Typical SWR meters don't measure forward power very accurately at the lower end of the scale, which can be a real problem at QRP power levels. Better power meters like the Bird model 43 are expensive, and the calibration of used models can be very uncertain. However, I've heard that an RF probe, a dummy load, and a multimeter can be used to make a power meter. I've also heard that such a power meter can be more accurate, if an oscilloscope is available for calibration.

How can one make an RF probe, and how does one use an RF probe for accurate power measurement?

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    $\begingroup$ what is a QRP power level? Can you put numbers to that? $\endgroup$ Mar 25, 2020 at 18:01
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    $\begingroup$ The usual definition for QRP is < 5 W. $\endgroup$
    – rclocher3
    Mar 25, 2020 at 20:27
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    $\begingroup$ Wow! That is low power? You can measure whether something is 1 W or 1.2 W pretty solidly by terminating it with a resistor, and measuring that resistor's temperature rise... $\endgroup$ Mar 25, 2020 at 21:06
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    $\begingroup$ Check out this classic by Hayward: qsl.net/sz1a/download/build%20an%20rf%20power%20meter.pdf. Calibration can be tricky (the usual method involves using a 10MHz CMOS oscillator). $\endgroup$
    – Buck8pe
    Mar 26, 2020 at 14:08
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    $\begingroup$ ...and here's the accompanying calibration project (sp-hm.pl/attachment.php?aid=1043) by Kopski. Also look for the revised version of this for newer versions of the AD8307. $\endgroup$
    – Buck8pe
    Mar 26, 2020 at 14:29

6 Answers 6


A basic probe looks something like this:


simulate this circuit – Schematic created using CircuitLab

Ideally the diode has a low voltage drop, like a Schottky diode. But any ordinary silicon diode will work in a pinch.

The capacitor and resistor values aren't really critical as long as the time constant (the resistance times the capacitance) is much greater than the period between cycles. Something like 1nF and 10kΩ would be fine. I usually use whatever I have laying on the bench.

The circuit works because when the input voltage is positive, C1 can charge through D1 up to the peak RF voltage (less the forward voltage drop of D1). But when the input voltage is less, C1 can discharge only relatively slowly through R1.

To use, set up your transmitter to dump power in a dummy load or whatever, and then place this probe in parallel. VM1 can be an ordinary digital multimeter because this voltage changes only as fast as the RF output power, which for an unmodulated carrier, will be DC.

It's important that the leads on everything left of C1 are kept short. Otherwise the added inductance of the long leads will introduce significant errors in the measurement. This becomes more critical with increasing frequency.

To convert the measured voltage to power:

  1. add the forward voltage of D1
  2. divide by $2 \sqrt 2$, converting peak-to-peak to RMS
  3. use Ohm's law to calculate power for this voltage, assuming some load impedance

Assuming a 50Ω load impedance, the calculation is:

$$ \text{power} = { \left(V_\text{measured} + V_{D1} \over 2\sqrt 2 \right)^2 / 50\Omega } $$

For low power levels where the forward voltage drop of D1 represents a significant fraction of the output voltage of the transmitter, accurately characterizing that voltage drop is critical for accuracy.

The simplest method for determining the voltage drop is to read it from the data sheet, but behavior at RF may not be accurately characterized.

A more sophisticated approach could bias the diode to compensate for the forward voltage drop:


simulate this circuit

To use, temporarily short the input terminals, providing an input of 0V. Adjust the pot for minimum voltage, then turn it up to the point where the meter just starts to measure some voltage. You've now biased the diode so it will turn on at 0V, at least for DC.

The bias can be further tuned if you have or can build an attenuator of known value. Say you have a 3dB attenuator: this represents a halving of power, and because power is proportional to the square of voltage, a reduction in voltage by a factor of $\sqrt 2$.

Perform one measurement, then again with the same input power but with the attenuator. If the diode is properly biased, the voltages measured by the meter should differ by a factor of $\sqrt 2$: any difference is likely due to uncompensated diode voltage drop. Correct it by adjusting the pot or by manually adding a correction factor in the calculation.

Accuracy could be further improved by adding a transformer on the input to increase the voltage, thus rendering error due to the diode less significant.

