Determine current and voltage for the given diode circuit

[Anwesa Roy]

Q. Cutin voltage for each diode=0.7V.

Determine current throught D1 and Vo for

(a) R1=10 kohm, R2 =5 kohm. Ans(I(D1)=0.93 mA, Vo=-15V)

(b) R1=5 kohm, R2 =10 kohm. Ans(I(D1)=1.86 mA, Vo=-15V)



My approach

(a)First considering both the diodes connect.

Let the voltage of the node where n terminal of the diode D2 is connected be v. For each diode there is a forward voltage drop is 0.7 v

=>v=0.7v

I(10k) =(10-0.7)/10 mA=0.93 mA.

I(D2)=(-15-0.7-0.7)/5=-3.28mA

=> D2 is off

I(D1)=I(10k) + I(D2)=0.93-3.28=-ve

=> D1 is off too.

But according to book D1 is on D2 is off. Where have I gone wrong.

(b)Followed the same procedure as above for (b) and got the right answer.

Still showing the steps:

First considering both the diodes connect.

Let the voltage of the node where n terminal of the diode D2 is connected be v. For each diode there is a forward voltage drop is 0.7 v

=>v=0.7v

I(10k) =(10-0.7)/5 mA=1.86 mA.

I(D2)=(-15-0.7-0.7)/10=-1.64mA

=> D2 is off

I(D1)=I(10k) + I(D2)=1.86-1.64=+ve

=> D1 is on.

Therefore

I(D1)=(10-0.7)/5=1.86 mA Vo=-15V

Which matches with the answer.

Please tell me if my procedure is wrong. As by following this procedure I am getting the right answer for (b) but not for (a)

 

[CataM]

(a)First considering both the diodes connect.
=>v=0.7v

I(10k) =(10-0.7)/10 mA=0.93 mA.

I(D2)=(-15-0.7-0.7)/5=-3.28mA

=> D2 is off

I(D1)=I(10k) + I(D2)=0.93-3.28=-ve

=> D1 is off too.

But according to book D1 is on D2 is off. Where have I gone wrong.

D2 OFF => current through it = 0.

I(D1)=I(10k) + I(D2)=0.93+0=0.93 mA

- - - Updated - - -

b)....

I(D1)=I(10k) + I(D2)=1.86-1.64=+ve

=> D1 is on.

Therefore

I(D1)=(10-0.7)/5=1.86 mA Vo=-15V

You have a contradiction here. First you say "I(D1)=I(10k) + I(D2)=1.86-1.64=+ve" and then by magic "I(D1)=(10-0.7)/5=1.86 mA"

Problem is in the first calculation of I(D1) because I(D2)=0.

 

[Akanimo]

Q. Cutin voltage for each diode=0.7V.

Determine current throught D1 and Vo for

(a) R1=10 kohm, R2 =5 kohm. Ans(I(D1)=0.93 mA, Vo=-15V)

(b) R1=5 kohm, R2 =10 kohm. Ans(I(D1)=1.86 mA, Vo=-15V)

View attachment 135809

My approach
...
=>v=0.7v

I(10k) =(10-0.7)/10 mA=0.93 mA.

I(D2)=(-15-0.7-0.7)/5=-3.28mA ... (there is no I(D2))

=> D2 is off
...
Where have I gone wrong.
...

Any how you want to look at it, D2 is off.
Now, look at it this way:
Since 10V is positive, "v" (node D1.anode) can only be less positive (but not negative) if some voltage is dropped across R1.
With a positive voltage connected to the cathode of diode D2, D2 is off (open). ... point 1
Since -15V is -ve, Vo can only be less negative (and not +ve) if some voltage is dropped across R2.
With a -ve voltage connected to the anode of D2, D2 is off (open). ... point 2

Putting points 1 and 2 together, with D2 off in this circuit and there being an open at that point, the following plays out in the circuit.

--> No current flows through R2 and so the -15V appear at Vo (the open end of R2). The -15V supply just hangs there doing nothing.
--> Current only flows through loop 10V-supply ==> R1 ==> D1 ==> 10V-supply; That current is I(R1=10kohm) = I(D1) = 0.93mA just like you calculated.

 

[BradtheRad]

One trick you can try is to switch the positions of D2 & R2. They are in series therefore it's okay. Then the resistors form a divider. Calculate nodes.

 

[Anwesa Roy]

D2 OFF => current through it = 0.

I(D1)=I(10k) + I(D2)=0.93+0=0.93 mA

- - - Updated - - -



You have a contradiction here. First you say "I(D1)=I(10k) + I(D2)=1.86-1.64=+ve" and then by magic "I(D1)=(10-0.7)/5=1.86 mA"

Problem is in the first calculation of I(D1) because I(D2)=0.

I had calculated the first I(D1) to check if current through it is +ve or -ve. But that is perhaps not necessary and wrong as you pointed out that as we have determined D2 is off, I(D1)=1.86+0=1.86 mA in the first place only. Thank you sir for helping me with it. :grin: