# biased clipper

Discussion in 'Homework Help' started by electronicsbeginner12, May 7, 2015.

1. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
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1
Hi. I an confused on how the biased clipper works. I understand the ideal diode acts as a short when forward bias and acts as an open when reversed bias. In the attached image, in figure A, I understand that that the input voltage has to be at least equal to Vb to conduct the ideal diode as shown. But when the diode is reversed in figure B, Im confused on how it produced that clipping at and below positive 4 volts. How can I simplify my analysis? Thanks.

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2. ### MikeML AAC Fanatic!

Oct 2, 2009
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You said the right words, but drew the wrong picture for A.

If the instantaneous AC Input voltage is less than 4V, what is happening in the diode? If the diode is conducting, what is the output voltage?

What happens if the AC Input voltage is greater than 4V, what is happening in the diode? What controls the output voltage?

3. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
I don't see anything wrong with the circuit diagram for (A). If the signal source is above 4V, then the diode will turn on and clamp the output to 4V (for an ideal diode). Your waveform diagram for (A) is ambiguous. Usually the dashed line means a level that is NOT present (usually indicating a level that WOULD have been present if not for the circuit's particular behavior). So you horizontal line at 4V should be solid and the arch above it should be dashed.

Answer MikeML's questions and you will probably understand what is going on.

4. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
27
1
Ok. I reworked my problem. Please let me know if my explanations are correct. Thanks.

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5. ### WBahn Moderator

Mar 31, 2012
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When you say that your input signal is 5V, is that 5V amplitude? 5V peak-peak? 5Vrms? How are you going from that 5V to a 10V peak signal indicated in your waveform diagram. Same for part (b), only more so since you are saying that your signal is only 3V.

6. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
27
1
The input voltages i used in my examples are just the sine wave amplitudes in those particular instances.

7. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
So, again, how are you getting from the stated input waveforms to a 10V amplitude sinewave in your output waveforms? This is particularly relevant for the second case.

8. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
27
1
In my diagrams, the peak voltage is 10 volts, but in my explanations, the input voltages used are just amplitudes occurring at a particular instance in time to explain what happens to Vout when the diode is forward and reverse bias.

9. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
There is a disconnect between your circuit diagrams and your waveforms. What is the amplitude of the signal sources in your circuit diagrams?

10. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
27
1
Sorry. It is 20V peak to peak.

11. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
In that case, your waveforms are correct and you just need to label your circuit diagrams properly. You need to indicate 10Vp, 20Vpp, or 7.07Vrms (I'd recommend 10Vp).

12. ### electronicsbeginner12 Thread Starter New Member

Dec 14, 2014
27
1
Ok. How about the way I analyzed the circuits with my explanations? Is it logical and correct? Is this how you would explain how these circuits work? Thanks.

13. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
If you are going to talk about Vt, then you need to define what Vt is. Be sure that ALL of your terms of precisely defined, either in words or via annotations on the diagram.