ET 210 Quiz #3
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ET 210 Quiz 3
The theoretical material that covers problem 1
(circuit below containing E
S
, D
1
, D
2
, R
1
, R
2
, and R
3
) is contained in the Theoretical Discussion of ET 210 Laboratory Experiment 3
: Series-Parallel Diode-
Resistor Circuits and Kirchhoff's Laws. The third sheet of
this document
contains the beginning of the Theoretical Discussion of ET 210 Laboratory Experiment 3, which can be used as a reference
.
1)
Given the following circuit consisting of silicon diodes D1 and D2, and resistors R1, R2, and R3, along with Es voltage source:
Source
Voltage E
S
= 8 V (as shown in the schematic above)
Find
the following:
a)
Determine the bias
(forward-biased or reverse-biased) conditions of diodes D1 and D2. The bias D1 is negative (reverse) biased and D2 is forward biased with a voltage of 0.7V in the silicon diode.
Use Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, Ohm's Law and the appropriate diode approximations to calculate the circuit currents and
voltages in this problem. You will write a system of simultaneous linear equations and then solve the equations to obtain the currents though R
1
, R
2
, and R
3
. You will then calculate the voltages across R
1
, R
2
, and R
3
. The current through diode D2 and the voltage across Diode D1 can then be found. Refer to the Theoretical Discussion
contained in ET 210 Lab 3: Series-Parallel Diode-Resistor Circuits and Kirchhoff's Laws (by John Buoncora) for a similar analysis using KVL, KCL, and Ohm's Law directly applied to the approximate equivalent circuit
corresponding to the circuit of Lab 3 Figure-1 (in the Theoretical Discussion of Lab 3
).
b)
Find the currents
I
R1
, I
R2
, and I
R3
(that is, find the currents
through resistors R1, R2, and R3
KVL Loop 1: -ES+ 560IR1 + 220 IR3 + 0.7V = 0 VR2 - 8/560 + VR2/1000 + VR2 -0.7V/220 = 0 VR2 -
8/560 + VR2/1000 +VR2 -0.7V/220 = 0 VR2 { 1/560 + 1/1000 + 1/220 } = 8/560 + 0.7/220 VR2 = 2.38V IR1 = Es - VR2 / R1 = 8-2.38V / 560 = 10.03Ma IR2 = VR2 / R2 = 2.38V / 1000 = 2.38Ma IR3 =
VR2 - VD2 / R3 = 2.38V - 0.7V / 220Ohms = 7.63mA
c)
Find the voltages
V
R1
, V
R2
, and V
R3
(that is, find the voltages
across resistors R1, R2, and R3
VR1 = IR1 R1 = 10.03*10^-3 * 560 = 5.61V VR2 = 2.38V VR3 = IR3 R3 = 7.63 * 10^-3 * 220Ohms =
1.67V
d)
Find the currents
I
D1
and I
D2
(that is, find the currents
through diodes D1 and D2)
ID1 = 0A ID2 = IR3 = 7.63mA
e)
Find the voltages
V
D1
and V
D2
(that is, find the voltages
across diodes D1 and D2)
VD1 = VR2 = 2.38V
The quiz is continued on the next page
.
2)
Transformer and Half Wave Rectifier Circuit Problem:
Refer to the schematic below of the transformer and Half-Wave Rectifier for this problem.
The input voltage (Vin)
is a sine wave with an rms value (effective value) of 230 V (rms)
as shown in the schematic. The V
L
(Load Voltage) can also be referred to as V
OUT
(Output Voltage).
