect the hairpin wires to. You can only connect wires at their ends, even though they have multiple segments. So yes, in the new model the source is just in the middle of the wire which is the middle section of the D.E. Thanks for your clarification that confirms my expectation that I shall put the generator in the middle of the 10 cm-long center section of the D.E.. Everything works fine and, for a reference 200-ohm resistance, I get an SWR of almost 1 not too far away from the frequency at which it actually occurs. If I may take a bit more of your time, I have noted two inconsistencies: - with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I just delete the three hairpin segments leaving everything else unchanged, the gain grows up to 13.59 dB. I am surprised about that result, as I had understood that, in EZNEC, the antenna gain has nothing to do with the way the antenna is fed. - with the hairpin, at the antenna resonant frequency EZNEC shows an impedance of 200.5 +j2.1 ohm which obviously corresponds to an SWR of almost 1. If I remove the hairpin, at the same frequency EZNEC shows an impedance of 42.4 - 39 ohm which could seem reasonable, but it is actually not. As a matter of fact, by applying the appropriate series-to-parallel conversion formula, that impedance corresponds to the parallel of a 78.2 ohm resistance and a 85.1 ohm capacitive reactance. Resonating the reactive capacitance with an equal (though opposite) reactance hairpin, the remaining resistance would be just 78.2 ohm, that is far away from the 200 ohm EZNEC shows with the hairpin installed. May I have your opinion? 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
On Apr 10, 1:22 pm, "Antonio Vernucci" wrote:
I can understand your confusion and doubt. I hope this says it in clear terms: In the revised model with the hairpin done as three wires, the original D.E. single wire was replaced by three wires: a 10cm long center section which should have the source in its center, and a "left" and a "right" outer section. Those three sections are three separate wires. The reason to break the original single wire into three wires is so that there will be wire "ends" to connect the hairpin wires to. You can only connect wires at their ends, even though they have multiple segments. So yes, in the new model the source is just in the middle of the wire which is the middle section of the D.E. Thanks for your clarification that confirms my expectation that I shall put the generator in the middle of the 10 cm-long center section of the D.E.. Everything works fine and, for a reference 200-ohm resistance, I get an SWR of almost 1 not too far away from the frequency at which it actually occurs. If I may take a bit more of your time, I have noted two inconsistencies: - with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I just delete the three hairpin segments leaving everything else unchanged, the gain grows up to 13.59 dB. I am surprised about that result, as I had understood that, in EZNEC, the antenna gain has nothing to do with the way the antenna is fed. - with the hairpin, at the antenna resonant frequency EZNEC shows an impedance of 200.5 +j2.1 ohm which obviously corresponds to an SWR of almost 1. If I remove the hairpin, at the same frequency EZNEC shows an impedance of 42.4 - 39 ohm which could seem reasonable, but it is actually not. As a matter of fact, by applying the appropriate series-to-parallel conversion formula, that impedance corresponds to the parallel of a 78.2 ohm resistance and a 85.1 ohm capacitive reactance. Resonating the reactive capacitance with an equal (though opposite) reactance hairpin, the remaining resistance would be just 78.2 ohm, that is far away from the 200 ohm EZNEC shows with the hairpin installed. May I have your opinion? 73 Tony I0JX Hi Tony, Hmm...very interesting. I hope some others will check in on this with ideas too. My first thought is that the hairpin conductors are really more a part of the antenna than you may have thought. The current in that 10cm section of the hairpin, especially, is in a direction parallel to the D.E. and to all the directors and to the reflector(s). It will couple to them, and could be changing the gain and impedance some. I am also surprised by the high resistance you are seeing with the hairpin wires removed. From what you posted before, and from the 3- element NBS and the 6-element designs I ran in EZNEC, I would expect a lower feedpoint resistance than that. If I put a pure inductance in parallel with your 42.4-j39 ohms, I get the same answer as you did, that it moves to 78 ohms resistive. My offer to look at your EZNEC file is still open, of course. Maybe I could see something in it--I know that I can look and look at something and not see an obvious problem, even when I know perfectly well what to look for. Also--in the model before you split the D.E. into three wires, did you get a more reasonable feedpoint impedance (with no hairpin, just looking at the D.E. feedpoint)? I think we were expecting something like 19-j58 ohms. If you get that, then I would look for the reason things change (so very much!) when you split the D.E. into three separate (but connected, end-to-end) wires. I can understand a small change but not so large a change. Is the total number of segments for the three wires making up the D.E. still about the same as it was in the original version of the model, with just one wire for the D.E.? Cheers, Tom |
Efficiency of 200-ohm hairpin matching
I am also surprised by the high resistance you are seeing with the
hairpin wires removed. From what you posted before, and from the 3- element NBS and the 6-element designs I ran in EZNEC, I would expect a lower feedpoint resistance than that. If I put a pure inductance in parallel with your 42.4-j39 ohms, I get the same answer as you did, that it moves to 78 ohms resistive. My offer to look at your EZNEC file is still open, of course. Maybe I could see something in it--I know that I can look and look at something and not see an obvious problem, even when I know perfectly well what to look for. If your personal mail on the newsgroup is correct, I could send you the EZNEC file of my antenna. Also--in the model before you split the D.E. into three wires, did you get a more reasonable feedpoint impedance (with no hairpin, just looking at the D.E. feedpoint)? I think we were expecting something like 19-j58 ohms. If you get that, then I would look for the reason things change (so very much!) when you split the D.E. into three separate (but connected, end-to-end) wires. I can understand a small change but not so large a change. Is the total number of segments for the three wires making up the D.E. still about the same as it was in the original version of the model, with just one wire for the D.E.? I am also surprised as I would have expected a much lower resistance (as you suggested) I did the test you have suggested and I noted some change, but not much. With a "solid" D.E., i.e. not broken in three pieces (and without the hairpin of course), impedance is 43.27 - j 40.97 ohm, not terribly away from 42.4 - j39 ohm. In that test I configured EZNEC for 17 segments, whilst when the D.E. was broken in three parts I had used 17 segments for the two outer arms and only 5 segments for the 10-cm center section, due to its much shorter length (I see that result depend somewhat on the number of segments). 73 Tony I0JX |
Efficiency of 200-ohm hairpin matching
Antonio Vernucci wrote:
. . . - with the hairpin EZNEC gives an antenna gain of 12.76 dB but, if I just delete the three hairpin segments leaving everything else unchanged, the gain grows up to 13.59 dB. I am surprised about that result, as I had understood that, in EZNEC, the antenna gain has nothing to do with the way the antenna is fed. What you're probably seeing is a numerical problem in the NEC calculating engine. It's very fussy about the region near a source, and doesn't like small loops which include a source. You should run an Average Gain check (see "Average Gain" in the EZNEC manual index), which will reveal whether this is the problem. The double precision calculating engine in EZNEC+ is considerably more tolerant of small loops, but can still have problems with average gain for other reasons. Roy Lewallen, W7EL |
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