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