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Left Coast, CA (USA)
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Measurement changes behavior
 « Reply #15 on: March 14, 2012, 03:24:45 pm » Bigger Smaller Reset

Lefty, I have a couple of questions on your RF load.  You have folks adjust the voltage up by .4 to allow for the diode, doesn't it have a .7V forward drop?  Also, you show two diodes in the picture and only talk about one in the description.  I realize that .3V or even 1V with two diodes isn't statistically significant in the scheme of things, but it might confuse someone (besides me).

The author of the article told that he used two series diode because that is what he had on hand and needed the PIV value that two series diodes provides. I assume if you were to use the one in the drawing it has a high enough PIV rating. And yes the diode drop will limit both the total accuracy possible and the minimum amount of RF power that can be measured. But for the purpose of measuring the nominal RF output power of a typical 100 watt ham transmitter, it is simple, cost effective, and that method has been used for many many decades.

I also wonder about the voltage reading since you only use a half wave to charge the capacitor and RMS calculations usually use the entire wave form, but then again this would only be a small error at HF and could probably be ignored.  The other question is the value of the capacitor.  At 3MHz it's impedance should be (if my math isn't too rusty) less than an ohm rising to around 5ohms at 50Mhz.  What am I missing?

Well to review, a half wave filted rectifier with charge up to the peak value (not peak to peak) of the RF signal (ignoring the diode voltage drop), the fact that it's half wave Vs full wave does not effect the measurement value as the meter used to read the value is required to have high (10megohms+) input impedance, so as to not 'load down' the measured value. The value of the cap has no bearing on accuracy as it will charge to the peak value of the RF voltage. Power can be calculated with this by multiplying the peak value by .707 to give the RMS voltage value. Power is then calculated by: (Vrms X Vrms) / 50 ohms. This assumes that the RF being measured is a sine-wave. Keep in mind this kind of measurement circuit is mostly used to just allow a visual reference to be watched as one makes tuning adjustments on a transmitter looking to adjust things for a maximum value. The fact that one can calculate the approximate power is usually just a secondary purpose or goal. Many ham operators certainly use other more accurate watt meters, but it's hard to beat the bang for the buck that this simple diode method allows. I personally didn't build that measurement circuit into my version of the dummy load. I rather just hook my scope to the power input and ground via a BNC T adapter and measure the RF signal directly on my analog oscilloscope as a AC signal.

Lefty
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 « Reply #16 on: March 14, 2012, 11:00:01 pm » Bigger Smaller Reset

But, the capacitor is directly across the terminals from center through the diodes to ground, so it's in parallel with the resistors.  Measuring across it would see the combination of the impedance of the capacitor (around 1 to 5 ohms depending on frequency) and the reisistance of the load resistors.  Something a lot lower than 50 ohms.

I now realize that you didn't write the article, so you can't speak for the author, and I understand that you don't use his measurement techniques; just the load itself.  I started wondering when he missed on the diode forward bias drop value and cited how accurate it was.  Which it could be at higher wattage levels where a lower value won't matter, but not for a 1W handheld.  Now, I'm trying to understand how it can work.  Every calculation I do show essentially a dead short through the .01 microfarad capacitor.  Even when I consider that the frequency is cut in half by the diode in series.

I bet he meant to type .01pF or something similar, which would make more sense.  Or maybe I'm missing something.

You're right about how simple the load portion of this project is though.
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Measurement changes behavior
 « Reply #17 on: March 15, 2012, 01:30:04 am » Bigger Smaller Reset

But, the capacitor is directly across the terminals from center through the diodes to ground, so it's in parallel with the resistors.

Measuring across it would see the combination of the impedance of the capacitor (around 1 to 5 ohms depending on frequency) and the reisistance of the load resistors.  Something a lot lower than 50 ohms.

No, the resistors (the 50 ohm load) wire to the BNC connector's center pin and ground. The cap is wired to the output side of the diode to ground. The cap just filters the halfwave rectified signal so as to hold a charge equal to the peak of the RF signal voltage. The diode effectivly isolates the cap from the load resistor(s).

I now realize that you didn't write the article, so you can't speak for the author, and I understand that you don't use his measurement techniques; just the load itself.  I started wondering when he missed on the diode forward bias drop value and cited how accurate it was.  Which it could be at higher wattage levels where a lower value won't matter, but not for a 1W handheld.  Now, I'm trying to understand how it can work.  Every calculation I do show essentially a dead short through the .01 microfarad capacitor.  Even when I consider that the frequency is cut in half by the diode in series.

I bet he meant to type .01pF or something similar, which would make more sense.  Or maybe I'm missing something.

Why don't you check out this link about building and using simple low power measurement circuits using diodes. It's better written then that first link I used. The circuit there called 'peak voltage detector' is the exact equivalent of the first one I linked.  http://epic.mcmaster.ca/~elmer101/rfpower/rfpower.html and a construction page at http://epic.mcmaster.ca/~elmer101/rfpower/projvi.html

Now if your only interested in measuring low power RF signals then you might look into the chip I built a small RF milliwatt meter with. It has a frequency range of DC to 500Mhz and a power measurement range from -75dbm to +15dbm and is very accurate, which is total of 90db of range, a feat just short of amazing. It's based on the Analog Devices AD1307 logrithmic amp in a 8 pin DIP package and is powerd with just +5vdc. I paid about \$15 maybe 10 years ago for the chip, but in my opinion worth every cent as accurate RF power levels to 500Mhz is not a trivial thing. It's output analog DC voltage would be perfect for wiring to a arduino analog input pin for display or logging functions.

http://www.analog.com/static/imported-files/data_sheets/AD8307.pdf

Lefty

You're right about how simple the load portion of this project is though.
 « Last Edit: March 15, 2012, 02:11:36 am by retrolefty » Logged

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 « Reply #18 on: March 15, 2012, 02:28:24 am » Bigger Smaller Reset

Wait, I finally figured out where I was having problems.  I was thinking impedance which is a result of an AC current flowing through the capacitor.  Not really flowing through since the alternate voltage will suck the power off the cap as the wave form varies.  The diode doesn't really isolate the cap, it changes the RF into pulsed DC and there is no bottom side of the wave form to affect the cap.  This means that the cap will charge up to peak voltage in a few cycles and essentially stay there.

None of the sites clarifies this simple little item.  I even looked at a bunch of power supply calculation examples trying to understand why the impedance of the capacitor wasn't a factor other than filtering the pulse by discharging along the current path.  My kind of thinking (related to AC impedance) comes from calculating the proper capacitor value to control a ceiling fan since a resistor would have to be too large and a triac introduces way to much noise into the motor assembly (causing a buzz).

Sigh

I was right about the diode forward voltage drop though
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Measurement changes behavior
 « Reply #19 on: March 15, 2012, 11:21:43 am » Bigger Smaller Reset

Quote
I was right about the diode forward voltage drop though

There is a pretty easy method to 'compensate' for the diode voltage drop using a simple op-amp circuit that uses a second 'matched' diode in it's feedback circuit to factor out the voltage drop and allow small millivolt measurements. One of those last links shows and tells about it.

Lefty

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 « Reply #20 on: February 08, 2013, 02:19:11 pm » Bigger Smaller Reset

In winter holidays we decided to work on our dummy load.

We posted some pictures of the design and the completed device.

http://www.binarytaskforce.com/Weblog/2013/02/08/dummy-load/

Cheers

Team BTF
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