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I have a question about speed at detecting peaks.

Essentially, my question is that I have a precision rectifier with a capacitor equipped to the end of it to smooth the incoming full bridge rectified wave, which works 100%.

however I am concerned, as the application is to play audio through this and capture the current peak. I am worried that the peaks won't be captured in time as audio changes very quickly worst case 1/20kHz.

I understand I need a resistor to discharge the cap so it can change on the next cycle, however when I put a low value resistor (For a faster discharge time) it doesn't reach my peak value anymore oppose to using a big resistor or no resistor which is important as the next stage is a comparator with a fixed Vref to compare to.

How do I balance this? I need a resistor value that can discharge the capacitor fast enough for incoming music in order to capture the peaks of that instant moment. Or is that the trade off?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Must you TRACK the peaks, or simply find the next and higher peak? \$\endgroup\$ – analogsystemsrf Jul 10 '18 at 17:14
  • \$\begingroup\$ Tracking the peaks would be preferable \$\endgroup\$ – Leoc Jul 10 '18 at 17:57
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C2 is too big for this sort of application. The OPA350 is only capable of 80mA of output current and it takes a long time to charge the 680uF cap. You then also need then a very small R3 value to discharge it rapidly between peaks.

Set C2 at about 2uF and R3=100k Ohm

That will give you reasonable peak capture and a decay time in the 200mS range. If you need to make this voltage available elsewhere you could add another OPA350 as a buffer.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I shall try it out ! Give me a minute \$\endgroup\$ – Leoc Jul 9 '18 at 22:08
  • \$\begingroup\$ Datasheet sheet says around 40mA? and I did what you said, it ripples a lot oppose to before where it was straight \$\endgroup\$ – Leoc Jul 9 '18 at 22:14
  • \$\begingroup\$ Yup, the circuit feedback is loading the peak capacitor. If you eliminate R3 and replace R1 and R2 with 100k Ohm you should get the right result. Most peak detectors are arranged a little differently, see here (page 32): ti.com/lit/an/snla140b/snla140b.pdf \$\endgroup\$ – Jack Creasey Jul 9 '18 at 23:33
  • \$\begingroup\$ I see I see, would you be able to explain the logic on why eliminate R3 and replace R1? I would imagine you need a discharge resistor for the cap to be empty for the new audio signal, cause what if the cap is charged to 1.96V due to the previous and the next audio amplitude is coming in however the cap is still at 1.96V and this amplitude is 0.5V wouldnt the cap still be charged at 1.96V then? \$\endgroup\$ – Leoc Jul 9 '18 at 23:43
  • \$\begingroup\$ I think its personally working more accurate to be honest, but I have no idea why those changes :/ \$\endgroup\$ – Leoc Jul 9 '18 at 23:45

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