Circuit with diode

  • Thread starter pyroknife
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  • #1
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I attached the circuit.
I need to find I_1, Voltage drop across R1, voltage drop across R2, and I_2. We're assuming the diode is 'not' ideal.

Vin=5V. R1=R2=1kohm
R3=2kohm


voltage across R2=Vin=5V
I2=5/1kohm=5mA

The real diode drops about 0.7 V. So using voltage divider gives voltage across R1=(R1/(R1+R3))*(5-0.7)=1.43V

Then I1=1.43/1k=1.43mA


Is this the right idea?
 

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  • #2
phinds
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without comment on the correctness of your specific numbers, let me ask you this:

Do you feel that I2 = 5m in the same sense that I1 = 1.43ma. That is, other than magnitude, would they have the same waveform if graphed?
 
  • #3
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without comment on the correctness of your specific numbers, let me ask you this:

Do you feel that I2 = 5m in the same sense that I1 = 1.43ma. That is, other than magnitude, would they have the same waveform if graphed?
I guess I don't see what you're asking.
This is a DC signal.

For an AC signal, the would not have the same wave form due to the diode.
 
  • #4
phinds
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I guess I don't see what you're asking.
This is a DC signal.

For an AC signal, the would not have the same wave form due to the diode.
Hm ... why do you suppose a DC source is labeled "Vin, VAC ?"
 
  • #5
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Hm ... why do you suppose a DC source is labeled "Vin, VAC ?"
Ignore the AC part, that's for another problem.
 

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