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An Accurate Power Meter For QRP Ops

Contributed by Alex Mendelsohn, AI2Q

Have you ever wondered what the real power output of your QRP rig is? How about being able to measure QRPP?

Here's a circuit using two junkbox transistors and a garden variety op-amp that will let you accurately measure your rig's flea power down to milliwatts. It makes use of your ordinary shop multi-meter. If your DMM or analog voltmeter's accuracy is good, your results will be too. It works because power is proportional to the square of the voltage across a resistor, assuming the resistance stays the same.

Although it can't measure a rig's output power while operating into an antenna, it will tell you how much power your QRP rig puts out into 50-ohm resistive loads like a well-matched coaxial feedline. The best thing is that it's accurate!

Simply Use A Bulb

The basis of the circuit is an ordinary filament type bulb. If you build a small attenuator as well, you can measure up to 5 W. This circuit was originated by G4COL. My version uses parts available here in the US. The semiconductors can be obtained from Radio Shack. You can build it on a piece of perf-board, or Manhattan-style (more about that in a later article), and then subsequently mount it in a small cabinet.

As you may know, the resistance of a bulb changes depending on how much current goes through it (how hot it gets). This simple circuit allows you to adjust the resistance by passing a DC current through the bulb. The bulb's brightness is monitored by an infrared phototransistor.

The phototransistor's output feeds an op-amp that keeps the current constant. The pot R8 sets the brightness. The brightness of the lamp is always held constant because of the closed feedback loop; transistor Q1 is fed by the op-amp. How much it conducts sets the current through the bulb.

The current flows from ground (I prefer the old fashioned electron flow convention), up through the RF choke, then through R2/R3 combo, through Q1 from collector to emitter and finally back to the positive side of the 12 V power supply. To see this, trace the current flow with a colored pencil. If you study it, you'll see that the bulb brightness is always the same, providing a constant impedance match over all the bands from 160 meters up through ten meters.

Your RF is fed into capacitor C1 and is coupled to the bulb. The bulb is also fed with DC, although the RF is blocked from Q1 by the RF choke (it passes only DC). When the bulb lights, its light causes current flow in the phototransistor Q2, causing a voltage drop across R5. This voltage is "felt" by the non-inverting input of the op-amp, and is amplified and appears as output.

The output voltage of the op-amp will change (although it happens too fast for you to see on a voltmeter) until it is equal to whatever level is set by the three-resistor voltage divider across the +12V line (R7, R8 and R9 in series). The lamp is always at constant brightness and therefore at constant
impedance. That's the neat part!

Ohm's Law Applied!

Power and lamp resistance is monitored by your external meter. With S1 in the lower position and no RF applied, the voltage across the bulb is measured. Since R2 and R3 in parallel equal 50 ohms, and the current through the two is the same current through the bulb (they are in series after all) the voltage across the
50-ohm resistor pair can be set to equal the voltage across the bulb. You can check this by flipping the switch to the upper position and then to the lower. Then you set the pot R8 until they're equal. Easy, eh?

DC power is found by using an ordinary calculator. Measure the voltage, square the value found, and divide that result by 50. With no RF applied, this is the maximum power that can be read.

When you apply RF, the lamp will absorb it and the amount of DC required to maintain brightness will, naturally, be less! A new lower voltmeter reading will be obtained. The difference between it and the previous (no RF applied) reading is used to quickly calculate your actual applied RF power.

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