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Wattmeter Location
Hi,
I am contemplating a pep reading wattmeter so I can check the output of my HB amplifier. Consequently, I am putting it right on the output of the amp. The output of the amp should always see 50 ohms because it will be feeding either a 50 ohm dummy load, or an ATU tuned to 50 ohms. With that in mind, I am simply using a resistive voltage divider, to get a voltage sample, and squaring it with an AD633 multiplier. This seems like a simple, cheap way to get watt info. assuming you will always be working into a near 50 ohm resistive load. Anyone see any reason why this will not give you a pretty good indication of your power output? I realize that the load may not always be exactly 50 ohms, and that there are losses in the ATU. The reason I have not tried getting a current sample and using the conventional VI COS Theta with the multiplier is due to the additional complexity of circuitry. It is also difficult to get accurate current samples over a wide frequency range. By making everything resistive it somewhat takes the frequency dependency out of the problem. Thanks. 73 Gary N4AST |
The most accurate way of measuring power is to measure volts across a known
resistance. |
JGBOYLES wrote:
Hi, I am contemplating a pep reading wattmeter so I can check the output of my HB amplifier. Consequently, I am putting it right on the output of the amp. The output of the amp should always see 50 ohms because it will be feeding either a 50 ohm dummy load, or an ATU tuned to 50 ohms. With that in mind, I am simply using a resistive voltage divider, to get a voltage sample, and squaring it with an AD633 multiplier. This seems like a simple, cheap way to get watt info. assuming you will always be working into a near 50 ohm resistive load. Anyone see any reason why this will not give you a pretty good indication of your power output? I realize that the load may not always be exactly 50 ohms, and that there are losses in the ATU. The reason I have not tried getting a current sample and using the conventional VI COS Theta with the multiplier is due to the additional complexity of circuitry. It is also difficult to get accurate current samples over a wide frequency range. By making everything resistive it somewhat takes the frequency dependency out of the problem. Thanks. 73 Gary N4AST Good luck getting a straight answer here, and I won't try one, because every technical answer related to power transfer is always wrong by someone elses measure on this list. That being said, I'd do it the way you propose, since it's simple and repeatable, and probably close enough. tom K0TAR |
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Why complicate matters?
All Gary needs is a diode, a capacitor, a resistor and a micro-ammeter. |
Why complicate matters?
All Gary needs is a diode, a capacitor, a resistor and a micro-ammeter. Hi Reg, Thanks for the reply. That will certainly work, but I would need a Log scale (or is it square) on the micro-ammeter. I want to be read 0-1500watts pep. Down around the 100 watt level things would get crowded. I bought some 3-1/2 digit multimeters from a company here in the US for $3 each. Bought 11 of them. Or I have a 0-1 ma movement that I have re-labeled the meter face to read watts, but it is a linear scale. 73 Gary N4AST |
If you are thinking of E²/R, you should note this is division, not
multiplication. Thanks Richard. What I am thinking is: assuming a perfect 50 ohm load, scale the voltage divider rectifier combo to give 10.0 VDC at full scale watts, 1500 in my case. I now have a relative power indicator as suggested by Reg. I can take care of the squaring and division by putting a Log scale(watts) on a 0-10V analog meter movement. But, what I intend to do is run the voltage divider signal to a multiplier whose output will be V**2/10. If V is 10. volts for 1500 watts then the multipier output will be 10.0 volts. let 10.0 volts equal 1500 watts on the meter scale. Checking another point, assume 500 watts, then the output of the divider will be 5.76 volts, and the output of the multiplier will be 3.33 volts, which is 33% of 10 volts. The wattmeter will read 33% of 1500 or 500 watts. The scale will be linear. I have not seen it done this way before, although I am sure it has. For this method to accurate your load must be near 50 ohms. The AD633 is a cheap, easy-to-use multiplier chip. Using a +/- 15 VDC supply, it looks like it will have the dynamic range I need. 73 Gary N4AST |
