Peak Voltage Detector Circuit…
- Find your transmitter's output wattage by
using these simple circuits and your multi-meter, and begin to
learn the fundamental dynamics of Radio Frequency energy by trying out
these enlightening experiments.
Originally published in The
Printed Circuit, Newsletter of the Tallahassee Amateur
Radio Society, April 2013, page 14
Edited/Updated December 2023
You Short A Watt Meter?
you're into homebrewing
your own ham radio transmitters, low-power operation ("QRP",) or have
lack for a watt meter, there's a simple solution to finding the output
wattage of your transceiver ("rig") by just by simply
measuring for the voltage through a peak voltage detector
with your multimeter and using a little Ohm's law math. Yes,
rig’s signal going out to the antenna is more than just an invisible
force – it starts off with voltage and current components well before
it leaves your antenna as magical electro-magnetism. .
How much voltage is in my
transmitter's signal? Well, for example: most garden
transceivers will deliver 100 watts (a.k.a. "barefoot" power)
average power to an ideal impedance load of 50 ohms (50 ohms for an
ideally 'matched' antenna system,) provided that your rig is outputting
its RF signal with as a perfect sign wave (in CW-mode.) This
“100” watts, as considered, is actually the Root Mean Square (RMS) of
its output power. Even though “power is power” which is the
whether in AC or DC form; RF power, as with its characteristic sign
wave pattern oscillating back and forth in polarity at its
tuned frequency, is not constant when measured at any given
There is an instant within its wave
cycle when there
no power (zero-crossing) and at its maximum at the wave peaks.
The ham radio General Class License manual goes into detail
about RMS and how to calculate it, but we can find the equivalent
RMS Voltage (E) for our signal if we know the resistance (R) of the
transmitter and associated antenna system, which in most general cases
is designed to operate at 50 ohms. Using the Ohm's law
voltage (E) = the square root of the wattage (P) times resistance (R)
we can deduce that our 100 watt signal contains a voltage of 70.7 volts
Any radio experimenter
familiar with using a multimeter may consider the obvious option of
using one to read this 'AC' voltage. Yes, that is plausible,
only as long as the AC frequency is lower than a few hundred kilohertz
as most budget multimeters were not designed to read input signals at
higher AC frequencies, and if you look at the specifications for your
meter, you'll see that as the frequency goes up, the accuracy goes
down. Most meters stop listing specs at more than 20 kHz
just barely out of the human-perceived audio range, let alone radio.
A purpose-built frequency counter, vector network analyzer
or oscilloscope are apt choices, but even many oscilloscopes have
frequency range maximums, and all of those options are not likely
designed to accept direct power from any transmitter over a few
So I present three
variations of a simple detector circuit to convert the RF voltage of
your transmitter's signal to a measurable DC voltage readable by any
voltmeter. All you have to do is a little math to derive the
power level. . AN IMPORTANT NOTE:
These experimental circuits and methods presented here are
intended for academic use and are not designed to give crucially
accurate readings, as mentioned further. Radio Frequency
can be harmful and even fatal if improperly handled so this article
intended only for those legally licensed as amateur radio operators
and/or those with a fundamental understanding of electronics, radio
theory and any applicable safety measures involved. The
not liable for any misuse or mishap associated with this article.
Safety first! It
would be in line with ‘best practice’ to build these circuits,
and any dealing with RF on copper-clad PC board. I was able
to use my solderless prototyping breadboard with success.
You only need to worry about losses on the RF-side of the
circuit. All you'll need is a 50 ohm resistive dummy load,
either a commercial one or an actual 50 ohm resistor(s) with a
(combined total) wattage handling rating over that of your test range,
and you'll need a BNC splitter-tee with terminals. Having an
actual watt meter will be good comparison reference if experimenting.
. For The Converted.
First, a little setup. Make
rig's transmitter (TX) output is connected to a 50 ohm
'dummy load' that is rated to handle the range of wattages you want to
test. A 50 ohm dummy load is typical because most
finals are designed to work into an antenna system that
has 50 ohms of impedance. BNC connectors with some kind of
thinner coax cable like RG-158 are easier to work with and smaller
connectors and adapters are easy to find on Amazon and like sources.
Make sure to order a few SO-238 to BNC adapters.
the most standard female connector on the back of most rigs and PL-259
are the male counterparts, but, as BNC is easier to work with for small
experiments then you'll need to convert. .
Next, you'll need a way to sample the power from this line.
can use a BNC female 'T' splitter... actually, a bunch of adapters that
ended up making a 'T' was what I used and that was terminated by a
center connector for a dipole. You can use “alligator” clips
connect the terminals to the circuit. Using an actual RF
connector with very short wire lengths overall is preferred.
