What is RF current?

Answer 1

Measuring RF Current

Once again, our multimeter on AC scales fails at measuring radio-frequency sinewaves. Current measurement is a bit more difficult than voltage measurement shown above. We shall use the same peak-detector circuit, measuring the voltage drop across a small sampling resistor. This resistor will drop a small voltage as RF current flows through it. The peak detector will measure this voltage drop. Ideally, a very small sampling resistor is best: it must be much smaller than the 50 ohm load. If not, the combination of 50 ohm load and series sampling resistor will increase the load seen by the transmitter.

If we were to use a one-ohm sampling resistor (1st circuit), the voltage available (for a 5W output) would be 0.447 peak volts. Not very much: errors due to diode-drop will be high, and even worse at lower power.

Let's use a transformer to step up the available voltage going into the peak detector (2nd circuit). We'll use a 1:10 turns ratio on our transformer, and use a ferrite core to make sure all the flux links to every winding. If the ferrite has high enough permeability, the primary winding can be one turn, while the secondary winding can be ten turns. It needn't be a big core, since very little power is going into the peak detector.

Rather than place a one-ohm resistor at the primary side, its better to add a 100-ohm resistor on the secondary side (3rd circuit). Since impedance is transformed by the turns-ratio-squared, the primary will still see a one ohm impedance. The diode measures the peak voltage across this 100-ohm resistor. Now instead of 0.447 volts peak, we'll get 4.47 volts peak - a substantial improvement in sensitivity.