1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V. As you already have 10V by divider action, what do you need to multiply? Hi Richard, I want 10.00 VDC to indicate 1500 watts on my 0-10v meter with a linear scale. If I run the 10 volts thru the multiplier (squaring) circuit I get (10)**2/10=10 volts. So far so good, and you don't need a multiplier. 500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain 5.77V. OK If I run 5.77v to my 10 volt full scale meter it will read 5.77/10*1500=865.5 watts which is not 500 watts. However, if I run 5.77 thru my squaring circuit I get (5.77)**2/10=3.33 volts. Now my meter reads 3.33/10*1500=500 watts which is correct. If you plot the voltage across a 50 ohm resistor vs power V**2/50=power, you get a squared relationship. You can use volts to represent watts only if you scale the meter face correctly. I have seen wattmeters that did this. I believe the Drake model did. 0-100 watts took up the first half of the meter, and 100-1000 watts took up the other half. If you linearize the voltage by squaring and scaling you get a nice readout with 100 watts being on the left hand side of the meter, and 1500 on the right hand side, and linear in between. The AD633 probably uses log amps in realization of multiplication, I haven't looked at the internals. I have done multiplication with Log amps. I have to disagree on the AD538 being a better solution. One could use the 538, but the 633 is much simplier to use. Also I already had some AD633s 73 Gary N4AST |
"JGBOYLES" wrote in message ... 1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain 10V. As you already have 10V by divider action, what do you need to multiply? It all works for me in Excel. Squaring the voltage eliminates the need for any log stuff. E = Root (P*R) Scaled V is V / 273 but you'll have to go further to stay in the dynamic range of the multiplier. V^2 should be around 10V or whatever the mult can output. W V scaled V V^2 V^2/10 Output ratio 1500 273.9 10 100 10 1 500 158.1 5.77 33.33 3.33 0.33 100 70.71 2.58 6.67 0.67 0.07 -- Steve N, K,9;d, c. i My email has no u's. |
"JGBOYLES" wrote in message ... 1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain 10V. As you already have 10V by divider action, what do you need to multiply? Hi Richard, I want 10.00 VDC to indicate 1500 watts on my 0-10v meter with a linear scale. If I run the 10 volts thru the multiplier (squaring) circuit I get (10)**2/10=10 volts. So far so good, and you don't need a multiplier. 500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain 5.77V. OK If I run 5.77v to my 10 volt full scale meter it will read 5.77/10*1500=865.5 watts which is not 500 watts. However, if I run 5.77 thru my squaring circuit I get (5.77)**2/10=3.33 volts. Now my meter reads 3.33/10*1500=500 watts which is correct. If you plot the voltage across a 50 ohm resistor vs power V**2/50=power, you get a squared relationship. You can use volts to represent watts only if you scale the meter face correctly. I have seen wattmeters that did this. I believe the Drake model did. 0-100 watts took up the first half of the meter, and 100-1000 watts took up the other half. If you linearize the voltage by squaring and scaling you get a nice readout with 100 watts being on the left hand side of the meter, and 1500 on the right hand side, and linear in between. The AD633 probably uses log amps in realization of multiplication, I haven't looked at the internals. I have done multiplication with Log amps. I have to disagree on the AD538 being a better solution. One could use the 538, but the 633 is much simplier to use. Also I already had some AD633s 73 Gary N4AST Gary, Comment I have is in your resistive divider. To keep the power dissipations in the resistors low, you will have to use large resistor values, but at RF these can be off 50 % or more. So, you will want to build what looks like a 'scope probe, with each of the resistors shunted with a capacitor, where the two RC products are the same. For starters, try something like 2 - 5 PF for the smaller capacitor. Do not use ceramic capacitors, except for NPO. I would build the thing with at least 2 scales, say 0 - 150 and 0 - 1500 W. Tam/WB2TT |
E = Root (P*R)
