Given a full wave rectifier (http://www.electronics-tutorials.ws/diode/diode_6.html) with an RC filter on the DC output, how can I measure the efficiency of conversion?


simulate this circuit – Schematic created using CircuitLab

I've seen the formulas to compute the efficiency. I'd like to actually verify this in the full wave rectifier that I built.

The DC side seems simple. Measure the voltage across the load resistor. That gives me \$V_{DC}\$. Knowing the resistance, I can calculate \$i_{DC}\$. Then \$P_{DC} = (V_{DC}) x (i_{DC}) \$.

The AC side confuses me. \$V_{AC}\$ is easy to measure with a scope. Is the only option to calculate \$i_{AC}\$? or simulate the circuit? (I don't have an AC current probe, only a DC current probe) Can a load or shunt resistor be added to measure the current?

The application of this rectifier will help explain the goal. My son is trying to convert acoustic energy into electrical energy (it is a very low yield process). He has already built a speaker array, connected them in series and connected them to this rectifier. He chose Schottky diodes to reduce the bias voltage below the typical Zener diode.

What he has found is that the process is a lot less efficient than he expected. I'm trying help him measure the electrical power conversion efficiency. We're using a Function Generator for the AC source. In the actual application \$V_{DC} < 500 mV\$ even with (x6) 10 inch speakers.

I've been trying to help him understand where he is losing all his power. One way to investigate this is to evaluate the rectifier with a Function Generator. He's also built a Spice model (LT Spice) and is trying to compare his measurements / characterization with that model.

  • \$\begingroup\$ Yes, you could use a shunt resistor to measure the current on the AC side. Note that since it's AC think about how you want to express the value of the current and voltage, I would use RMS voltage / current as that gives you the power at the AC side. \$\endgroup\$ Feb 25, 2016 at 8:49
  • \$\begingroup\$ I should reassure everyone that the AC voltage here is quite small ( < 5 V) and the AC current is small as well ( < 100 mA). \$\endgroup\$ Feb 25, 2016 at 21:21
  • \$\begingroup\$ Large or small is relative, I design with transistors that would not survive 5 V and for which 100mA is a very large current :-) Your addional comment does not change much, you can still use a shunt resistor to measure the current. Maybe you need to make it a 1 ohm resistor so that you get 100 mV across it for 100 mA. 100 mV is a voltage you can easily measure on your scope. \$\endgroup\$ Feb 25, 2016 at 21:56

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

From your answer I have understood you have a scope. Put shunt resistor before the diode bridge (between source and bridge). Then with scope measure current trough it and also voltage over diode bridge. Product of those two is loss (Ploss). Both these values are variable (AC) so you can multiply them on scope.

Because you already have Pdc, efficiency is Pdc/(Pdc+Ploss).

  • \$\begingroup\$ Be very careful to include the requirement for the AC voltage to be fully uncoupled from power ground, or the scope can cause serious problems measuring across a shunt resistor. Differential measurement will not offer decent enough resolution. On a side note I doubt the scope will be able to multiply such that you get power, since the load is largely reactive (capacitors) and Irms and Vrms will not be in sync. \$\endgroup\$
    – Asmyldof
    Feb 25, 2016 at 10:59
  • \$\begingroup\$ Agree. When working with voltages above 50V extra care should be taken. \$\endgroup\$
    – Darko
    Feb 25, 2016 at 19:46
  • \$\begingroup\$ Won't the load of the entire circuit change when I add the shunt? Do I need to make sure that R_Shunt << R1? \$\endgroup\$ Feb 25, 2016 at 21:26
  • \$\begingroup\$ I am concerned about the grounding of the scope. We ran into trouble assuming the scope was floating. Differential measurements can be tricky with small voltages / currents. \$\endgroup\$ Feb 25, 2016 at 21:26
  • \$\begingroup\$ And you should be concerned about the grounding ! But if the rectifier + capacitor and load resistor are floating (like in your schematic) you can ground on the AC side only and that will be OK. If this is not possible then let the AC side float and ground only the DC side. You can then still measure on AC side but you must use 2 probes, connect signal across the resistor and subtract the signals. Scope ground must then be connected to ground on DC side ! \$\endgroup\$ Feb 25, 2016 at 22:00

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