Electronic – Unexpected rectifier waveform

^{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).

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) Reverse conducting diode. (b) Symmetrical LED.

The problems seem to be caused by the asymmetrical current path caused by the in-line diode. Since you are using and LED to LDR opto-isolator you can use the LED as the rectifier. Figure 1a shows the circuit with D2 providing protection for reverse voltage (since the audio signal is AC).

Figure 1b shows a pair of LEDs in back to back arrangement. This has the advantage of symmetrical loading of the AC. If you can find a package with both LEDs mounted back to back (internally) it means your LDR will be illuminated on both positive and negative audio signals. Failing that you could try to mount both LEDs on the one LDR or use circuit of Figure 2a.

^{simulate this circuit}

Figure 2. (a) Pair of opto-isolators with parallel LDRs. (b) Bridge rectifier feeding LED.

For completeness, Figure 2b shows one LED fed by a bridge rectifier. This has the disadvantage of higher voltage drop due to two series diodes at all times.