Scaled V is V / 273 but you'll have to go further to stay in the dynamic range of the multiplier. V^2 should be around 10V or whatever the mult can output. Thanks for checking my calcs. Steve. I had to do what your spreadsheet did by hand. I should note that since I have to convert the voltage to the multiplier to DC, at 1500 watts we are working with 273*SQRT 2 or 386 volts. I size the divider so that 386 this gives 10.0 volts to the multiplier. With a dual polarity 15VDC supply, the multiplier has enough dynamic range. 73 Gary N4AST |
Comment I have is in your resistive divider. To keep the power dissipations
in the resistors low, you will have to use large resistor values, but at RF these can be off 50 % or more. Well, I hadn't thought of that, so I'm glad I asked the question. I had planned on using a 200k 1/2 watt carbon and about 5k 1/2 watt for the divider, keep the leads short, and locate it a close as possible to what I think is 50 ohms in my system. Do you think it would be worth the time and effort to do an rf frequency sweep of the divider to see how accurate it is? I realize this method is not going to be dead accurate, but I was hoping for better than 50%. 73 Gary N4AST |
Hi Gary,
I like the topic of discussion. I've been working with the AD633's for several years now. I'm wondering why you vered away from the V*I idea. That would make your calculation good for any load. Instantaneous V times instantaneous I will give the answer whatever the relative phase happens to be. Maybe I'm missing something? Hi Jim, this didn't show up on my ng reader for some reason, hope you don't mind me replying on the ng. The reason I didn't the V*I route, which as you say, is the obvious way to go: I did not feel I could get accurate V and I samples over a wide range of RF frequencies, with my limited knowledge and resources. I thought a simple resistive voltage divider would make it real simple, and somewhat frequency independent. I am now finding through the responses I have gotten, that resistors will be frequency dependent as well. Darn. I recently tried to use 633s in an application at 60hz to detect when a very large generator went from producing power (generator) to absorbing power (motor). I found that the potential and current transformers I was using introduced an unacceptable phase shift from primary to secondary. I tried several different types but finally took another approach. This is probably another reason that steered me away from the V*I solution. If you have a simple way to get accurate V and I samples for say 3-30MHZ I am all ears, and forever in your debt. 73 Gary N4AST |
"JGBOYLES" wrote in message ... Comment I have is in your resistive divider. To keep the power dissipations in the resistors low, you will have to use large resistor values, but at RF these can be off 50 % or more. Well, I hadn't thought of that, so I'm glad I asked the question. I had planned on using a 200k 1/2 watt carbon and about 5k 1/2 watt for the divider, keep the leads short, and locate it a close as possible to what I think is 50 ohms in my system. Do you think it would be worth the time and effort to do an rf frequency sweep of the divider to see how accurate it is? I realize this method is not going to be dead accurate, but I was hoping for better than 50%. 73 Gary N4AST Hi Gary, I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz. This is not a precision instrument, but shows a trend. Nominal Measured 5.6 K 0 - j586 220K 0 -j 600 1.8K 99 - j539 (convert this to parallel form) As a sanity check 11 Ohms 12 + j4 (some lead inductance here) What this is tending to show is that the resistors are showing a shunt capacitance whose reactance is about 600 Ohms at 30 MHz. That is about 9 PF, which seems high. I was expecting more like 1 PF. I want to redo this at a higher frequency, might be out of range for the MFJ. I notice my Kenwood power meter uses a capacitive divider for the voltage sample. A friend of mine built a meter along the lines of what you want to do. I will ask him what he did. Tam |
Hi Gary,
I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz. This is not a precision instrument, but shows a trend. Hi Tam, I can't thank you enough for your efforts. I will be waiting for your results. Unfortunately, I am waiting for Hurricance Ivan also, so I won't be in the shop for a few days. 73 Gary N4AST |
It's much more likely that the shunt capacitance is in the MFJ.
