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Improve bandwith of peak detector
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marcoccio:
Hi, i'm in the process of dessigning a peak detector to feed and adc in arm chip for an univercity proyect and have encountered a wall. I need to detetect with 2% accuracy voltages between 50mV and 5V of an AC sinusoidal or triangular waveform with frequency ranging from 5 to 500kHz. For this i have dessigned a peak detector ussing an LM393 comparator. The peak detector will be driven by a pga MCP6S21 controlled by the arm chip to improve the sensibilty.  There are two problems i have encountered so far:

1. At low frequency, less than 1Khz, the cap C2 can't hold it's voltage for long enouth. My solution to this would be to enlarge R11 and add another (smaller) resistor in parallel controled by the arm chip using a jfet. This resistor would switch on at a predefined frecuency so that the cap C2 is able to  rapidly follow the changes in the voltage, frecuency would also be measured by the arm chip.

2.  At frequencies higher than 200Khz the charge in C2 starts to fall bellow my 2% allowable drop. And despite my efforts i haven't been able to improve the frequency responce of the circuit.

Heres a photo of the circuit:



Any input would be appreciated!
Audioguru again:
Your second comparator does nothing because it is missing an output pullup resistor. If you add the pullup resistor then the negative feedback will cause the comparators to oscillate because they do not have frequency compensation like opamps have. The pullup resistor on your first comparator has a value too low (too much current) for some of them but its current is not enough to quickly charge the capacitor. 

Use one very fast opamp with the diode inside its negative feedback loop. The opamp will provide a high output current to quickly charge the capacitor. 
David Hess:
A completely different circuit topology would be appropriate.  Some operational amplifier peak detectors can be fast enough.  The example below is way faster.  Replacing the diode with a transistor allows charging the hold capacitor with a much higher current without lowering the value of the pull-up resistance which applies to your existing circuit.

marcoccio:
Thank you! I will explore those options and see what i can do.
David Hess:

--- Quote from: David Hess on July 24, 2019, 03:09:41 am ---Replacing the diode with a transistor allows charging the hold capacitor with a much higher current without lowering the value of the pull-up resistance which applies to your existing circuit.
--- End quote ---

In case I was not clear enough about this, replace your diode D2 with a transistor with its collector tied to the positive supply.  Now the current available from R7 is multiplied by the transistor's current gain to charge the hold capacitor faster.  The example I gave uses the same idea.

The reverse breakdown voltage of the base-emitter junction will limit the maximum range of operation to about 5 volts peak-to-peak however.  If this is a problem, then a diode placed in series with the base can be added.
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