# Diode clipping circuit

1. Sep 10, 2010

### novop

1. The problem statement, all variables and given/known data

Hi. I'm being asked to make a circuit to take the input waveform, on the left, and extract the output waveform on the right. In my circuit, I can only use basic diodes, resistors, and DC supplies of +10 and -10V.

3. The attempt at a solution

Last edited: Sep 11, 2010
2. Sep 10, 2010

### Staff: Mentor

If you can assume 0.6V drop across a diode, then just stack them up to make 10V one way and 10V the other way -- you wouldn't need the power supplies in that case. Effectively you make a 10V bi-directional clamp with the two diode strings...

3. Sep 10, 2010

### novop

I only added the 10V supplies because I can't make 10V using 0.6V drop diodes, the closest I can get is 10.2V

4. Sep 10, 2010

### Staff: Mentor

Can you assume 0V forward drop diodes? If so, you can use a slightly modified version of your solution circuit. Look at the direction of the diodes compared to the polarity of the suppllies....

5. Sep 10, 2010

### novop

I can only "use" 0.6V drop diodes. That being said, is my first solution the closest I can get?

6. Sep 10, 2010

### Staff: Mentor

To be honest, I'm having trouble understanding how your circuit would work. Have you tried simulating it?

7. Sep 10, 2010

### novop

8. Sep 10, 2010

### Staff: Mentor

But in that circuit, the power supplies keep the clipper diodes reverse biased until the voltage across the diode + power supply is enough to forward bias the diode, causing clipping at that voltage.

In your circuit, I think I see that you are trying to subtract 10V from the 15V total drop across the diode strings, but I don't think that will work (I could be wrong).

I think I know how they want you to do this, so I'll try a hint to see if it clicks for you.

Look at that 3rd circuit, and imagine what you would get if the diode reverse-biasing supplies were 10V, and the diodes were 0.6V (they use 0.7V in their circuits, but whatever). What would that circuit give at its output? How far off would the output be from what your problem is asking for? Then, can you think of a way of adjusting the output voltage to get it to match your problem?....

9. Sep 10, 2010

### novop

If the supplies were 10V then the circuit would clip the signal at +/- 10.6V (even though the input in this particular problem is +/- 10V). So, I'm 5.6V off from where I want to be. The last part is where I am stuck, I can't think of a way to adjust the output voltage!

Last edited: Sep 10, 2010
10. Sep 11, 2010

### Staff: Mentor

You're only 0.6V off in each direction.....

11. Sep 11, 2010

### novop

In the diagram in the first post, the output is clipped at at 5V/-5V.

Last edited: Sep 11, 2010
12. Sep 11, 2010

### vk6kro

You need to clip at +/-5 volts.
You could get close to this with 9 diodes for positive clipping and 9 for negative clipping. This would give you (9 times 0.6) or 5.4 volts. Or you could use 8 diodes to get 4.8 volts.

You have to have a series resistor to avoid excess current flow while the clipping takes place.
Something like 330 ohms.
Clipping occurs when the input exceeds 5 volts in either direction.

To get exactly 5 volts, you could take the 5.4 volts and arrange a voltage divider. This is two resistors in series from the output to ground. They need to be large compared with the 330 ohm resistor.

Clipping with diodes would give a "rounded top" type of clipping, which is useful for some applications.

13. Sep 11, 2010

### mmmboh

So there is no place one can put a capacitor to fix the clipping voltage problem? And how did you get 330 ohm's for the resistor?

Last edited: Sep 11, 2010
14. Sep 11, 2010

### novop

I don't understanding where to make the voltage divider in this case. If, say, we take 9 diodes in each direction, for 5.4V clipping, how can I reduce this to 5V given the circuit?

Last edited: Sep 11, 2010
15. Sep 11, 2010

### vk6kro

[PLAIN]http://dl.dropbox.com/u/4222062/clipper%203.PNG [Broken]

You would put the resistors across the output as shown above.

Since this is the homework section, I have not shown the value of the lower resistor (although I did calculate it. :) )

The 330 ohm resistor was a bit arbitrary. I just guessed a value of 15 mA peak for the diodes.
This is a value I often use for LEDs.

Last edited by a moderator: May 4, 2017
16. Sep 11, 2010

### Staff: Mentor

Good solution, vk6kro. I did have a math error in my proposed solution.

17. Sep 11, 2010

### novop

Thanks very much vk6kro. If I understand this correctly, the resistor R's value is arbitrary, it's purpose is a voltage sink after the diodes start conducting, right?

Assuming the above to be true, the bottom resistor should have a value of 12.5k ohms to extract the right waveform. Am I making sense?

18. Sep 11, 2010

### vk6kro

Yes, that is right. In practice, you would know the load that existed after the output and could design around that.
In this case, the only load is the voltage divider.

R isn't entirely optional. You need a substantial current to ensure that the voltage drop of the diodes is about 0.6 volts.
If you chose 15 K and kept the same voltage divider, the diodes would never turn on because the voltage across them would not be greater than 4.7 volts.

Yes, I agree about the 12.5K. Good.

19. Sep 11, 2010

### novop

Thanks very much.

20. Sep 12, 2010

### skeptic2

I programmed the circuit into PSPICE and ran a simulation. The attachment shows the circuit and the simulation input and output. Also included is an alternate circuit with its simulation run. The second circuit is what I thought you were thinking about when you were talking about using a 10 V supply.

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