simple Thevinin question

Thread Starter


Joined Sep 20, 2008
The title is : simple Kirchof question


I have a loop that consists of 2 voltage sources and 3 resistors.
The voltage sources cancel one another (i.e. they have the same value but in opposite direction).

can I say directly that the current in that loop is equal to zero ?
or it is not necessarily zero ?
Last edited:

Thread Starter


Joined Sep 20, 2008
The problem is to find the current through Diode 2 ( the middle one)

The instructor started his solution by considering the first loop only ( he considered the current through the second loop to be zero), but I don't understand why.


The instructor probably considered the current in the second loop to be zero based on a very ideal model of the diode. If Vf < 0.6 Volts then the diode is OFF. Vf = 0.6 Volts when diode is ON. (This is a very ideal model of the diode that does not reflect reality.)

Given that model, the center diode has to have 0.6 Volts across it, which limits the voltage across the resistor and diode in series on the right to 0.6 Volts. That also means the voltage across the right most diode must be less than 0.6 Volts and therefore OFF.

This answer is assuming the very simple diode model... Is that the model of diode that you are currently working with?


Joined Apr 26, 2005
The instructor is making alot of assumptions. Rf at 15 ohms would require 40 mA of current to achieve the 0.6 "turn on" voltage. Unfortunately the 20k and 5k will limit, at best, the current from the 50 V supply to 2 mA. You come up 38 mA short of turning on the circuit.

Had the Rf of 15 ohms not been listed as part of the problem, you could "assume" there would be (Vs - 1.2) voltage drop across the two resistors in the first loop and the second loop wouldn't be conducting. As it is now, there is not enough voltage across either diode in the first loop to "turn the diodes on" So, with roughly 2 mA (1.9976 mA), there is 30 mV across each diode in loop one ... which is not the conducting condition of 0.6V. The voltage drop across the diode in the second loop is roughly 200 uV (certainly not conducting) and (30 mV - 200 uV) across the resistor.

There is a reason the diagram included the Rf of 15 ohms.
Some of these "simplified" diode models can be kind of strange. You are not necessarily trying to produce the 0.6V over the 15 ohm resistor but may have a 15 ohm resistor in series with a 0.6V drop when the diode is ON...