  • $\begingroup$ Niice answer! Especially like the how-to-use section! $\endgroup$ Mar 26, 2020 at 9:40
  • $\begingroup$ Is your explanation really how it works, though? In my head, it's not, at all! You're arguing the diode effectively works as a half-wave rectifier, right? (<-- honest question, I might be misinterpreting you.) But: the bias should always have the diode in forward operation, so what the diode actually does is introduce a nonlinearity (which in the end is a mixer, mixing the input signal to 0 Hz ($f_{in}-f_{in}$) and harmonics $f_{in}+Nf_{in}$), and the low-pass kills all the harmonics, so that the scope measures the 0 Hz part. $\endgroup$ Mar 26, 2020 at 9:43
  • $\begingroup$ @MarcusMüller I don't know, maybe that's a valid frequency domain explanation of the same thing? In my head, it's simply a peak detector, and the bias only compensates the forward voltage drop of the diode. Once the circuit has reached equilibrium, the diode spends most of its time reversed biased or below the forward voltage. $\endgroup$ Mar 26, 2020 at 18:09

How to make an RF probe for accurate power measurement at QRP power levels?

                               (emphasis mine)

Now, accurate measurement of RF power is actually pretty hard once you cross over into bands that your oscilloscope can't measure directly anymore. Say, you want to measure what your 21 cm LNA does.

Sure, you'd rectify the signal, first, like in Phil's excellent answer, but that assumes you know the frequency behaviour of your diode – which you really do not at higher frequencies. If you even could know the frequency-dependent behaviour of all your semiconductors, you'd have little need to measure the output power of your amplifier, you could just calculate it on paper (it's really nothing but a single transistor, seriously).

Minicircuits LNA

This LNA is really just a transistor, plus a few passives to reliably bias it, plus a few passives to impedance-match. If we knew how semiconductors like diodes and transistors behave at high frequencies, there'd be no need for the manufacturer to measure and calibrate these.

So, long story short, you need to build a detection system (like Phil's!) and then calibrate it; get the nonlinear effects out of there, and the effects that depend on frequency:

A diode is not only the perfect diode you've read about ("current through the diode is exponential to voltage across the diode yadda yadda"), but when looked at with dynamic signals, also a capacitor, meaning that it does, in fact, pass higher-frequency AC signals pretty well, especially in low-voltage situations, where charge carriers get "dangerously close" to each other and the diode "looks" like a plate capacitor with very close-by plates. (By the way, that's what a varicap is: a voltage-adjustable capacitor, which is just a normal diode in reverse operation, optimized for high capacitance shift.)

Diode Capacitance over reverse voltage

This diode has a bit of a voltage-dependent capacitance and hence acts differently on different frequencies, depending on the voltage, just to make things less easy

Soooo, you'll go, build your first measurement circuit, buy a USD 100,000 spectrum analyzer and start calibrating. Or maybe you just buy a used one, spend a lot less, not like you need to go up to 60 GHz or anything, but still


New: a proper boat. Used: A car. Or the financial equivalent.

Then you go get a coffee, leave the lab door open, the room cools down, you go back in, notice that everything changed, sigh, add a temperature probe to your measurement circuit board, note down the temperature, and spend the next four days calibrating your board. Then a friend comes over, sees what cool stuff you can do with your self-built power meter, and asks that you make him one, too. So, you take your spare parts, assemble another one, and sit another three days in the lab, calibrating that board, because at microwave frequencies no two semiconductors are identical, and especially not yours...

Diode Capacitance Change over Temperature

The same diode obviously also has a temperature dependency. You get the feeling physics doesn't like you... and that's a diode already partially optimized for temperature-stable behaviour

Doesn't sound tempting, right? If only there were companies that sold small circuits that they built, and tested, and calibrated themselves for this purpose...

Good news, everyone!

They do. For example, Analog Devices (ADI) has a few of them. This demo board stuck out:

Analog Devices CN0399

ADI CNO399 evaluation board for the ADL5904 power meter IC

What it does is relatively straightforward:

  1. There's an envelope detector, like in Phil's answer, but with a lot of compensation built-in
  2. There's an analog-to-digital converter, because, well, something's actually got to measure the output
  3. and there's a bit of clean, stable voltage supply circuitry and input matching on there

Eval board block schematics

CN0399 Block diagram

If you can shell out USD 150, just buying that eval board, and then talking SPI to the ADC (e.g. from a microcontroller board like an Arduino, or from a Raspberry Pi), that's a working solution; the "what the ADC says vs what was on the input curves" do look pretty nice:

ADC output over input power graph

CN0399 ADC output over input power graph

Of course, you can also just download the board layout files, have them manufactured for cheap, buy the components, solder them on and probably come out cheaper (but not as well-tested, probably, and on a different FR4 substrate, so the matching might be slightly off).

The purpose of such boards, however, is not production usage; it's to encourage you to use the schematics to build your own power meters from. Maybe the ADC is not what you're looking for? You'd be better off with an Opamp non-inverting amplifier that drives a simple analog voltage gauge? Sure! Just do that, replace the ADC, design your own board. Really, the ADL5904 isn't such a complicated beast to use in a minimal RMS detection setup:

fig. 44 from ADL5904 Datasheet

From the ADL5904 datasheet: Figure 44, Basic Connections for RMS Power Measurement

Counting that: One IC, four capacitors, two resistors; one coax connector for the input, and a stabilized 3.3 V supply IC, plus probably two more caps for that. Done!

As a starting point, this will be quite a bit more accurate over quite a larger frequency range than a much simpler single-diode envelope detector. If you can actually verify your input matching and use your oscilloscope on a low-frequency oscillation to input a known power to the detector, you'll be able to put a high degree in certainty in your power measurements.

  • $\begingroup$ I think your answer is closer to the mark for smaller signals. What about the abundance of AD8307 DIY power meter projects? $\endgroup$
    – Buck8pe
    Mar 26, 2020 at 13:54
  • $\begingroup$ @Buck8pe didn't know these! Generally, I pick components based on availability, purpose, price and ease of usage, and the ADL5904 fit these criteria. "Existing DIY projects" isn't one of my criteria! $\endgroup$ Mar 28, 2020 at 13:39
  • $\begingroup$ @Buck8pe generally, "the manufacturer has a tested and simple eval board" trumps "there's a lot of amateur work that uses this chip", usually; one tested-by-who-makes-these-chips design is better than 100 this-works-according-to-the-beardy-guy-in-the-club circuits :) $\endgroup$ Mar 28, 2020 at 13:41
  • $\begingroup$ However, "ease of use" VERY much is a criterion! So, referring to existing reference DIY projects usually is a good idea! I just find it often pretty hard to assess the quality of my references in the amateur radio "sphere" – I mean, there's literally people recommending to use Germanium diodes, in 2020. Hams use AX.25 for their communications to satellites they're shooting to space NOW, where every communication engineer can tell you something about how stupid that frame format is on a link that needs FEC to be reliable. The amateur radio "existing solutions" world is basically a minefield. $\endgroup$ Mar 28, 2020 at 13:44
  • $\begingroup$ wow, the AD8307, being an older device, covers much less frequency range, and costs, when bought somewher you can buy less than 1000 of them, about twice as much than the ADL5904. So, that kind of settles that for me :) $\endgroup$ Mar 28, 2020 at 13:48

First off, a 1 Watt transmitter puts 7.07 volts RMS across a 50 ohm load, or 10 volts peak. So most any diode detector circuit using a silicon signal diode (perhaps a 1n4148) would be plenty accurate for ham uses. For a 0.1 Watt QRP transmitter (3.16 volts peak), a schottky 1n5711 diode would still keep it accurate within 10%. If the voltmeter has a 10meg impedance (to keep diode currents low), you could accurately measure down to 0.01 Watts using that 1n5711 and a calibration chart. Note that the voltage drop across a diode is a logarithmic function of the current, that drop can be astonishingly low when currents are under a microamp. Germanium diodes can have lower voltage drops than schottky diodes, though are hard to obtain and are more sensitive to temperature. Many shops selling cheap 1n34a's these days actually ship a schottky instead.

Phil's diode probe looks good, though the meter sees peak, not peak-to-peak, so should divide by sqrt(2) to obtain the RMS voltage, not 2*sqrt(2). I would do away with R1, relying instead on the meters input resistance to perform that function, keeping diode currents to a minimum. The bias source may not be as useful as you might think, since larger signals will create larger diode currents and thus larger diode drops.

Jim's classic 1n34a probe is similar to Phil's, but a bit harder to understand. Not obvious to me why either topology is inherently better. Jim does show the cap and diode in their correct positions. The cap does double duty, both DC blocking of the incoming RF, and it gets charged up whenever the diode's anode is above ground. The result is that the diode's anode has an RF signal equal to that being measured, but with a DC shift such that the bottom tip of that RF signal is at ground. The voltmeter (he shows a scope) is assumed to have sufficient input capacitance to create a low pass filter with the 4.7meg resistor. The 4.7meg resistor in series with the assumed 11meg internal resistance of the voltmeter (typical of many DVM's, and VTVM's too if you remember what those were) scales the voltage reading down to give volts RMS.

The AD8307 is easy to work with, can be bought on ebay for well under a dollar, good to 500mhz (which is plenty for most of us), has a 90dB dynamic range, and is easier to dead bug than an ADL5904. That would be my second choice if the classic 1n34a/1n5711 diode probe was not sensitive enough. You can buy a complete AD8307 RF probe on ebay for $10.

Getting back to basics, I have measured power output from transmitters by putting a finger on the 50 ohm load to see how hot it is getting as suggested by Marcus. Compare the rate of temperature rise to a similar load driven from an adjustable DC power supply, adjust the DC voltage until the temperature rise is identical.


Especially for narrowband (by modern standards, I'd call that anything < 50 kHz bandwidth) signals, the oversampling that you can do with even the cheapest SDR devices (e.g. RTL-SDR dongles) gives you excellent detectability, down to thermal noise. (Make sure you turn of any AGC, though.)

So, assuming you just want to know what the power of that single tone is, which has low phase noise (and hence very low actual bandwidth), you'd tune your SDR such that the LO of the mixer doesn't lay over your signal of interest, then filter your signal drastically in the digital domain to restrict the observed power to really your band of interest.

Then magnitude-squaring and averaging gives you the digital power.

But, what's the physical power?

With the SDR alone, you couldn't say – SDR devices are not calibrated measurement devices.

However, if you have a source of known physical power, you can measure that, and calculate a factor between digital and physical power. Easy!

Now, the task has just been reduced to:

  1. Build a digital-power-detection system: RTL-SDR, and GNU Radio companion (osmocom Source -> complex to mag^2 -> moving average -> Qt number sink; done!).
  2. Build a way to switch between your reference power source and your device under test; in a pinch, reliable screw connections like SMA can be unscrewed and rescrewed without incurring significantly different losses
  3. Build a reference source of power.

The third point is probably the only thing you'd need to build; in interest of good performance, a lab would usually buy a calibrated noise source (examples), or use a calibrated spectrum analyzer to calibrate one, but seeing you probably are running low on lab-grade RF equipment...

Now, that's not your only option. For example, an oscillator can relatively reliably be built with a known output power in the HF ranges; you'd only need an oscilloscope to trim the gain on an opamp until the amplitude over a 50Ω load is exactly, say, 5 mV, and the rest is math :)

For higher frequencies, there's really no way around building something with a constant power, and getting it somewhere that can measure it. If you're building a semiconductor-based 800 MHz noise source to compare your other devices too, but it turns out your build is 3dB off in power, then all your measurements would be invalid.


Most homebrew RF probes use a diode with low voltage drop to rectify the RF. You then measure the resulting pulsed DC voltage to determine the power. Most people use a germanium diode like a 1N34, but these can be hard to find. You can use a schottky diode instead. You want the voltage drop across the diode to be as low as possible. A diode with a high voltage drop like the 0.7 volts of most silicon diode will distort the reading.

Add a high value resistor, 4.7 megaohm works well, and small capacitor, 0.01 microfarad for example. Connect them up as shown.

There is still going to be distortion cause by the voltage drop of the components, especially the diode. This will be noticeable at QRP power levels.

Some dummy loads include a rectifying diode that lets you measure the voltage to determine the power going into the dummy load. They basically are including a RF probe in the dummy load.

enter image description here

  • $\begingroup$ I think you have the capacitor and diode flipped. $\endgroup$ Mar 25, 2020 at 20:53
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    $\begingroup$ "Most people use Germanium": yeah, but just because someone told them "most people use...", not because it's a good idea. So, your Schottky recommendation is spot on. $\endgroup$ Mar 25, 2020 at 21:07
  • $\begingroup$ Maybe you can add a word or two on how measuring using this works? I can't figure it out $\endgroup$ Mar 25, 2020 at 21:09
  • $\begingroup$ It's really only "most people" in the 50s to the 70s that would have used germanium. These days, most people can't even buy a germanium diode and there's really no reason to use them. ham.stackexchange.com/questions/9133/… $\endgroup$ Mar 25, 2020 at 21:42
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    $\begingroup$ @rclocher3 True, ebay sells all kinds of obsolete electronics. But seeing as how the entire electronics industry has moved on to higher performance, cheaper components 50 years ago, perhaps it's time the amateur radio community does, too? $\endgroup$ Mar 26, 2020 at 0:00

Building is educational and fun or you could just bye the W1 from Electraft (disclosure: I own one and love it) https://elecraft.com/collections/test-equipment/products/w1-100w-wattmeter-kit-1


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