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[SOLVED] what is the voltage across the diode "considering the complete model"

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MSAKARIM

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Is this solution right?
1.PNG
2.PNG
 
Last edited:

Solution is wrong. About 0.86 mA If.

No "complete model" is given in your post.
 

    M

    MSAKARIM

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I mean when I replaced the diode with ( small resistance rd'=10 and battery of=0.7v). The diode is forward biased
FvM said:
Solution is wrong. About 0.86 mA If.

No "complete model" is given in your post.
Click to expand...
 

That's all correct, but the current is wrong. You ignore the current from the +10V source.
 

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    MSAKARIM

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You can replace the diode with a 10 ohm resistor and a 0.7 volt power supply. Then a equation should be solved to find the Voltage across the diode which is:
(-690) / 998 = -0.69138276553
 

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    MSAKARIM

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aminpix said:
You can replace the diode with a 10 ohm resistor and a 0.7 volt power supply. Then a equation should be solved to find the Voltage across the diode which is:
(-690) / 998 = -0.69138276553
Click to expand...
Two rough assumptions (without explanation) - and a result with a pico-volts accuracy.´?
No - without any further information, we have no other chance than to ASSUME 0.7V across the diode and find the current through the diode (method of node voltages): Id=0.86mA.
 

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    MSAKARIM

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You can calculate the Thevenin equivalent of the circuit without the diode.
From that you can calculate the current through the diode when it is connected.
 

LvW said:
No - without any further information, we have no other chance than to ASSUME 0.7V across the diode and find the current through the diode (method of node voltages): Id=0.86mA.
Click to expand...
Yes. The OP is apparently assuming 0.7V with 10 ohm series resistance as model, without stating this explicitly or even a substantiation. In this case, the exact current would be
4.3 V/5.01 kOhm = 0.858 mA, negligible difference to the simplified calculation without series resistance. Post #5 is wrong in any case.
 

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