Question

Refer to the following circuit, assuming Is = 1.0 PA: D, is a generic rectifier diode, and it is at temperature of 50°C. Calculate the voltage drop VR across R, for (a) R = 1.09 (b) R, = 102 (c) R = 1002 (d) R, = 1.0kg R1 1.00 D1 V1 -1.0V

Answer 1

Here Source voltage 1 V will get divided among resistor and diode. Since diode is at 50 degree celcius, we can understand that it is dicipating power. So there is a voltgae drop across diode and we are going to find that in each case. After that voltage drop across Diode will be subtracted from source voltage to get voltage drop across resistor.

Here Is=

1012A

and V1= 1 V

1.1$\Omega$

R= 1$\Omega$

Here

R= V1 - Vd Is

So $Vd=1-(1*10^-12)$

So we got voltage drop diode.

Now voltage drop across resistor is,

Vrl= V1- (1 - 10-12) = 1-(1 – 10-12

=

=1 pV

2.10$\Omega$

Here

R= V1 - Vd Is

So $Vd=1-(10*10^-12)$ =$1-(10^-11)$

So we got voltage drop diode.

Now voltage drop across resistor is,

$Vr1=V1-(1-10^-11)=1-(1-10^-11)$

=

3.100$\Omega$

Here

R= V1 - Vd Is

So $Vd=1-(100*10^-12)$ =$1-(10^-10)$

So we got voltage drop diode.

Now voltage drop across resistor is,

$Vr1=V1-(1-10^-10)=1-(1-10^-10)$

=

4.1 K$\Omega$

Here

R= V1 - Vd Is

So $Vd=1-(1000*10^-12)$ =$1-(10^-9)$

So we got voltage drop diode.

Now voltage drop across resistor is,

$Vr1=V1-(1-10^-9)=1-(1-10^-9)$

==1 nV

Thus we got voltage drop across resistor for all resistor values.