Calculate the parallel impedance of 11 + j0 (the presumed resistor) and 0 - j600 (the shunt C) and you'll see that you wouldn't be able to see the shunt C when making the 11 ohm "sanity check". Even at HF, measurements aren't nearly as simple as they sometimes seem they should be. Roy Lewallen, W7EL Tam/WB2TT wrote: Hi Gary, I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz. This is not a precision instrument, but shows a trend. Nominal Measured 5.6 K 0 - j586 220K 0 -j 600 1.8K 99 - j539 (convert this to parallel form) As a sanity check 11 Ohms 12 + j4 (some lead inductance here) What this is tending to show is that the resistors are showing a shunt capacitance whose reactance is about 600 Ohms at 30 MHz. That is about 9 PF, which seems high. I was expecting more like 1 PF. I want to redo this at a higher frequency, might be out of range for the MFJ. I notice my Kenwood power meter uses a capacitive divider for the voltage sample. A friend of mine built a meter along the lines of what you want to do. I will ask him what he did. Tam |
Roy,
You are right (as per usual). I didn't check the MFJ269 open circuit readings. At 30 MHz, with the N to UHF adapter installed, open circuit Z is 0 & j656. Since my capacitor readings were about 10% less than that, my guess is that the resistor reactances were actually about 1/10th of that (~1 PF). If I remove the N/UHF adapter, x rises to 1389. So, most of the C is in the adapter, which was in place. I think the 12 Ohm data is OK. The capacitance washed out, and it showed a series inductance of 21 nH. As I said, this is not a precision instrument. Unfortunately, it is less precision than I thought. I hope that somebody with access to an HP or similar instrument will feel inspired to measure some resistors at RF. Meanwhile, I am going to look at some resistor manufacturers web sites. Tam/WB2TT "Roy Lewallen" wrote in message ... It's much more likely that the shunt capacitance is in the MFJ. Calculate the parallel impedance of 11 + j0 (the presumed resistor) and 0 - j600 (the shunt C) and you'll see that you wouldn't be able to see the shunt C when making the 11 ohm "sanity check". Even at HF, measurements aren't nearly as simple as they sometimes seem they should be. Roy Lewallen, W7EL Tam/WB2TT wrote: Hi Gary, I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz. This is not a precision instrument, but shows a trend. Nominal Measured 5.6 K 0 - j586 220K 0 -j 600 1.8K 99 - j539 (convert this to parallel form) As a sanity check 11 Ohms 12 + j4 (some lead inductance here) What this is tending to show is that the resistors are showing a shunt capacitance whose reactance is about 600 Ohms at 30 MHz. That is about 9 PF, which seems high. I was expecting more like 1 PF. I want to redo this at a higher frequency, might be out of range for the MFJ. I notice my Kenwood power meter uses a capacitive divider for the voltage sample. A friend of mine built a meter along the lines of what you want to do. I will ask him what he did. Tam |
All radio people suffer from delusions of measuring accuracy.
RF power measurements are the most inaccurate of all. The accuracy of measurements are a function of the instrument user. They who attempt to grasp support by stating the manufacturer's type number of the instruments used are most in need of the self-confidence it falsly generates. Either that or the statements are gratuitous adverts. How cynical can one get at this hour of the day? |
All radio people suffer from delusions of measuring accuracy. RF power measurements are the most inaccurate of all. The accuracy of measurements are a function of the instrument user. They who attempt to grasp support by stating the manufacturer's type number of the instruments used are most in need of the self-confidence it falsly generates. Hi Reg. What exactly are you talking about? I had a few minutes in between Hurricane Ivans wrath to get the Emergency generator cranked up and had a chance to read this. Lucky you don't have these things in the UK. 73 Gary N4AST |
Reg wrote
