Physics P-n Junction Diode Problem

[Avi1995]

Q:-Find the current through the resistance R in the figure:-

, if the value of R is:-
a.12Ω
b.48Ω.
Assume resistance offered by diode is 0 in forward bias and ∞ in reverse bias.

Homework Equations

Ohms Law:-
V=Ri

The Attempt at a Solution

I tried using kirchhoffs law but we don't know its in reverse bias or forward.
Then I imagined the circuit without the diode wire so as to calculate the potential drop b/w the ends(this would tell whether its Forward bias or reverse) but realized that resistance would change if we connect the wire again? Plz help!

 

[gneill]

Hi Avi1995, welcome to Physics Forums.

You had a good idea, calculating the voltage that would appear across the open terminals of the diode location.

If you remove the diode from the circuit and find the potential that would appear at its open terminals you can determine whether the diode should be considered to be forward or reverse biased. Once you know that you can replace the diode with the appropriate open circuit or short circuit and proceed to find the current through R.

So, what are the open circuit voltages that appear for each case (be sure to specify the polarity!).

 

[Avi1995]

Sorry for the delay but here is my attempt:-
a)R=12Ω
After removing the diode and applying kirchhoffs law:-
-10+12i+12i=0
24i=10
i=5/12=0.43A
Potential drop btw A and B is:-

Vb-Va=6-5=1v
Hence in reverse bias we can disconnect the diode ans is 0.43A, WHICH IS CORRECT!
b)R=48Ω
Proceeding same as before we get
:-
i(without the diode)=1/6
Vb-Va=6-8=-2V
this Means Va is at higher potential hence forward bias.
Circuit is:-

Solving we get(I am short of time but I solved on paper)
current through resistor is
i=0.13A Correct!
Thanks very much but I have one last ques.
If we remove the diode, what about the 1 Ω resistor, won't it cause any change in potential drop, is it safe removing the diode?

 

[gneill]

If we remove the diode, what about the 1 Ω resistor, won't it cause any change in potential drop, is it safe removing the diode?

If no current flows through a resistor then there is no potential drop across it. By removing the diode an open circuit results, so no current can flow. Thus the "diode end" of the resistor must be at the same potential as its other end, namely at the same potential as your node B.

 

[asteriatic]

I would approach this problem by first understanding the basic principles of a p-n junction diode. A p-n junction diode has two regions - a p-type region and an n-type region. The p-type region is positively charged and has an excess of holes, while the n-type region is negatively charged and has an excess of electrons. When these two regions are brought together, they form a depletion region where the charge carriers (holes and electrons) recombine and create a potential barrier.

In forward bias, the positive terminal of the battery is connected to the p-type region and the negative terminal is connected to the n-type region. This reduces the potential barrier and allows current to flow through the diode. In reverse bias, the positive terminal of the battery is connected to the n-type region and the negative terminal is connected to the p-type region. This increases the potential barrier and prevents current from flowing through the diode.

Now, in this problem, we are given two different values of resistance for R - 12Ω and 48Ω. In order to find the current through R, we need to determine whether the diode is in forward bias or reverse bias. We can do this by considering the value of R in comparison to the resistance offered by the diode.

If R is 12Ω, then it is much smaller than the resistance offered by the diode in forward bias (which is assumed to be 0). This means that the diode will be in forward bias and current will flow through it. In this case, the current through R will be the same as the current through the diode.

If R is 48Ω, then it is much larger than the resistance offered by the diode in reverse bias (which is assumed to be ∞). This means that the diode will be in reverse bias and no current will flow through it. In this case, the current through R will be 0.

Therefore, the current through R will depend on the value of R and the resistance offered by the diode. If R is 12Ω, the current through R will be the same as the current through the diode. If R is 48Ω, the current through R will be 0.