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We have the circuit in the figure.I have to find I and V. The solution manual of my book says :

V= 1 V .D1 is conducting and D2 is cut off.

I dont get this,why is V=1 V and not 3 Volt? Also why is D1 conducting and D2 cut off? I know that a diode is conducting when anode voltage>cathode voltage but what are the anode voltages in each diode?

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  • \$\begingroup\$ Has your book introduced the concept of "forward voltage drop" yet? \$\endgroup\$ Commented Nov 26, 2013 at 20:06
  • \$\begingroup\$ @ScottWinder: the arrow in a diode symbol points towards the cathode. The voltages at the left of the drawing are the voltages on the cathodes. The voltage at V (junction of resistor and diodes) is the anode voltage for the diodes. \$\endgroup\$ Commented Nov 26, 2013 at 20:25
  • \$\begingroup\$ @PeterBennett: You're of course right, I had them reversed. (Prior comment deleted.) (chanting to self) ACID, ACID, ACID, ACID... \$\endgroup\$ Commented Nov 26, 2013 at 20:30

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D2 has 3V on its cathode and for it to conduct there has to be more than 3V on its anode. By simple inspection, the cathode voltage cannot be greater than 3V therefore D2 is not conducting.

D1 has 1V on it's cathode and because D2 is not conducting we can forget about it as having any significant role in the circuit. Therefore the remaining supply voltage being 3V via a resistor to D1's anode means D1 has to be be forward biased and conducting.

Why would D1's anode (the net labelled "V") be 1V? It won't be - it will be somewhere between 1V and 3V - it has to be greater than 1V because D1 has to be conducting and the forward volt drop for a forward conducting diode will be somewhere between 0.2V and 1V (a wide range of diodes and possible currents are included in this estimation).

So if the person giving you the question believes the voltage at "V" is only 1V they are misled - liklihood being, that for a silicon diode it will be about 1.5V to 1.7V.

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    \$\begingroup\$ Or the book has specified that they consider "ideal diodes" with 0 forward voltage. Maybe not something you or I would recommend when teaching this stuff, but probably not uncommon. \$\endgroup\$
    – The Photon
    Commented Nov 26, 2013 at 20:16
  • \$\begingroup\$ @ThePhoton probably although it's hardly a big deal saying forward diodes drop (say) 0.6V but read the small print! \$\endgroup\$
    – Andy aka
    Commented Nov 26, 2013 at 20:19
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I dont get this,why is V=1 V and not 3 Volt?

Suppose V was 3 V.

Then D1 would be strongly turned on and many amps of current would flow through it. But this current would have to come through the 2 kOhm resistor. Which would thus have a huge drop across it.

Since the voltage at V is 3 V minus the drop across the resistor, this is a contradiction, so we know this is not the correct solution to the circuit.

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