Exam revision, no solutions to check answers against.

[Macka]

Hi Everyone,

I am currently studying for an exam (on semi conductors) and have come across a question that I can not complete, nor do I know if what I have done so far is correct.

The lecturer will not give us solutions for tutorial questions or past exams, so I have no way of checking my answers. Many of the exam questions are taken from our text book, however this one does not appear to have come from there.

What is the power dissipation in the zener diode (Vzener = 15V, rz=0) in this circuit for:
a) RL = ∞
b) RL = 75 Ω
c) RL = 50 Ω

My solutions:
a) RL = ∞ = open circuit
so V(150Ω) = 50 -15 = 35V
Id = 35/150Ω = 200mA
P = VI = 35V*200mA = 3.5W

b)
Approach 1:
It = I total
I2 = current through RL

L1: 50V = It*150Ω + 15V
L2: 50V = It*150Ω + I2*75Ω
It*150Ω + 15V = It*150Ω + I2*75Ω
15V = I2*75Ω
I2 = 15/75 = 1/5 = 200mA Alarm bells! surely this can't be right?

Approach 2:
V2 = Vt*(rz//RL) / (150Ω + (rz//RL)) Voltage divider rule
15V = 50V*(rz//75Ω) / (150Ω + (rz//75Ω))
50*(rz//75Ω) = 2250 +15*(rz//75Ω)
35(rz//75Ω) = 2250
(rz//75Ω) = 64.28
rz = 0.0155 Alarm bells! surely this can't be right? rz should be 0

c)

Same as b).

I am completely lost, a prod in the right direction would be much appreciated.

 

[dougy83]

I would first work out what the voltage at the 150R/RL junction would be without the zener connected. If it's >15V, then the zener is conducting; the current through RL is then 15/RL & the current through the zener is (50-15)/150 less the current through RL. If the voltage at the resistor junction would be <15V, then the zener would not conduct & therefore has no power dissipation.

 

[Macka]

Thanks for that Doug, I am getting much nicer results now!

 

[mneary]

I would read the question more carefully. It asks for the power in the zener.