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  #1  
Old 03-20-2012, 06:01 PM
ronaldocavalcante ronaldocavalcante is offline
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Default Help with clipping circuit

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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  #2  
Old 03-20-2012, 08:30 PM
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bertus bertus is offline
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Hello,

Have a look at this page of the eBook:
http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

Bertus
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  #3  
Old 03-21-2012, 02:16 PM
ronaldocavalcante ronaldocavalcante is offline
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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
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  #4  
Old 03-21-2012, 04:38 PM
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Ron H Ron H is offline
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Are both resistors 5.6kΩ?
This looks like homework to me. We don't give answers to homework.
Show us your work and the answers you got, and maybe we can help you.
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Old 03-21-2012, 11:58 PM
ronaldocavalcante ronaldocavalcante is offline
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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
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  #6  
Old 03-22-2012, 03:56 AM
ronaldocavalcante ronaldocavalcante is offline
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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 by ronaldocavalcante; 03-22-2012 at 04:04 AM.
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  #7  
Old 03-22-2012, 02:44 PM
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Ron H Ron H is offline
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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.
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  #8  
Old 03-22-2012, 10:16 PM
ronaldocavalcante ronaldocavalcante is offline
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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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