ET 210 Quiz 4
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ET 210 Quiz 4
1) Given the circuit drawn below, containing a Zener diode
along with the other components shown:
Find the following (
only
in
parts a
through
e
, the
Load Resistance RL = 160 Ω
):
a) Load Voltage V
L=5.1V
b) Load Current I
L=31.8 mA
c) Current I
Rs
(which is the current through resistor R
S
)=
55.18 mA
d) Zener Diode Current I
Z=23.38 mA
e) Zener Diode power
dissipation P
Z=119.23 mW
f
) Assume that the Load Resistance is changed to
RL = 70 Ω
and find the new values of: V
L = 4.11 V I
L = 58.8 mA
I
Rs
= 58.8 mA
I
Z = 0 A
g
) Assume that the Load Resistance is now
RL = 120 Ω and
the power rating
of the Zener diode is P
Z (max specified)
= 400 mW
Now also assume that the input voltage Vin
is a variable
quantity (
V
in
is not
20 V in this part g
).
Find the minimum
input
voltage V
in (min)
at which regulation will still be maintained = And:
Find the maximum
input
voltage V
in (max)
before the Zener diode is damaged.
(That is, you will determine the range of input
voltages
from minimum to maximum where the circuit will regulate properly (maintain constant output Load
voltage
based on the Zener diode) and
the Zener diode will be not
be damaged).
RL = 120Ω IZMAX= PzMAX / Vz = 400mW/5.1V = 78.43mA IRSMAX= IZMAX + IL = 78.43mA + 42.5mA = 120.93mA IL = 5.1V/120 Ω = 42.5mA
The quiz is continued
on the next page
.
2) Given the circuit drawn below, containing a Zener diode
along with the other components shown:
Find the following:
a) Load Voltage V
L = VZ = 9.1 V
b) Load Current I
L = 9.1 V / 75
Ω = 0.121 A
c) Current I
Rs
(which is the current through resistor R
S
)
IRS = VS - VRS/150 Ω = 36V-9.1V/150 Ω = 0.179A d) Zener Diode Current I
Z
IZ = IR - IL = 0.179A - 0.121A = 0.0583A = 58.3mA e) Power
P
Z
dissipated in Zener Diode Pz = IZ * VZ = 58.3mA * 9.1V = 0.531W f) Powers
P
Rs
and P
RL
dissipated in resistors R
S
and R
L
respectively.
PRL = IL * VL = 0.121 A * 9.1V = 1.101W PRS = IR * VR = IR(Vs - Vz) IR(36-9.1)
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