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[Edited]Diode in series with a resistor

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tamilarasantpvm

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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. Untitled.jpg
 

KlausST

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Hi,

Your headline is confusing.
And you missed to ask a question.

Klaus
 

crutschow

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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.
 

amiyapbehera

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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
 

LvW

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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.
 

Audioguru

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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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