How Do You Find and Draw V(out) for a Sine Wave in an Ideal Diode Circuit?

AI Thread Summary
To find and draw V(out) for a sine wave input in an ideal diode circuit, start by analyzing the input voltage (V(in)) as it varies from -10V to +10V. Identify the different regions where the diodes switch between off and on states, which will affect the output voltage. The approach involves solving for each region separately and then plotting the output response to the sine wave. Even with ideal diodes, the input impedance and transfer function will vary with the input voltage due to the bias voltages in series with the diodes. This method will yield a complete understanding of the circuit's behavior.
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for the following circuit find and draw V(out) if the V(in) signal is a sine wave with amplitude of 10V (Consider all diodes to be ideal)




This is the problem , could anyone please solve it 4 me ?:?:

10x
 
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The policy is that we don't do homework for you. Would you please explain how you have tried to solve this problem so far?
 
ok , no problem

i've solved it but i don't know if it is true

i have taken two values

one as 10 v

and other as - 10 v

and i have got as result two out put voltages (one for 10 v and one for

- 10 v)

and solved the problem normally after that

but , is my solving correct ? is this how i should solve the problem ?
 
Just start by drawing the Vo versus Vi graph for a linearly increasing Vi from -10V to +10V. You wll find different behavior regions where the various diodes go from Off to On. Figure out what these voltages are where the diode behaviors change (how many regions are there?), and solve each one separately. Then you can plot the Vo response to a Vi sine wave to finish answering the homework question.
 
No berkeman, the OP said to consider all diodes to be ideal. Eng_Electo, your approach is correct. Tell us what your answers are.
 
Averagesupernova said:
No berkeman, the OP said to consider all diodes to be ideal. Eng_Electo, your approach is correct. Tell us what your answers are.
Even with ideal diodes, it seems like the input impedance and transfer function will change with input voltage, based on the bias voltages in series with the diodes. Pushing the input voltage high is a lot easier, for example, than pulling it low where the first clamp stage will be operating.
 
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