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Improve bandwith of peak detector
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MasterT:

--- Quote from: David Hess on July 24, 2019, 03:19:48 pm ---
--- 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.

--- End quote ---

The problem with BJT is high reverse leakage current, that is just cross out any gain in charge current it's introduced.

To OP: What is the arm chip? If it has an ADC, may be you don't need peak detector after all, sampling input fast enough by 12-bits ADC and running peak detector in software.
LF398 data sheet is good place to start in case hardware solution preferable.
SiliconWizard:

--- Quote from: MasterT on July 24, 2019, 04:52:50 pm ---To OP: What is the arm chip? If it has an ADC, may be you don't need peak detector after all, sampling input fast enough by 12-bits ADC and running peak detector in software.

--- End quote ---

Well... you're right. The OP talks about 500kHz max. Many recent MCUs have several-MSPS 12-bit ADCs, so that should be doable in software without much problem.
Now if a 12-bit ADC would not have enough resolution for the OP's needs, then I think they would be dreaming if they think they can get better than the equivalent of 12-bit with a simple, fast and pure analog peak detector...

David Hess:

--- Quote from: MasterT on July 24, 2019, 04:52:50 pm ---
--- Quote from: David Hess on July 24, 2019, 03:19:48 pm ---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.
--- End quote ---

The problem with BJT is high reverse leakage current, that is just cross out any gain in charge current it's introduced.
--- End quote ---

Common transistors have lower base-emitter leakage than gold doped diodes like the 1N4148.  They just do not bother testing it below the 50 or 25 nanoamps specified in the datasheet because nobody cares and the tests at low currents take significant time on the test machine which increases costs quickly.  The input bias current of the second stage is greater and the 1 megohm shunt resistor is greater yet.
MasterT:

--- Quote from: David Hess on July 24, 2019, 11:49:43 pm ---
--- Quote from: MasterT on July 24, 2019, 04:52:50 pm ---
--- Quote from: David Hess on July 24, 2019, 03:19:48 pm ---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.
--- End quote ---

The problem with BJT is high reverse leakage current, that is just cross out any gain in charge current it's introduced.
--- End quote ---

Common transistors have lower base-emitter leakage than gold doped diodes like the 1N4148.  They just do not bother testing it below the 50 or 25 nanoamps specified in the datasheet because nobody cares and the tests at low currents take significant time on the test machine which increases costs quickly.  The input bias current of the second stage is greater and the 1 megohm shunt resistor is greater yet.

--- End quote ---
Probably, they don't test reverse current of the common BJT because they have no idea that someone is gonna to build a peak detector with B-E junctions. 100 nA is enormous current in such application. Same time circuit and components selection shown in the original post is also wrong, and indeed BJT wouldn't make much harm.
 There is example out of TI data sheet,

though TI didn't bother to draw reset switch. Resistor connected to uCPU will work as reset, just changing pin mode from Input to Output all it would takes to do resetting.
radix:

--- Quote from: MasterT on July 25, 2019, 12:29:28 am ---100 nA is enormous current in such application.
--- End quote ---
*proceeds to post a schematic with a schottky diode that has max. leakage specified at 250 nA at 3 V reverse voltage at room temperature*
David Hess also stated that a BJT has lower leakage than the 1N4148 (specified at 25 nA at 20 V reverse voltage at room temperature).

On a more constructive note: You can also substitute the opamp with a few discrete transistors, although I suspect there is going to be way more offset in such a circuit. Description in the attached pdf document.
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