They who attempt to grasp support by stating the manufacturer's type number of the instruments used are most in need of the self-confidence it falsly generates. Hi Reg. What exactly are you talking about? I had a few minutes in between Hurricane Ivans wrath to get the Emergency generator cranked up and had a chance to read this. Lucky you don't have these things in the UK. 73 Gary N4AST ============================ Gary, we have heard the news over here about the devastating Hurricane Ivan. We get them here at about 1/2 strength of yours only once every very few years. And even then it's only over a relatively small area. If you can't understand what I am wittering about then its due either to the storm stress you are under or because you are one of those suffering from delusions of accuracy. In your case I prefer the former excuse. I hope your generator started up OK and that you and your family suffer the bare minimum of danger and damage. At least communications between us are still intact. My best wishes. --- Reg, G4FGQ |
"JGBOYLES" wrote in message ... All radio people suffer from delusions of measuring accuracy. RF power measurements are the most inaccurate of all. The accuracy of measurements are a function of the instrument user. They who attempt to grasp support by stating the manufacturer's type number of the instruments used are most in need of the self-confidence it falsly generates. Hi Reg. What exactly are you talking about? I had a few minutes in between Hurricane Ivans wrath to get the Emergency generator cranked up and had a chance to read this. Lucky you don't have these things in the UK. 73 Gary N4AST Gary, I saw an interesting curve at a resistor manufacturer's web site. It plotted resistor error as a function of F(MHz) x R(Meg) for 1/4 W carbon resistors. To make a long story short, the resistor error will be about 20% where the Megahertz x MegaOhms = 1. That means the resistor value will be 1/ Frequency. So, at 30 MHz, the resistor will be in error by 20% if it is bigger than 1/30 =.033 Meg, or 33K. That, I believe ignores capacitive effects. Personally, I have never tried to put RF through a resistor bigger than a few hundred Ohms. It occurs to me that you can ignore capacitive effects if you make all resistors identical. For instance, if you want a 3:1 divider make the series resistor 10K, and the shunt resistor two 10K resistors in parallel. Of course, you will need a high impedance load on it. Let's see if anybody shoots this down. Tam |
On Thu, 16 Sep 2004 21:26:48 -0400, "Tam/WB2TT"
wrote: To make a long story short, the resistor error will be about 20% where the Megahertz x MegaOhms = 1. And not so curiously Trc = 1 MOhm · 1 pF = 10^-6 F = 1 / T F = 1 MHz perhaps the product rule should be: Megahertz x MegaOhms x picoFarads = 1 The 20% error is, of course, simply the rolloff response at the RC inflection point described by 1/Trc. Let's see if anybody shoots this down. Hi All, I think chipping at the clay feet of saints is more appropriate metaphor. What is the saint? The RF response of the resistor. It should be suspect right out the gate. Being suspect, you employ the conventional techniques already evidenced even by the cheapest Power Meter builder (MJF) by swamping the stray capacitance with series capacitors (paralleling the resistors). One capacitor is either variable, or further paralleled with a trimmer. The saint is also the unspecified requirement: is this divider BEFORE OR AFTER the detector? If before, and thus subject to RF, the simple RC compensated divider has served for eons. If after, and thus subject to only DC - who cares? The one clay foot of the discussion. The other clay foot of the discussion is that for placement before OR after the detector, ALL ratios are post-hoc determinations (in other words, design with variable components fully expecting you WILL be wrong). Further, ALL descriptions to this point have been of normalized levels. With the RC compensated divider, you are throwing the knee if rolloff into lower frequencies so that ALL frequencies of interest reside on the same slope. Hence the common "calibration" procedure has you adjust the resistors for the low frequency readout, and the capacitors at the high frequency readout. This "calibration" is simply distributing the error so that it doesn't accumulate outrageously. The greater challenge is how do you know how much power you are setting your meter to read? Compounding errors are common in RF. 73's Richard Clark, KB7QHC |
