homework help > electrical engineering > electrical engineering q&a board > topic: zena diode...

Q BgQuestion:

      
Novice
Karma Points: 25
Respect (83%):
posted by  hooky on 11/6/2009 6:48:24 PM  |  status: Closed  |  Earned Karma: 25

zena diode

Course Textbook Chapter Problem Needs by
Electrical Engineering Microelectronic Circuits (5th) by Sedra, Smith Diodes N/A 11/1/2009 at 3:00:00 PM
Question Details:

A student designs a zener shunt regulator to produce a regulated output of 10 V using a zener diode specified to have a 10V drop at a test current of 25mA. Incremental resistance of the zener is 10 ohm and minimum diode current =2mA. Given that supply voltage is 20V ±25%, and load current can vary between 0mA and 20mA,  find

(a) Vzo for the zener diode                                                                             

(b) Required series resistance R for zener diode                                               

© Line regulation                                                                                            

(d)Change in Vout corresponding to ±25% change in unregulated supply      

(e)Load regulation                                                                                          

(f) Change in Vout corresponding to 0-20mA change in load current             

(g) Zener diode power rating to handle the worst case power dissipation     

Bonus Point Alert! Earn +7 additional karma points for helping this gold member.

AAnswers:

Answer Question Ask for clarification
Sage
Karma Points: 5,343
(IIT Kharagpur)
posted by Chelsea_fan(MNK) on 11/7/2009 1:47:52 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
for a zener diode,
Vz=Vzo+IzRz
a)10=Vzo+25*10
 so we gret,
Vzo=9.75 V
b)now voltage drop across zener diode is 10 V, so the voltage across series resistr R is (20-10) = 10 .....
also current flowing thru zener is also equal to currrent flowing thru R...
R=voltage acrooss R/ current thru R=10/25*10-3=.4 kΩ

c)applying voltage divider rule,
ΔVo=ΔVi*Rz/R+Rz=±1*10/400+10=±24.4 mV

so line regulation=ΔVo/ΔVi=24.4 mV/V


d)25% of supply voltage is 5V means change is ±5V

so applying voltage divider rule,
ΔVo=ΔVi*Rz/R+Rz=±5*10/400+10 =±121.9 mV

e)now change in o/p voltage is given by,
ΔVo=Rz*ΔIz=10*(±1*10-3)=±10 mV
load regulation=ΔVo/ΔIz=±10 mV/mA


f)load current is increased by 20 mA zener current decreases by samee

ΔIz=-20 mA
so ΔVo=Rz*ΔIz=10*(-20*10-3)=-200 mV

g)maximum power disssipation takes place when current thru zener diode is maximum
here max current is 25 mA

so power dissipation is,
P=I2*Rz=25*10-3*25*10-3*10=6.25 mW
Answer Question Ask for clarificarion
Ask New Question

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »

Need Help In

Useful Links