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My doubt is a trivial one, can I take VD1 and VD2 to be V1? If so doing that doesn't make sense to me 'cause when the current flows VD1 is certainly not equal to V1. enter image description here

This isn't a Homework question, I found a similar question in my textbook with D2 reversed (Given above) and there I can understand why VD1 equals V1, as for the off condition no current flows through the 10K resistance but here this isn't the case. If I'm right then the following solution is wrong, how should I proceed then?

Any help is highly appreciated.

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  • \$\begingroup\$ In your notes D1 and D2 are in the same direction but in Figure E4.26 they're in opposite directions !!! In note's your circuit (diodes in same direction) you could simplify the circuit to one diode, a 5 V battery and a 5 kohm resistor. Your calculations do assume diodes as in Figure E4.26. I would like to suggest that before throwing calculations at the problem, first think and envision what the circuit does. \$\endgroup\$ – Bimpelrekkie Apr 18 at 6:57
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In the textbook when the input is between +5 and -5 both diodes are reverse biased so they do not conduct and the signal crcuit is just a 10K resistor,

in the case when either diode is forwards biased the signal now encounters a resistive divider.

So you get \$ V_i + 5 \over 2 \$ or \$ v_i + -5 \over 2 \$ depending on which diode is conducting

In your circuit,

the the diodes are pointing upwards so vith voltages above that from the most positive source (the right leg) wich is 5V neither diode will conduct and input will equal output, with voltages below that first the one diode will conduct (the right one) but as the voltage gets more negative it's not until -15V in that the output gets to -5 and only then doe the other diode conduct.

with this type of circuit the transfer function should be continuous, that's a simple check you can apply to your answers.

for your circuit one way to derive the transfer function is to consider how much current must flow through the input resistor to produce each output voltage, by considering it that way you can see that -5V out needs -1ma and thus -15V

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