# Help with clipping circuit

Discussion in 'General Electronics Chat' started by ronaldocavalcante, Mar 20, 2012.

1. ### ronaldocavalcante Thread Starter New Member

Mar 20, 2012
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0
Hi, all. I'm from Bahia-Brasil and this is my first post.

Please, I need help to calculate currents across two ideal diodes D1 (direct) and D2 (reverse) on clipping circuit: (see circuit)

Main source: Vs = 7.5 peak (symmetrical triangle wave) with resistor (series) of 5.6 kohm

Diode D1 with DC source +3.25 V on katode
Diode D2 with DC source -3 V on anode
RL = 5.6 Kohm

I used electronics Workbench but the currents was wrong.

thanks for help

ronaldo

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3. ### ronaldocavalcante Thread Starter New Member

Mar 20, 2012
5
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Hi, Bertus

I read the site and can't find the currents across D1 e D2, yet.

The V-out is OK, but the currents is always wrong.

ronaldo

4. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Are both resistors 5.6kΩ?
This looks like homework to me. We don't give answers to homework.

5. ### ronaldocavalcante Thread Starter New Member

Mar 20, 2012
5
0
Yes, both are 5.6K.

I don't want the result! I'd like a hint to help me solve the question.

The Vout is simple: V1 and V2 (positive and negative) + 0,7V (diode). In this case (ideal diode) level of the clipping voltage, V1 and V2.

But to determine currents I tried thevenin and superposition without success.

Steps: open V2, short in V1 determines Vo1 and IR1a using Vin

Open V1, short in V2 determines Vo2 and IR1b using V1.

Vo = Vo1+Vo2 and IR1=IR1a+IR1b.

When I use the simulation (EWB) Vo is OK and currents not.

I repeat that not need results, but help the way to try solve the problem.

regards

ronaldo

6. ### ronaldocavalcante Thread Starter New Member

Mar 20, 2012
5
0
Ok.
Tomorrow I'll make this experience. I've got a scope 60MHz and 2 true rms multimeter.

As I don't have a function generator, I'll try find here a triangle wave oscilator circuit.

Maybe my calculation error is found.

thanks

Last edited: Mar 22, 2012
7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Convert Vs and the two resistors to the Thevenin equivalent. Then the solution will be simple.

Is Vs ±7.5V? You said 7.5V peak. I'm assuming that means 15V p-p.

8. ### ronaldocavalcante Thread Starter New Member

Mar 20, 2012
5
0
Solved!!!

It was my problem. Caculation of decimals was wrong
Also, I forgot that the function generator of EWB generates the RMS and oscilloscope uses peak voltage.

thanks for all

ronaldo