ac signal peak detector

here is circuit diagram of a peak detector i need its explanation. i want to know its workng. whats each component is doing here eg capacitor diode resistor, mode of op amp?
and what if i use LM741 op amp and 1n4001 Diode?

kindly reply.
thanks

Op amp is in a none inverting feedback mode.
Consider the circuit without the diode or capacitor it will just be a buffer with gain x1
Add the diode and when the input goes negative the opamp output goes negative (to max value), when the input goes positive the the opamp output goes positive until the diode turns on and closes the loop and vout =vin
The capacitor holds the voltage at the peak value just measured and R3 slowly discharges this value.

Just for more information if the feedback resistor went to the opamp output the circuit would measure the peak voltage less the diode drop as the diode would be outside of the feedback loop.

741 would not work as their bias currents are too high, need something similar to given MOSFET input op amp.
1N4001 are power diodes working up to 1A and not good at the low currents in this circuit, hence why they chose a signal diode.

consider two cases
case 1 :Op amp is used as a comparator and it compares between the vin and the capacitor voltage
if the vin is larger than the capacitor voltage then op amp output will be the vin and the diode is forward biased (anode voltage larger than cathode which is capacitor voltage) then the capacitor will charge to the output of the diode and the output is capacitor voltage --peek detected

case two : but if the vin is less than capacitor voltage then the output of the opamp is less than the capacitor voltage the diode is reverse biased then the output voltage Vout is still the same voltage as the previous case not effected by the output voltage of the opamp

note the opam here is working as comparator

The peak detector is a circuit that "remembers" the peak value of a signal. As shown in Fig. 9-a, when a positive voltage is fed to the noninverting input after the capacitor has been momentarily shorted (reset), the output voltage of the op-amp forward biases the diode and charges up the capacitor. This charging last until the inverting and noninverting inputs are at the same voltage, which is equal to the input voltage. When the noninverting input voltage exceeds the voltage at the inverting input, which is also the voltage across the capacitor, the capacitor will charge up to the new peak value. Consequently, the capactor voltage will always be equal to the greatest positive voltage applied to the noninverting input.
Once charged, the time that the peak detector "remembers" this peak value is typically several minutes and depends on the impedance of the load that is connected to the circuit. Consequently, the capacitor will slowly discharge towards zero. To minimize this rate of discharge, a voltage follower can be used to buffer the detector's output from any external load, as shown in Fig. 9-b. Momentarily shorting the capacitor to ground will immediately set the output to zero.

http://www.electronics.dit.ie/staff/...20Tutorial.htm


Best regards,
Peter

I would still describe the opamp mode as an "Op amp is in a none inverting feedback mode" and probably add the term "Non-linear".

It is performing a comparitor like function, but it is working in a linear region and as described provides an analogue output equal to the positive peak, not a discrete level as provided by a comparator.
When the input is less than the peak detected the opamp goes to its max negative output level as it is now working at its open loop gain, but this does not change the output level, hence my description of the opamp mode.

actauly i made a circuit using lm741 and 1n4001 that peak detector circuit output moves to adc0804 and that works well. my microcontroller work fine. i just changed capacitor 200nF and its working fine.

just tell me one thing when voltage decreses after going to peak capacitor will hold peak value or the low value.
i need peak voltages at various time intervals. as my voltage source goes up and down. practicle circuit measures the next lower voltage but simulations dont.

Your original circuit will be like a saw tooth shape, quick to rise to the peak and a slow discharge from the resistor R3 and whatever else you connect across it.

If you want true sample and hold at peak you need a switched discharge such as provided by tpetar. You could problably replace the switch with a fet for micro control.

There used be a great chip PKD01 from memory for this purpose which I used with GAP01 for much of my analogue processing before I started doing it in code!



Just to add, if you consider the diode, C and R circuit this is similar to what is used in simple amplitude demodulation in an AM radio, the diode charges up a small c at every peak of the AM carrier wave and the discharge is set to match the highest frequency of audio of interest. So the R is setting the discharge time along with the C. Just FYI, nothing more!

I'm not sure what you mean with "the next lower voltage". Can you sketch a typical waveform?

The most serious deviation from ideal peak detector behaviour will be certain peak overshoot depending on input voltage risetime and voltage level.

i got what i was asking thanks for replying.
with the next lower voltage i mean that input signal voltage decreses after going to high value.