Measuring Ultra-Low AC Input Power?

[nikhil9dua]

Hi there,

I have an AC-DC rectifier for which I'm trying to measure efficiency for. The input to the rectifier is a 13.56 MHz signal which hovers around 2.8V Rms. The output is just essentially a DC Rail which sources a certain amount of current.

My goal is to measure overall efficiency (DC_Power_Out / AC_Power_In). I am able to measure DC_Power_Out without a problem using a current source circuit, multimeter and a current measurement circuit. However, I'm not sure what the best way is to measure AC_Power_In

The rectifier takes in <2mW of AC power. What is the best way to measure this AC power going into the rectifier? I have tried using a CT-1 AC Current probe however this is quite sensitive since input currents are in the range of hundreds of uA to few mA. Another issue is the timing between the voltage and the current due to phase lag doesn't allow for accurate AC power measurement.

Any ideas & solutions are much appreciated!

Thanks,

Nikhil

 

[barry]

I don't understand the statement "however this is quite sensitive". Don't you WANT it to be sensitive? Why is this an issue? If you're using an oscilloscope, it should be a simple matter to capture the current and voltage waveforms and then calculate the "real" power from the two traces.

 

[nikhil9dua]

Hi Barry,

Appreciate the response. By sensitive, I meant that since currents can go as low as a few hundred uA's, the current probe doesn't produce accurate readings at this level. I have tried this out with , say a resistor and a few capacitors to understand if low level current are measurable and there is some error. The probe has 5mV / mA gain. Perhaps this can be gained up with a low noise amplifier to reduce noise at low current levels?

I have tried that method out with current and voltage waveforms, however, there is an inherent propagation delay to each current and voltage probe that messes up the measurement (eg: MATH on CH1 and CH2)

Best,

Nikhil

 

[betwixt]

By current probe what exactly do you mean?
You can add a resistor in series with the source and measure the voltage drop across it then multiply by the voltage from the source.

Brian.