Diode Ciruit problems 2

[paulmdrdo]

again i need some help this one

determine the voltage across the diode using complete diode model. r'd=10 ohm(forward resistance) r'R= 100 M ohm(reverse resistance).

please where do Should I start? I know that the diode is forward-biased.

 

[sankar.m8]

Total Effective resistance on 30V is ~4.852k (Assuming the diode as 10 ohm resistor on forward bias). Total circuit will draw a current of 6mA. On the diode wing, the current drawn is 3mA (Current division rule). So the voltage across the diode is 30.879mV.

Please note that the forward resistance of a diode is not constant.

 

[paulmdrdo]

the answer on my book is 0.731V. why is that?

 

[sankar.m8]

is that the forward voltage of that diode?...then it is right.

 

[paulmdrdo]

0.731 v is the voltage across the diode. can you tell how do you get the 30.879mV?

 

[sankar.m8]

30.879mV i calculated by assuming the diode as a 10ohm resistor since you given the forward resistance. I guess, the forward voltage of the diode is 0.7V. after the 0.7V only, the forward resistance will come into picture. So the total voltage across the diode will be 0.7+voltage drop due to forward resistance. ie. 0.7+0.030879 = 0.731V.

 

[FvM]

I guess, the forward voltage of the diode is 0.7V. after the 0.7V only, the forward resistance will come into picture. So the total voltage across the diode will be 0.7+voltage drop due to forward resistance. ie. 0.7+0.030879 = 0.731V.

Yes, apparently. Unfortunately the claimed "complete diode model" is missing in the original question.

 

[paulmdrdo]

i still don't understand how did you calculate that 30.879mv across the diode.

 

[sankar.m8]

Effective resistance = 1k + 1.5k + (4.7k || ( 4.7k+10ohm))=4.8525kohm
Current drawn from source= 30/4.8525k = 6.182mA
Applying current division rule, current through diode wing = 6.182mA*4.7k/(4.7k+4.71k) = 3.087mA
Voltage across the forward resistance of diode = 3.087mA*10ohms= 30.87mV.
Total voltage across the diode= forward voltage+voltage across forward ressitance= 0.7+0.03087= 0.731V

 

[paulmdrdo]

now I get it. anyways can we use kvl here?

 

[sankar.m8]

Yeah...you can use .

 

[paulmdrdo]

can you check my equations

Ia = current through 2500 ohm resistor
Ib = current through 4700 ohm resistor
Ic = current through 4700 ohm resistor and 10 ohm diode


30-2500Ia-4700Ib=0

30-0.7-10Ic-4700Ic-2500Ia=0

4700Ib-0.7-10Ic-4700Ic=0


are they correct?

 

[sankar.m8]

There is only two currents...
1. Through 1k, 1.5k, 4.7k
2. Through 4.7k, diode, 4.7k

 

[paulmdrdo]

is the current through 1k, 1.5k and 4.7k the same? in the diagram they are not in series. there's junction between 1k, 1.5k and 4.7k current will be divided. how about that?

here's how I redraw the circuit. is this correct?

 

[FvM]

There is only two currents...

Means you put ia = ib+ic into the equations.

 

[paulmdrdo]

2500Ic+2500Ib+4700Ib=30
2500Ic+7200Ib=30 ----1

4710Ic+2500Ic+2500Ib=29.3
7210Ic+2500Ib=29.3----2

now are they right?

 

[sankar.m8]

is the current through 1k, 1.5k and 4.7k the same? in the diagram they are not in series. there's junction between 1k, 1.5k and 4.7k current will be divided. how about that?

here's how I redraw the circuit. is this correct?View attachment 101176

This is not KVL... This is KCL

 

[paulmdrdo]

can you show how you would set up the proper kvl equation for the circuit using only two currents?