Help with clipping circuit

Thread Starter

ronaldocavalcante

Joined Mar 20, 2012
5
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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Thread Starter

ronaldocavalcante

Joined Mar 20, 2012
5
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.

However, tahks for your attention.

regards

ronaldo
 

Thread Starter

ronaldocavalcante

Joined Mar 20, 2012
5
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:

Ron H

Joined Apr 14, 2005
7,063
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.
 

Thread Starter

ronaldocavalcante

Joined Mar 20, 2012
5
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
 
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