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The problem states that the diodes are ideal. That must mean that the diodes have no forward voltage drop and no reverse current. So in one circuit D1 will have no voltage across it and in the other D1 will have no current flowing through it.
As far as I can recall, you're right - an ideal diode has no voltage drop and no reverse current.
Imagine that diode D1 was not connected and calculate the voltage at the anode of D2.
If the result is positive, then D1 will conduct when it is re-connected.
If the result is negative, then D1 will be off and so no current will flow through it when it is re-connected.
Thus the currents can be calculated very easily.
Imagine that diode D1 was not connected and calculate the voltage at the anode of D2.
If the result is positive, then D1 will conduct when it is re-connected.
If the result is negative, then D1 will be off and so no current will flow through it when it is re-connected.
Thus the currents can be calculated very easily.
Are you sure that this method is right?
Becuase I proved this but in some cases doesn't give the right results...
The general technique is to replace the diode with either a short circuit or an open circuit then look for a logical inconsistency. Does the short circuit current flow in the forward direction? Does the open circuit voltage reverse bias the diode? If one of these conditions is false, then that is the wrong answer! Therefore the other condition must be true.
Read my very first suggestion and just do the calculations for a voltage divider using just the resistors.
Calculate the voltage at the mid point.
Now presume there is a diode between the junction of the resistors and ground. Will current flow through the diode in either case?
If so, what will the voltage be at the bottom of the top resistor? And therefore what will the current through the top resistor be? Now add the second diode. Given that you know the voltage at the bottom of the top resistor in this case, what current will flow through the diode between the resistors?
If the diode to ground does not conduct, then the circuit is effectively a voltage divider. Those calculations are trivial.
Bonus question... What effect does the diode between the resistors have?
Calculate the voltage at the mid point.
Now presume there is a diode between the junction of the resistors and ground. Will current flow through the diode in either case?
If so, what will the voltage be at the bottom of the top resistor? And therefore what will the current through the top resistor be? Now add the second diode. Given that you know the voltage at the bottom of the top resistor in this case, what current will flow through the diode between the resistors?
If the diode to ground does not conduct, then the circuit is effectively a voltage divider. Those calculations are trivial.
Bonus question... What effect does the diode between the resistors have?
This is an old post.
1. In case A, both diodes 1 and 2 are conducting due to the forward bias of +5 Volts. The -5 Volts supply cannot reverse bias diode 1 because its series resistance of 10k is more than the series resistance of the +5 Volts supply. Therefore, the output voltage is zero (0).
2. In case B, diode 1 is reverse biased and diode 2 is forward biased for the same reason above. This makes all voltage sources and all resistances in series. Why? because you can just remove diode 1 in the circuit.
1. In case A, both diodes 1 and 2 are conducting due to the forward bias of +5 Volts. The -5 Volts supply cannot reverse bias diode 1 because its series resistance of 10k is more than the series resistance of the +5 Volts supply. Therefore, the output voltage is zero (0).
2. In case B, diode 1 is reverse biased and diode 2 is forward biased for the same reason above. This makes all voltage sources and all resistances in series. Why? because you can just remove diode 1 in the circuit.
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