Build one of these circuits and connect your multimeter
Volts. Make sure that your meter is set to a higher voltage
first if it doesn't have an auto-range feature as you'll want
surprises that would zap a good meter. Once set up – You
now have an RF probe!
measured voltage will be a constant DC signal and our input RF (AC)
signal is an RMS sign wave, how do we convert one to another?
We'll use the following formula to convert this DC voltage
reading to RMS Power by using the Ohm's law formula: .
RMS conversion is another thing.
For true RMS
conversion, especially with non-sign wave signals, the math alone would
cover half this page, but we hams don't really need to worry about all
that, as we, for the most part, can transmit pure sign wave signals,
which requires much less math to. In this case, we'll be
with CW (continuous wave) and not an SSB voice signal. Since
Root Mean Square of a sign wave can be found by getting the
root of '2', then we'll want the inverse of that to go from
value to RMS: 1 / √ 2. Simply put, for a pure sign wave, hams
use the multiplier: 1.414 for RMS and 0.707 for the
So when we read the DC voltage from our voltmeter using our
Voltage detector circuit, we can take that reading and multiply it by
0.707 to get our RMS voltage. With that, we can then
our RF Power figure (which is in RMS) using Ohm’s Law. Don't
worry, the formulas are to follow, and maybe the Digi-Key
Ohm's law calculator may be a help. . Reading
RF Output in the QRP Range: 1 - 10 Watts
The base of this circuit is
as about as simple as it gets; it only requires a diode and capacitor.
If you're into experimenting with
transmitter circuits or enjoy building QRP (low-power operation)
transceiver kits, the measurement of your output signal power
eventually becomes a necessity! Without an oscilloscope or
QRP-level watt meter you might be left guessing. Try this
watt version of the PVM circuit shown here: .
One common popular switching
diode, a 1N4148, rectifies the incoming RF (AC) voltage, which should
be a pure sine wave, and passes that on to charge a small capacitor.
Once the capacitor fills to a nominal peak input
voltage level, a more stable voltage is available to be sampled by any
standard voltmeter. The resultant measured voltage will be
the DC equivalent to the RMS voltage of the sampled RF output.
To take our measured DC voltage to find the RF output power
(wattage) of our transmitter, we first figure the RMS voltage by
multiplying the DC voltage times a factor of 0.707. So for
instance, if your voltmeter reads 21.1 volts; 21.1 times 0.707 gives us
14.92 volts RMS. Finding the wattage of our signal is now a
simple application of Ohm's law. .
Use the Ohm’s law chart to
help. So power (P) = voltage (E)-squared divided by
resistance (R). P=E2/R
The resistance is of course, the 50 ohm dummy load connected
to our transmitter output. By the way, because this is
resistance load is measured when RF power is present, your dummy load
my not read 50 ohms when power is not present, depending on the type of
load, but that's beyond the scope of this article. In my
experiment, the dummy load, along with the splitter detached from the
rig actually reads 72 ohms, but it sees 50 ohms when running.
So using the formula we find that (14.92)2/50 = 4.45 watts!
Now, with lead lengths and stray capacitance, the subsequent
losses on the line may cause your reading to not be very accurate, but
it should be close. I measured 4.3 watts with the commercial
watt meter. . Considering
Because we're working with
RF energy, fast-switching diodes such as the germanium type,
Schottky, or other RF-rated diodes should be used in this type of
circuit. The ubiquitous 1N4148 switching diodes
should work fine on frequencies up to 30 mHz, but above that, other
diode types will have to be considered. Forward and reverse
voltage rating maximums should be considered; after about 20 watts, the
RMS voltage across these small diodes will surely breach their max
voltage ratings. Larger power rectifying diodes such as the
1N4000 series are not really useful over their intended design
frequency (i.e. 60 Hz), so you should avoid those for RF work.
Diodes do not linearly
conduct within its voltage range! Even if you use a circuit
that samples the RF via transformer action such as commercial watt
meters, this will be an issue. To solve this, watt meter
designers will coordinate the needle dial graphic with the actual
observed results. With microcontroller controlled watt
meters, a lookup table, formula or mapping structure will be used to
convert the read voltage to a digital number. Diodes also
have a voltage drop (typically around 0.6 volts) that is also skewed by
current flow through the diode, and this also has to be considered.
