2. NOTE:
So fromthe curve we conclude that for aftera particularvalue of supplyvoltage (forwhich
minimumZenercurrentrequiredtoensure the Zenerdiodetooperate inZenerbreakdownregion) the
outputvoltage isclampedto4.5 volt.Itdoesnot change eventhe supplygoesonchanging.Thisparticular
behaviouriscalledlineregulation.
DISCUSSION QUESTION:
1. Compare the simulationstudyof Zenerdiode andPN junctiondiodeonthe basisof theirV-I
characteristics.
TakingD1N4002 as a general PN junction,the VIcharacteristicis
NOTE:
BREAKDOWN VOLTAGE= 100 volt
FORWARDVOLTAGE = .47 volt
For zenerdiode ,the VIcharacteristicis
V1(V1)
-10V -8V -6V -4V -2V 0V 2V 4V 6V 8V 10V
V(D1:2)
-1.0V
0V
1.0V
2.0V
3.0V
4.0V
5.0V
TRANSFER FUNCTION
V1(D6)
-100V -80V -60V -40V -20V -0V-110V 15V
I(D6)
-60mA
-40mA
-20mA
-0mA
20mA
40mA
60mA
(-99.754,-29.426u)
(468.715m,131.285u)
3. NOTE:
ZENER VOLTAGE = 4.46 volt
FORWARDVOLTAGE = .7 volt
COMPARISION:
PN JUNCTION ZENER
Same forwardcharacteristicas Zener Same forwardcharacteristicas general PN
junction
Breakdownvoltage isveryhigh Breakdownvoltage (Zenervoltage) is
comparativelylow
Dopingiscomparativelylow Dopingiscomparativelyhigh
Irreversible breakdown Reversible breakdown
CONCLUSION:
General diode isdesignedtoworkinforwardbiasing conditionwhilethe Zenerisforoperationin
reverse region
2. A 10 V Zenerdiode (type 1N4740) placedinserieswitharesistor anda fixed12 V powersupply.
What isthe currentflowingthroughthe circuitif the value of the resistoris400 ohm?
V(V0)
-5.0V -4.0V -3.0V -2.0V -1.0V 0V 1.0V 2.0V 3.0V 4.0V 5.0V
I(D1)
-15mA
-10mA
-5mA
0A
5mA
10mA
15mA
(-4.4657,-1.1719m)
(713.191m,1.2868m)
4. Simulationcan’tbe done asthe componentmodel isnotavailable indemoversion
Kvl equation:
12-400*I-10 = 0 (asVin> Vz=10 V)
I = 5 mA > 1.17mA
So the correct ans is5 mA as itis greaterthanthe minimumcurrentrequiredtoputthe diode in
breakdownregion.
3. Why Zenerdiode iscalledasvoltage regulator?How the ratingof Zenerdiode isdecided?
It can be explainedbyconsideringthe followingcircuit
So fromgraph it isclearthat whenthe supplyvoltage ismore than12V the outputdoesnot change
evenif the supplyvoltage changes.The outputisfixedat4.59 V.it iscalledline regulation.
V1(Vin)
0V 2V 4V 6V 8V 10V 12V 14V 16V 18V 20V
1 V(V0) 2 D(V(V0))/ D(V1(Vin))
0V
1.0V
2.0V
3.0V
4.0V
5.0V
1
0
1.0
2.0
3.0
4.0
5.0
6.0
2
>>
(12.540,4.5957)
OUTPUT LOAD VOLTAGE
REGULATION
5. So fromgraph it isclearthat whenthe loadresistance ismore than500 Ohmthe outputvoltage
remainconstantirrespectiveof increase inloadresistance ordecrease inloadcurrent.Thisiscalledload
regulation.The minimum500 Ohmvalue ensuresthe Zenertoworkinbreakdownregion.
NOTE:
load regulation specification defines how close the series resistance ofthe output is to 0 ohms - the series
resistance ofan ideal voltage source.
SELECTING ZENER DIODE FOR APPLICATION:
IS = ID + IL
Zener voltage = voltage rating of the load
As we know in the process ofoperation if the load impedance increase and hence the load current decreases
the current from the supplyremains same and hence the Zener current increases.
Maximum Zener current >= off load current in Zener
Minimum Zener current= supplycurrent-full load current
Power rating = Zener voltage* Maximum Zener current
4. Obtainthe I-v characteristicof the Zenerdiode (e.g.D1N750 isa 4.7 V Zener).The voltage
regulatorcircuitshown belowhasR= 1 KW.
VIcharacteristic
OHM
0.1K 0.2K 0.3K 0.4K 0.5K 0.6K 0.7K 0.8K 0.9K 1.0K
1 V(V0) 2 D(V(V0)) / D(I(R2))
0V
2.5V
5.0V
1
0
0.5K
1.0K
2
>>
(560.163,4.5927)
RL
1k
IS
D7
D1N750
ID
V1
16Vdc
R2
1k
IL
6. a. Findthe outputvoltage forV1 = 16 V.
ans: 4.67 V
b. What isthe outputvoltage whenthe supplyvoltage droppedtoV1= 8 V?
ans: 4.597 V
V(V0)
0V 1V 2V 3V 4V 5V 6V 7V 8V 9V 10V
- I(D1) (16-V(V0))/1000 (8-V(V0))/1000
0
4m
8m
12m
16m
(4.5966,3.3834m)
(4.6694,11.251m)