Diode Limiter Circuit: Input-Output Graph & Voltage Output

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SUMMARY

The discussion focuses on the analysis of a diode limiter circuit powered by a 16V sine wave at 1kHz. The user correctly identifies that during the positive half-cycle, diode D1 conducts, resulting in a voltage output of 3.7V, while during the negative half-cycle, diode D2 conducts, yielding an output of -3.3V. The expected output graph is a sine wave clipped at +3.7V and -3.3V, confirming the user's understanding of the circuit's behavior.

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francisg3
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Hello,
I have recently had a lab on diode limiters and am a bit confused by the results. The circuit is seen in the picture below and I have obtained theoretical values for the Voltage output. The circuit is feed by a 16V sine wave with a frequency of 1kHz.

Now I know when the sine wave is positive, D1 conducts and D2 is cut-off. So the voltage output should be 3V + 0.7V = 3.7V right?

When the sine wave is negative I know that D1 is cut-off and D2 conducts. This means that the output voltage should be -4V + 0.7V = -3.3V correct?

I need to sketch the input and output so would the graph would be a sine wave clipped at +3.7V and -4.7V ?

Thanks for the help!
 

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would the graph would be a sine wave clipped at +3.7V and -4.7V ?

Yes, it would be like this.
 

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