The voltage drop should be somewhat consistent over a range
of frequencies. Do a little research with diodes of
different specifications and you may find a type with a lower voltage
drop or more consistent conduction/current relationship. The
following circuit will address this to some degree. . Reading
RF Output in the Barefoot Range: Above 10 Watts
If you use the circuit
described above, once the input power gets into the 20+ wattage range,
you'll surely destroy your diodes. By using a simple resistor
divider network pad to scale the input voltage down, you can spare your
diodes an eminent death and still have something useful to read by your
meter. Use can use this useful Voltage Divider network
Calculator from Digi-Key [HERE] to find the best
resistor match necessary to derive a proportional voltage you can work
with. If you use higher resistor values for the network, then
lower current will be drawn away from your input power which
would skew your voltage reading. In this circuit, basically,
we’re using two values of resistors configured together to scale the
output voltage down to a 10th of its input to make the math easier,
i.e.: 50k ohms / 5k ohms. I've used the values of 47k ohms
and 5.1k ohms here because that's what I had on hand and it's close
A few caveats* to consider
when using a resistance dividing network in this PVM circuit:
resistors have manufacturing tolerances and do not generally
contain all the resistive goodness advertised on its package.
A network will adversely scale these inaccuracies, so a
little adjustment will need to be made when building this circuit.
Consider using a potentiometer (in a network
configuration) or try reading DC voltages with just the divider portion
of the circuit in a simple DC circuit to adjust one or both of the
resistors so that you'll know you're going to get a 10 to 1 reduction
of input voltage. Since this is an RMS/RF voltage on the
divider, if you use a potentiometer to fine tune the divider, make sure
that the leads are short and that the potentiometers are not an
capacitive type. In my experiment, my resistors were way off
from the get-go, so I changed the resistor values and my final reading
was within just a watt or two even when testing up in the 100 watt
Of course, this is an
academic exercise and anyone outputting over a few watts could simply
use any readily available watt meter, but what about the experimenters
trying less-than-a-watt communications (known as “QRPp”)? The
following circuit may be the most valuable here! .
RF Output on a Flea-Powered Transmitter: Less-Than a Watt
At less than a watt of power
output from a QRPp transmitter, your multimeter should have no problem
giving you accurate low-voltage measurements, but now the common 0.6 to
1.6 voltage-drop on some diodes becomes a prohibitive issue.
It generally worsens with smaller voltages. In the following
circuit, I added another diode in parallel to the source to give
full-wave rectification, adding power coming from the other direction
on top of the capacitor charge and we double the voltage! .
This configuration is known as a Charge
Pump and is the
staple of boost converters, switching power supplies, flyback
transformers and Jacob's Ladders. So you can get a double
voltage boost from this configuration, but don't forget to divide the
measured DC voltage in half before entering it into your RMS/power
formula. Perhaps subtracting the voltage drop. .
. . Considerations.
A Peak-Voltage Meter circuit
has many useful radio applications, especially when it’s impractical to
use a dedicated watt meter. The DC output voltage could be
useful to trigger comparator circuits and for Tx/Rx switching relay
circuits. If you're working with live RF into an antenna
system, your results should still be useful as long as the impedance
impact of your circuit along with the transmitter/antenna system is at
a known value, such as 50 ohms. Since you’ll have the voltage
and resistance as known factors, you’ll also know the current of you
signal as well. .
A Loss... But Wait, There's More.
Though, since this circuit
reads Peak Voltage only, you'll not be able to measure directional
current, reactance/admittance or any present standing waves (SWR) of
your antenna system. That would require a circuit using a current
transformer. And for UHF/VHF frequencies, unless you're using
diodes rated for these higher frequencies, anything over the HF range
might not be accurately readable. You’ll also most likely see
substantial system power losses due to all the connectors and
added wire lengths. A circuit like this is not able to read
the complex impedances, capitances and reactances involved.
wait, I've only presented a small introduction into understanding what
comes out of your rig. For most "appliance operator" hams,
they may care about is the SWR and output watts. .
days, there are tools and devices that show you which used to take
Smith charts and dozens of calculations, and their inexpensive
highly suggest learning about using a Vector Network Analyzer (VNA) to
discover the wonders and nuances of your antenna system and any
experimental RF circuit. A NanoVNA can be had for
$50 and I suggest watching IMSAI Guy's very informative YouTube series
on the NanoVNA
With a NanoVNA you can uncover the complex impedences and capacitances
of your antenna system, see the losses and electrical length of your
coax, find the resonant frequencies of your antenna, find the
inductances of your hand-made toroid coils, and much more!
There's also a NanoSA Spectrum Analyzer for cheap to show you
the harmonics in your system. YouTube also has quite a few
tutorials on building transformer-based RF power meters, especially for
QRP levels. .
I provide this article only
as an academic exercise, and perhaps a way for some of you hams to get
to know a little bit more about what comes out of the back of the thing
you call a "rig". I had fun with this and I’ve been able to
better tweak the output of my current QRP rig. I didn’t even
need to warm up my soldering iron!
Please see the following
video from IMSAI Guy on YouTube for more insight on deriving power
information from your rig without a dedicated watt meter: