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Problem 4.4 from Microelectronics circuits by Sedra and Smith

Find voltage V and current I. As per the book, answer is (e) 3 mA, +3 V; (f) 4 mA, +1V . But I don't know how we got there.

Edit: For figure (e):

I understand that diode connected to +3V supply will be forward biased because the anode of this diode is connected to maximum positive voltage in the circuit. But, I don't understand why diodes connected to +2V and +1V is reverse biased.

For figure (f):

Diode connected to +1V supply is forward biased because cathode of this diode is connected to least positive voltage in the circuit.But, I don't understand why diode connected to +2V and +3V is reverse biased.

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  • \$\begingroup\$ Welcome to the site but we don't do your homework for you. If you edit your question to explain what you do and don't understand you will get good help. \$\endgroup\$ – Transistor Jun 10 '17 at 11:04
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    \$\begingroup\$ Well, this is not my homework I'm just brushing up my knowledge in electronics devices. And I've made the edit to the question what I do and don't understand. \$\endgroup\$ – Saurabh Jun 10 '17 at 12:02
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You have the answer already and have asked for guidance ("But I don't know how we got there"), which is all I can give for homework questions...

Your question uses 'perfect' diodes, so no voltage drop and no leakage currents amongst other things.

Perfect diodes do not conduct when they are reverse-biased. Therefore any path that has a reverse-biased diode is open-circuit. When you're assessing the circuit, pretend that reverse-biased diode paths are not there.

Look at each diode in turn. Seeing the voltage it puts onto the resistor bottom-end, consider the effect on the other diodes.

This should take you most of the way to the answer. Then its just Ohm's Law at the end.

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After thinking for a while I came up with the following explanation for the behaviour of the given circuit.

For figure (e):

Current from +3V supply, apart from flowing through 1k ohm resistor also try to charge +2V and +1V supply. But for this to happen current has to flow from cathode to anode of diodes connected to +2V and +1V supply which makes them reverse biased. Since now +2V and +1V supply are out of the circuit we can easily obtained the voltage and current in the circuit.

For figure (f):

Current from +5V supply will try to charge +3V, +2V and +1V supply through 1k ohm resistor. We know that current follow least resistance path or conversely we can say that a circuit is try to draw maximum current from the source. Maximum potential difference exist between +5V and +1V, hence maximum current will flow from this path. Since diode connected form +1V supply is forward biased, this will make the potential of anode of diodes connected form +2V and +3V equal to +1V which will make them reverse biased. Now current and voltages in the circuit can be easily calculated.

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    \$\begingroup\$ You've got it! You would see a similar situation in a hydraulic or pneumatic circuit if you replace the voltage sources with pressure sources and the diodes with non-return valves. \$\endgroup\$ – Transistor Jun 10 '17 at 12:37

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