Can anybody solve attached questions

Question 1:
(1)
Loop1:
6V = Vr + Vd1 + Vd2
= I1.1k + 2.Vf
I1 = (6 - 1.4) / 1k
= 4m6 A
Loop2:
Vd2 = Vr
= I2.1k
I2 = 0.7 / 1k
= 0m7 A
The current through D1 is I1 = 4m6 A.
The current through D2 is I1 - I2 = 3m9 A.

(2)
Loop1:
10V = Vr + Vd1 + Vd2
= I1.1k + 2.Vf
I1 = (10 - 1.4) / 1k
= 8m6 A
Loop2:
Vd1 = Vr
= I2.1k
I2 = 0.7 / 1k
= 0m7 A
Loop3:
Vd2 = Vr
= I3.500
I3 = 0.7 / 500
= 1m4 A
The current through D1 is I1 - I2 = 7m9 A.
The current through D2 is I1 - I3 = 7m2 A.

Question 2:
Ib = Vbb / Rb
= 5 / 47 k
= 106 uA
Ic = ∞
Of course, the C-E junction has just avalanched, shorting the B-E junction, and an infinite current is flowing through the collector. Why do they teach kids these inoperable and useless circuits ?
:D
Note: Theoretically, a current source can be put in series with a voltage source, so that the voltage source maintains a voltage across the current source, and the ideal current source maintains a current through the ideal voltage source.
:rolleyes:
 
Last edited:

kubeek

Joined Sep 20, 2005
5,660
Question 2:
Ib = Vbb / Rb
= 5 / 47 k
= 106 uA
Ic = ∞
Of course, the C-E junction has just avalanched, shorting the B-E junction, and an infinite current is flowing through the collector. Why do they teach kids these inoperable and useless circuits ?
Why do you think so? I would say that Ib=(Vbb-0.7)/Rb, and Ic=Ib*hFE and I can´t see any reason for CE to avalanche.
 

AnalogKid

Joined Aug 1, 2013
8,256
Question 2:
Ib = Vbb / Rb
= 5 / 47 k
= 106 uA
Ic = ∞
0 for 2.

Ib = (Vbb - Vbe) / Rb = 4.3 / 47 k = 91.5 uA
Ic = Ib x hfe = 91.5 uA x 80 = 7.32 mA

And by the way...

P = E x I = Vce x Ic = 9 x 7.32 mA = 66 mW

Depending on its package, the transistor probably is not overheating.

ak
 
You got me. Rash. But I have never seen a common emitter circuit without a collector resistor, but then again, even after years of uni, I don't think I understand that circuit. :rolleyes:
After thought, I kind of knew this was an exam question, and decided to 'throw a spanner in the works'. As I do ?
;)
 

AnalogKid

Joined Aug 1, 2013
8,256
Dangle-biasing always is the very first way taught to bias a transistor, and (hopefully) the first way used as a reason to do it some other way.

ak
 

JohnInTX

Joined Jun 26, 2012
3,947
You got me. Rash. But I have never seen a common emitter circuit without a collector resistor, but then again, even after years of uni, I don't think I understand that circuit. :rolleyes:
After thought, I kind of knew this was an exam question, and decided to 'throw a spanner in the works'. As I do ?
;)
How about you don't do that?
 
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