0
\$\begingroup\$

enter image description here

The above signal is fed to the following circuit -

enter image description here

It is given that RC>>T and diode is ideal.

Doubt - As RC>>T, the capacitor will not be able to charge during the positive half and output should be equal to the input. In the negative half cycle, the diode is open circuited and entire input voltage appears across it. Hence in the negative half cycle, output equals the input. So the overall output should be equal to the input signal. However I searched the internet and found this solution in the instructor's manual -

enter image description here

How is this waveform even possible? Surely the capacitor is not getting enough time to charge for the clamping to take place.

\$\endgroup\$
  • 1
    \$\begingroup\$ Have you tried simulating the circuit? That should allow you to play with the component values and see how it impacts the output waveform. \$\endgroup\$ – MrGerber Feb 14 '18 at 8:39
  • \$\begingroup\$ Don't think about 1 cycle think about what will happen after this waveform has been applied for a long time. \$\endgroup\$ – RoyC Feb 14 '18 at 8:51
3
\$\begingroup\$

Simulation with C=2u and R=1k gives ...

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Thank you. I finally understand it after looking at the waveform you provided. I didn't take into account the long term behavior of the circuit. \$\endgroup\$ – Rohan Singh Feb 14 '18 at 10:58
3
\$\begingroup\$

Your circuit is a modified diode clamp. A regular diode clamp looks like this for instance: -

enter image description here

If you look at the diode you can see that it limits the positive voltage on the output to about one diode drop (circa 0.7 volts) but has little restriction on what the signal does when it is negative because it is reverse biased.

This means that the whole of the input peak to peak voltage appears at the output but is "clamped" at the positive peak to about +0.7 volts.

With the added resistor, the clamping is initially less effective but it does eventually reach the situation seen above.

\$\endgroup\$
  • \$\begingroup\$ Thank you. As the capacitor has no discharge path, it slowly stores charge with each cycle and eventually the voltage across it reaches the peak value of the signal and the output becomes 0 (for an ideal diode) in the positive cycle. \$\endgroup\$ – Rohan Singh Feb 14 '18 at 11:04
  • \$\begingroup\$ Yes, there is no discharge path and it progressively stores charge. \$\endgroup\$ – Andy aka Feb 14 '18 at 11:06
-1
\$\begingroup\$

Here is the waveform required for C=2uF and R=1K and T=1ms. I have simulated in Proteus. Circuit

Output

\$\endgroup\$
-3
\$\begingroup\$

Here is link for online simulation of circuit, you can easily play with parameters.

\$\endgroup\$

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.