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My university asked to my do a report about a practice in the lab. One of questions it asks to us to describe the working of a peak detector, it's exactly like this:

Peak Detector Circuit

The output is like this:

OutPut Peak Detector circuit

For me it's strange this output behavior, I expected that the capacitor would only start to discharge when the input was below zero. Can someone explain why this thing is happening?

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3 Answers 3

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"I expected that the capacitor only start to discharge when the output were bellow of zero."

  • The capacitor and the output ARE the same (at the same voltage, in parallel, on the same leads/rails);
  • A capacitor discharges whenever an input (charging) voltage is below the capacitor's own charged voltage and a load is connected across the capacitor;
  • A capacitor is charging whenever the voltage applied to it is larger than its own charged voltage;
  • The diode in this circuit is there to allow charging of the capacitor when the input voltage (Vin) is above the capacitor's voltage, and to prevent the discharge of capacitor back to the input when the Vin is below the capacitor's voltage.
  • Your statement about capacitor discharging "only when the output is below zero" is wrong on at least 2 points. Where did you get "below zero"? Could you explain what do you mean by that?

UPDATE:

"Sorry, I committed a little mistake, the correct is : "whenever the input were bellow of zero", For me, this phrase means the same than " whenever V(input)<0". Is it correct?" – Lucas Vital

  • Your interpretation of that formula is correct, but the statement is still wrong.
    The input doesn't have to be below zero, only below the voltage to which the capacitor is charged.
    This means that as soon as the input voltage goes below the peak voltage to which the capacitor has charged up, the capacitor will start discharging.
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  • \$\begingroup\$ Sorry, I committed a little mistake, the correct is : "whenever the input were bellow of zero", For me, this phrase means the same than " whenever V(input)<0". Is it correct? \$\endgroup\$ Commented Apr 20, 2020 at 15:51
  • \$\begingroup\$ Your interpretation of that formula is correct, but the statement is still wrong. The input doesn't have to be below zero, only below the voltage to which the capacitor is charged. This means that as soon as the input voltage goes below the peak voltage to which the capacitor has charged up, the capacitor will start discharging. I am updating my answer with this information. \$\endgroup\$ Commented Apr 20, 2020 at 15:57
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    \$\begingroup\$ Good explanation! Now the things are very clear to me. When the V(input) <V(peak), the bias of diode is reversed, so the loop is broken, and the capacitor discharges through the resistor. Thanks so much! \$\endgroup\$ Commented Apr 20, 2020 at 16:06
  • \$\begingroup\$ You're correct! And you're welcome! I am glad I helped you understand. \$\endgroup\$ Commented Apr 20, 2020 at 16:09
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Once C1 becomes charged to the peak voltage, it can only receive charge again when: -

  • the input voltage rises to a higher peak voltage
  • the capacitor is discharged by the parallel resistor and another input peak comes along

So, if your input voltage peaked once and then went to zero volts forever, the capacitor (C1) will receive that peak input voltage and then will be slowly discharged by R1.

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When the diode is forward biased, you have a voltage follower configuration. The inverting input of the amplifier is connected to the output of the circuit. Thus, the op amp "changes" its output to ensure that the inverting input tracks the non inverting input. The op amp is only able to work as a follower if the diode is forward biased. If you reverse bias the diode, then you break the loop. This happens when the non inverting input voltage is lower than the inverting input. In this case the op amp output slews to the negative rail and the diode is reverse biased. The output capacitor then starts discharging through R1.

The output capacitor only starts discharging once you reach the peak voltage of your input signal because until you reach the peak the op amp is configured as a follower and he forward biases the diode. However, if you reach the peak and start reducing the input voltage, the op amp is no longer able to control the output voltage by reducing its output, since it will reverse bias the diode and break the loop.

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  • \$\begingroup\$ Nice! I got it. When V(input)< V(peak), the diode is reversed bias and then the loop is broken and the capacitor discharges through the 10KOhms resistor \$\endgroup\$ Commented Apr 20, 2020 at 16:10

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