"Richard Clark" wrote in message ... On Thu, 16 Sep 2004 21:26:48 -0400, "Tam/WB2TT" wrote: To make a long story short, the resistor error will be about 20% where the Megahertz x MegaOhms = 1. And not so curiously Trc = 1 MOhm · 1 pF = 10^-6 I think the curve ignores C, and is based on skin effect only. There is no explanation for the data. F = 1 / T F = 1 MHz perhaps the product rule should be: Megahertz x MegaOhms x picoFarads = 1 Go to http://www.xicon-passive.com/resistor.html and click on CC. There is also info on resistor performance vs frequency in the W6SAI book. He shows curves for 5 different carbon resistors vs frequency without identifying the resistor values. As a gross average, they show about 50% error at 15 MHz. The 20% error is, of course, simply the rolloff response at the RC inflection point described by 1/Trc. The curve goes from 0 - 100. I arbitrarily picked 20 % as being a point where there is apreciable error. Tam/WB2TT Let's see if anybody shoots this down. Hi All, I think chipping at the clay feet of saints is more appropriate metaphor. What is the saint? The RF response of the resistor. It should be suspect right out the gate. Being suspect, you employ the conventional techniques already evidenced even by the cheapest Power Meter builder (MJF) by swamping the stray capacitance with series capacitors (paralleling the resistors). One capacitor is either variable, or further paralleled with a trimmer. The saint is also the unspecified requirement: is this divider BEFORE OR AFTER the detector? If before, and thus subject to RF, the simple RC compensated divider has served for eons. If after, and thus subject to only DC - who cares? The one clay foot of the discussion. The other clay foot of the discussion is that for placement before OR after the detector, ALL ratios are post-hoc determinations (in other words, design with variable components fully expecting you WILL be wrong). So true. I notice the series C in the Kenwood meter is variable. Further, ALL descriptions to this point have been of normalized levels. With the RC compensated divider, you are throwing the knee if rolloff into lower frequencies so that ALL frequencies of interest reside on the same slope. Hence the common "calibration" procedure has you adjust the resistors for the low frequency readout, and the capacitors at the high frequency readout. This "calibration" is simply distributing the error so that it doesn't accumulate outrageously. The greater challenge is how do you know how much power you are setting your meter to read? Compounding errors are common in RF. 73's Richard Clark, KB7QHC |
Oh, yea! I forgot the bit about Peak...silly me.
73, Steve K9DCI "JGBOYLES" wrote in message ... E = Root (P*R) Scaled V is V / 273 but you'll have to go further to stay in the dynamic range of the multiplier. V^2 should be around 10V or whatever the mult can output. Thanks for checking my calcs. Steve. I had to do what your spreadsheet did by hand. I should note that since I have to convert the voltage to the multiplier to DC, at 1500 watts we are working with 273*SQRT 2 or 386 volts. I size the divider so that 386 this gives 10.0 volts to the multiplier. With a dual polarity 15VDC supply, the multiplier has enough dynamic range. 73 Gary N4AST |
On Fri, 17 Sep 2004 10:35:25 -0400, "Tam/WB2TT"
wrote: I think the curve ignores C, and is based on skin effect only. There is no explanation for the data. Skin effect would tend to increase resistance which contradicts the trend. As for explanation: From "Electronic Components and Measurements," Wedlock and Roberge, 1969: "At high frequencies the performance of a resistor will depart from Ohm's law because of stray capacitance and lead inductance." [pg. 77] There is an identical curve to your reference shown in Figure 7.4, same page: "Change in resistance of a ½ Watt carbon-composition resistor as a function of frequency. Frequency in MHz times resistance in Megohms" In Chapter 18 "RF Impedance Measurements": "Such behavior is often termed stray capacitance or stray inductance. Because these effects are usually undesirable and serve to limit the high frequency performance of components, they are also called parasitic effects." [pg. 276] However, my expression of this being rolloff was too simplistic as the curve does not follow the typical 10dB/Decade characteristic. Rather, it shows a 6dB/Decade+. Some of this may be accounted for in lead reactance, but at the Megohm scale this is inconsequential for conventional leads. 73's Richard Clark, KB7QHC |
The other clay foot of the discussion is that for placement before OR
