What do the capacitors in this ducker circuit do?

May 22, 2007 10 Replies

Hello, about this picture:

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I understand that OA1 is a threshold circuit using an opamp precision half-wave rectifier and that OA2 is a polarity inverter and a gain control circuit combined into an inverting mixer configuration.



What I don't get is the purpose of capacitors C8 and C5. Can any one explain please? Gracias!



C5 with R14 is creating a lossy integrator to provide a DC control voltage.

C8 looks like a bodge to me !

Graham

Thanks for high lighting the capacitors, so I didn't have to look all over for them. :-)

C5 is easiest. It is connected as negative feedback across the feedback resistor, turning the amplifier into a low pass filter. Since the capacitor impedance has the same magnitude as the 12400 ohm feedback resistor at about 270 Hz, all frequencies higher than that are progressively rolled off.

C8 is somewhat similarly connected, but in a nonlinear situation. It may be there to reduce the overshoot of the output as one diode turns off, and the output slews, open loop, (going as fast as it can) to the voltage where the other diode turns on, reclosing the feedback loop. It has a very much higher frequency effect, because it is normally across either the impedance of a turned on D2, or across the turned on impedance of D1 in series with R2, except for the brief moments I mentioned, first. Both those paths are pretty low impedance, so the capacitor current is significant only when both diodes are turned off.

Hey Graham, Thanks! What is a bodge?

Hi John, Thanks! I'm curious about the low pass filter characteristics. I'm having a hard time understanding because I'm seeing the output of the RMS Detector 4301P us strictly DC. How does the low pass filter play in this DC environment? Thanks again!

Like kluge/kludge.

I've never used a cap there myself in a FW rectifier for audio. Maybe there was something not quite right with the performance and that fixed it.

Graham

The output of the RMS detector must not be pure DC, but varying, unidirectional voltage. The low pass filter smooths the bumps. Anything that changes faster than the RC time constant of the feedback pair is smoothed (12400*47n=.00058 seconds), and the faster it changes, the more it is smoothed.

A better question might be: If the RMS detector outputs a DC signal, why is it followed by a rectifier?

Hi John, I was looking at the rectifier as circuit that categorizes the voltage level. If the output of the RMS detector is above a certain value, then the output of the rectifier follows the detector. If the RMS detector output is below threshold, then the rectifier stays at 0V. Did I look at this correctly?

Thanks!

I haven't studied how the RMS converter works. But after some digging, I did find the data sheet:

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The RMS converter does not look very well defined, but it does include a low pass filter (time constant = .026*Ct/It). But it also shows a bidirectional output current of +- 100 uA.

So you may be exactly right, that the negative values for below threshold levels, are clipped off by the ideal diode.

Ridiculous. I'm not that clever with this stuff, but I recognize high frequency roll-off when I see it.

That IS NOT high frequiency roll-off.

Graham

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