0
\$\begingroup\$

The text says: "When Vin is high the diode is FB and capacitor charges up to peak. When Vin is low the diode is RB and the capacitor tends to discharge. The reason given for this is the diode leakage currents.."

But i don't understand one thing that is that as we know that the leakage current of diode is the current that the diode will leak when a reverse voltage is applied to it and the leakage current is entire function of the reverse bias voltage applied to it...so why the text tells us that this leakage current of diode is responsible for the discharge of the capacitor...??

or if i am interpreting it right then it might be inferred as that the diode due to its leakage current provides a path for the capacitor current to discharge and get dumped in the op amp....?? enter image description here

\$\endgroup\$
  • \$\begingroup\$ forum.allaboutcircuits.com/threads/… LOL \$\endgroup\$ – Fizz Oct 8 '15 at 12:32
  • \$\begingroup\$ FB is forward biased...and the text says that leakage current of the diode is responsible for the discharge of the capacitor... \$\endgroup\$ – partykid Oct 8 '15 at 12:34
  • \$\begingroup\$ Yes, I figured it out. Not a common abbreviation though. "FB" is normally the feedback pin of some [SMPS] controller chip. \$\endgroup\$ – Fizz Oct 8 '15 at 12:35
  • \$\begingroup\$ The figure is from Horowitz and Hill, by the way. But I don't see (in the 2nd edition) the text that you are quoting (even after accounting for your abbreviations). So I'm guessing that came from the forum link I gave above. \$\endgroup\$ – Fizz Oct 8 '15 at 12:51
  • \$\begingroup\$ actually it came from a video-->youtube.com/watch?t=947&v=jllsqRWhjGM.... \$\endgroup\$ – partykid Oct 8 '15 at 12:56
2
\$\begingroup\$

This is a peak detector circuit. The capacitor is indeed discharged by the diode leakage current but the discharge rate is very slow, hence a very small value capacitor should be used. Reverse-biased (voltage blocking) diodes have non-zero leakage current, typically measured in nA (1N4148 has ~ 25 nA leakage current). My suspicion is that the capacitor leakage current (self-discharge) is similar to the diode leakage current. This parameter really depends on the capacitor specs (http://www.murata.com/en-us/support/faqs/products/capacitor/mlcc/char/0039).

Look at this Microchip application note: http://ww1.microchip.com/downloads/en/AppNotes/01353A.pdf

It states that the only discharge path is through the diode leakage currents and opamp bias currents (page 9). Sometimes a parallel resistor is added as well as shown in the application note.

But leakage current is flowing inside the diode then how does it seems to discharge the capacitor

The opamp can sink the current flowing through the diode from the capacitor. Opamps can source or sink current through the output pin.

Below is a snapshot showing the circuit input and output.

enter image description here

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Minor quibble, the Microchip note actually says "The only discharge path for C1 isthrough R5, via leakage or op amp bias currents." Leakage here can actually refer to the leakage to rail of the opamp output stage as well. I think they wrote it that way because details would depend on the actual opamp used. \$\endgroup\$ – Fizz Oct 8 '15 at 13:13
  • \$\begingroup\$ @RespawnedFluff That's right. I think that they added R5 because otherwise the only way to discharge the cap is through its internal leakage or the diode reverse leakage current. \$\endgroup\$ – SunnyBoyNY Oct 8 '15 at 13:45
0
\$\begingroup\$

If the cap is charged and Vin goes low, then that opamp's output will also go low, which will reverse-bias the diode and allow the flow of charge (leakage current) into the opamp's output.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ yes but how come leakage current discharge the capacitor.....?? \$\endgroup\$ – partykid Oct 8 '15 at 12:57
  • \$\begingroup\$ If you read SunnyBoyNY's answer, it actually contains that as well. \$\endgroup\$ – Fizz Oct 8 '15 at 13:13
  • 1
    \$\begingroup\$ @Mr.Nuts: since \$Q=CV\$, \$ V= \frac{Q}{C}\$. Then, since Q is the number of charges in the dielectric of the capacitor, for every one that leaks/gets away, Q will decrease and so will V \$\endgroup\$ – EM Fields Oct 8 '15 at 13:23
  • \$\begingroup\$ @RespawnedFluff ... \$\endgroup\$ – partykid Oct 8 '15 at 13:29
  • \$\begingroup\$ ... and (assuming that was Mr.Nuts' mental block) what EM Fields said applies to any current not just to one conventionally called "leakage" (called so because it's small). \$\endgroup\$ – Fizz Oct 8 '15 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.