But you'll still have to discharge the cap once in a while. Mathematically, the charge on a cap is an indefinite integral, namely unknown if you don't know its full history. As a practical matter, the question is "peak voltage since when?"
And a 1N4148 willy typically leak a lot more than a decent fet opamp.
John
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I was planning a bleed resistor across the capacitor, as mentioned in an earlier post. Also, I did mention changing to a fast, low-leakage diode. ... Steve
The c-b junction of a small-signal transistor (2N4400 maybe) is a superb low-leakage diode, ans the gate junction of a jfet is even better.
John
I guess you mean to leave the emitter open? I'll check this out. Otherwise, I'm pretty sure I've got some suitable diodes here. .. Steve
looks good here except for the use of the diode with that diode you have selected the Vfm (Max Forward voltage) is aprox 0.65 like most diodes in that fam. this means the first 0.60 or so won't be seen in the circuit from the your (Vin) also, this will produce a 0.65 error in the calculation of the multiplier due to the drop. using a lower (Vfm) diode in my opinion would help things greatly. something like what a Germanium diode would do.(aprox 0.3 volts). also there is the problem of linearity, a little negative feed back via the same type of diode could counter act it.
just an observation.
-- Real Programmers Do things like this. http://webpages.charter.net/jamie_5
have it your way, i wouldn't use it.
-- Real Programmers Do things like this. http://webpages.charter.net/jamie_5
Graham was kind enough to carry this almost to completion. He's steering you straight here.
As long as you can guarantee your input voltage won't exceed either your + or - rail, you could do worse than go with the TI part TLV2374. It's a quad op amp that operates on a single +5V supply, has rail-to-rail inputs and outputs, 3MHz GBW with 2.4V/uS slew rate, has only 60pA input current with very low offset drift, and is available in quad DIP, SOIC and TSSOP packages from Mouser for less than a buck in
100 qty:If your power supply is going to turn off very quickly, I'd be thinking about a protection diode to discharge the cap (even at the risk of higher leakage current), or 2.2K series resistors at the inputs of the second op amp to safely limit discharge current. And the 10M bleeder resistor at the cap is a good idea.
By the way, on your original question, I don't see any advantage over the standard method here: | |Vin |\\ | o---|+\\ |\\ | | >->|-o----o-----|+\\ Vo | .-|-/ | +| | >---o----o | | |/ | --- .-|-/ | | | | --- | |/ .-. | | | | | | | | | | === | 10K| | | | | GND | '-' | '---------' | | | '--------o | | | .-. | | | | 10K| | | '-' | | | === | GND (created by AACircuit v1.28.6 beta 04/19/05
Stand-alone peak detector, followed by a stand-alone gain block of 2. Same number of parts, similar connections, no?
Good luck Chris
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Hello Chris, thanks for the input. I would have thought that accuracy would be improved by taking the feedback from the output of the second op-amp. Is there likely to be an advantage in changing to your circuit configuration? Incidentally, I've now added a voltage-follower transistor, (BC549C), to the input of the second op-amp, to avoid the input creeping toward the positive rail. This would, of course, still work with your suggestion.
... Steve
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What makes you think that the diode will introduce an error, equal to the diode's forward voltage drop?
... Steve
As your supply voltage is low enough (5V), just change the 1N4148 for a small signal NPN emitter follower. That'll nicely reduce the rectifier leakage.
-- Thanks, Fred.
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