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# [Edited]Diode in series with a resistor

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#### tamilarasantpvm

##### Newbie level 6
Hi all,
I am preparing for interview. meanwhile i came across this circuit and need the current in the circuit.
i attach the circuit here. please rply as soon as possible.

Hi,

And you missed to ask a question.

Klaus

I hope you are not going into an interview for an electronics job. :???:
You can readily solve that to a close approximation by using the fact that the typical forward Si junction diode drop is about 0.7V.
Why do you say "Diode in series with a normal diode" in the title? There's only one diode in series with a resistor.

Assuming diode forward bias voltage of 0.7V (Silicon Diode), Current I= (12V-0.7V)/12K=0.94mA

Hi all,
I am preparing for interview. meanwhile i came across this circuit and need the current in the circuit.
i attach the circuit here. please rply as soon as possible.View attachment 119899

All the calculations result in an approximate value only because they are based on an estimated value for Ud (for example 0.7V).
This is OK - unless the U-I characteristic of the diode is available (classical exponential shape).
In this case - and also for a better understanding - you can do the following:

Because of 12V=Ud+I*R we have Ud=12V-I*R .
Now you can add this resistor characteristic to the diode diagram because both elements have the same current (straight line with a negative slope).
The point where both curves meet is the actual current I.

Every diode and transistor has a detailed datasheet.
You can calculate the EXACT typical current if you look at the datasheet of the 1N4001 diode where it shows a graph of its typical forward voltage vs current. The graph does not go lower than 10mA (where V= 0.6V) but you can extrapolate to get about 0.96mA.

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• 1N4001.png
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