On the above datasheet, page 11th, there is a peak detector circuits. I will adapt it but I didn't figure out why they put two diodes like this and I don't know how the 1k ohm work between the two op-amps either.
Could you explain it to me, thanks a lot.
When the input voltage is positive, the right diode conducts, and the pair of opamps act, together to produce a follower (100% feedback to the first opamp) that drives the storage capacitor to follow the input voltage as it heads up. The resistor in series with the capacitor keeps the capacitor from causing enough phase shift to destabilize the pair of opamps in a feedback loop. The 1 k feedback resistor has no current passing through it, except the bias current of the first opamp, so it acts essentially like a wire during this phase.
When the signal is negative, the left diode conducts, keeping the first opamp from saturating its output all the way against the negative rail so that it will recover quickly when the signal goes positive, again. During that time, the right diode isolates the first opamp form the voltage that has been stored across the capacitor.
The 1 k resistor conducts current from the second opamp output back to the first. This current is wasted, but keeps from having to disconnect the the feedback path between peaks.
thanks. then, why not just put a wire instead of a 1 K ohm? and how to pick this resistor's value?
John P> >
Because if it were zero ohms, A1's more negative output would be fighting A2's more positive output. With 1k in between, the current is limited to (V(A2out)-V(A1out))*1mA/volt.
Cheers, Tom
to limit current, a larger value > 1K would be better, apparently there are other considerations like phase compensation?
Tom Bruhns wrote:
You can't tie the two outputs directly together with only a diode between them, and have them battle for control of the voltage. The value is not critical, but lower resistance looks more like a wire when the feedback diode is off, than a 10k resistor does (as far as driving the input and stray capacitance with little delay or phase shift. But it is about as low as you can go to keep the outputs from going into current limit when it is being used for isolation (of the two outputs) during the lower than peak parts of the waveform.
(snip)
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