after the detector, ALL ratios are post-hoc determinations (in other words, design with variable components fully expecting you WILL be wrong). Further, ALL descriptions to this point have been of normalized levels. Hi Richard, I haven't been able to keep up with this like I wished because of that pesky Hurricane. If you put the detector circuit before the voltage divider, then the resistors see DC which they are a lot happier with. The detector diode will have to be 700VDC PRV rating, and the filter cap. will have to be sized properly. I guess the diode will have some frequency dependent properties, but as long as it still acts like a diode, and the forward bias drop is around .6V it ought to work. This looks like good alternative to frequency dependent resistors. What say you? 73 Gary N4AST |
FYI
I have used either 470 or 510 ohm carbon comp resistors way back when as 20dB probes for a sampling scope and Spectrum analyzer work. If I recall, they were quite comparable to the scope's regular probes for pretty fast digital signals. I would, however, get nervous with the high values stated / needed here. At first I was thinking I'd series-up 500 or 1K resistors if I had to do this, but then strays start to become significant as has been well discussed. If you have the equipment, I like the frequency response / sweep idea. This way you may even be able to adjust the compensating caps for best high freq (10M) response. You could also use the calibrator on most good scopes to start out.... 73, -- Steve N, K,9;d, c. i My email has no u's. "Richard Clark" wrote in message ... On Fri, 17 Sep 2004 10:35:25 -0400, "Tam/WB2TT" wrote: I think the curve ignores C, and is based on skin effect only. There is no explanation for the data. Skin effect would tend to increase resistance which contradicts the trend. As for explanation: From "Electronic Components and Measurements," Wedlock and Roberge, 1969: "At high frequencies the performance of a resistor will depart from Ohm's law because of stray capacitance and lead inductance." [pg. 77] There is an identical curve to your reference shown in Figure 7.4, same page: "Change in resistance of a ½ Watt carbon-composition resistor as a function of frequency. Frequency in MHz times resistance in Megohms" In Chapter 18 "RF Impedance Measurements": "Such behavior is often termed stray capacitance or stray inductance. Because these effects are usually undesirable and serve to limit the high frequency performance of components, they are also called parasitic effects." [pg. 276] However, my expression of this being rolloff was too simplistic as the curve does not follow the typical 10dB/Decade characteristic. Rather, it shows a 6dB/Decade+. Some of this may be accounted for in lead reactance, but at the Megohm scale this is inconsequential for conventional leads. 73's Richard Clark, KB7QHC |
|
On Fri, 17 Sep 2004 16:25:56 -0500, "Steve Nosko"
wrote: At first I was thinking I'd series-up 500 or 1K resistors if I had to do this, but then strays start to become significant as has been well discussed. Hi Steve, This would probably improve the parasitic capacitance while increasing the parasitic inductance. Off hand, I think the inductance would probably be tolerable in the HF. 73's Richard Clark, KB7QHC |
A 1N4007 or any other silicon power diode is a poor choice as a
detector, for several reasons. The first is the large shunt capacitance. The second is the long reverse recovery time, which makes silicon power diodes look like a resistor rather than a rectifier at HF. High voltage diodes like the 1N4006 and 1N4007 contain an intrinsic layer to increase the voltage breakdown. This dramatically increases the reverse recovery time to the point that they can be used as PIN diodes at HF. Schottky diodes don't have the reverse recovery problem, but power diodes still have a lot of capacitance (and reverse leakage current). Only a signal diode should be used as an HF detector. If you can't find one with a sufficient reverse voltage rating, you'll have to use a divider of some sort. Roy Lewallen, W7EL Richard Clark wrote: On 17 Sep 2004 20:36:21 GMT, (JGBOYLES) wrote: What say you? Hi Gary, For a diode, the 1N4007 is rated at 1000V. However, you have to design with Peak voltage in mind, and then add 50% safety factor for good design. Further, feeding a capacitive filter requires you DOUBLE the PIV rating. All in all, this suggests your design choice should tend toward the divider before the detector. Then the question becomes, do you want a peak reading meter, or an averaging meter? A peak reading detector (AKA Clamp) will lightly load the divider whereas the averaging will load it more (and ruin any fixed divider ratio - which returns us to the variable component to be designed in). 73's Richard Clark, KB7QHC |
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