Find or perform the followin
g:
a)
Find the secondary rms
voltage, V
SEC (rms)
VPRIMARY/ VSEC = 4/1 VSEC = VPRIMARY / 4 230V/4 =57.5V = V SEC
b)
Find the secondary peak
voltage, V
SEC (peak)
Vrms = VPEAK / 1.4142 VPEAK = 1.4142 * Vrms VPEAK= 1.4142 * 57.5V =81.32V = V PEAK
c)
Sketch the secondary voltage waveform (V
SEC
plotted against time)
SEE SKETCHES BELOW
d) Find the Load peak voltage, V
L (peak)
-VSEC+ VSI+ VL = 0 VLPEAK = VSEC - VSI VLPEAK = VSEC - 0.7V VLPEAK = 81.32
e) Sketch the Load voltage waveform (V
L
plotted against time)
f) Find the DC component of Load Voltage (average value of Load Voltage), V
L (DC)
= V
L (average)
VDC = IAC * RL = 0.517 * 50 = 25.88V VDC = 81.32V VDC = 25.88V
f)
Find the peak
load current, I
L (peak)
ILOAD - 1m 1m = Vm / RL = Vm / 50 = 81.32 / 50 = 1.626A
g)
Find the average
load current, I
L (average)
ILOAD AVG= ILOAD DC = 1m
i) Find the PIV (Peak Inverse Voltage) across the diode when the diode is reverse biased
PIV = Vdiode = Vinput MAX PIV = Vm = 81.32V
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Refer to the next page of
this document
for the beginning of the Theoretical Discussion of ET 210 Laboratory Experiment 3
.
The
Theoretical Discussion of Lab 3 will be helpful as a reference when solving problem 1
of this quiz. The next six pages of this document contain material to be read for reference
. The next
six pages of this document do NOT
contain any quiz problems
. ET 210 Laboratory Experiment 3
: Series-Parallel Diode-Resistor Circuits and Kirchhoff's Laws
Series-Parallel Diode-Resistor Circuits and Kirchhoff's Laws (
reference reading material
)
Prepared by John Buoncora
This page and the next five pages of this document contain reference reading material.
This page and the next
five pages of this document do NOT
contain any quiz problems
. Components:
Switching Diode (1N4148, 1N914, or equivalent switching or small signal diode)
General Purpose Silicon Rectifier Diode (1N400X, where X is any digit from 1 to 7)
Resistors (rated 1/2 Watt or above): 560 Ω, 220 Ω, 1 k
Ω
Theoretical Discussion
:
Analysis of Figure-1: Series-Parallel Diode-Resistor Circuit
(E
S
= 5 v)
The Series-Parallel Diode-Resistor circuit of Figure-1 contains two diodes, several resistors and one independent DC voltage source. Each diode is either Forward-Biased or Reverse-Biased, which leads to four possible states of operation for the circuit: Switching Diode
1N400X Diode
Reverse-Biased
Forward-Biased
Reverse-Biased
Reverse-Biased
Forward-Biased
Reverse-Biased
Forward-Biased
Forward-Biased
Note: The name Switching Diode simply represents a typical application of this particular diode and is not relevant to the following analysis. The circuit of Figure-1
, which is shown on the next page
, is operating in the state where the Switching Diode is Reverse-Biased
and the 1N400X Diode is Forward-Biased
.
Es
5 V R1
560 Ω R2
1kΩ R3
220 Ω 1N400X
+
_
Switching
Diode
Figure-1: Series-Parallel Diode-Resistor Circuit (E
S
= 5 v) Since the circuit of Figure-1
contains only one
independent DC Source
, we can trace the direction of current throughout the circuit and determine the bias condition (Forward-Biased
or Reverse-Biased) of each diode. The result of this analysis leads to the conclusion that the Switching Diode is Reverse-Biased and the 1N400X diode is Forward-Biased (that is, the circuit is operating in the state given in the first (state) row of the table above. Therefore, we can replace the Switching Diode by an open circuit (approximately, neglecting reverse diode current) and assume that the voltage across the Forward-Biased 1N400X Silicon Diode is approximately 0.7 V (that is, V
1N400X_Diode
≈ 0.7 V), if the 1N400X diode is operating above the knee. The following approximate equivalent circuit is obtained (where the voltage across the 1N400X diode will be denoted as V
D
to simplify the notation):
Es
5 V R1
560 Ω R2
1kΩ R3
220 Ω 1N400X
+
_
Open
Circuit
+
_
VD
Approximate Equivalent Circuit of the Series-Parallel Diode-Resistor Circuit (the actual original schematic was shown in Figure-1)
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OR Simply:
Es
5 V R1
560 Ω R2
1kΩ R3
220 Ω 1N400X
+
_
+
_
VD
Approximate Equivalent Circuit of the Series-Parallel Diode-Resistor Circuit (the actual original schematic was shown in Figure-1)
The voltage across the Forward-Biased 1N400X Silicon diode can be approximated as
V
D
≈ 0.7 V (assuming operation above the forward knee, which is an assumption that we will check after finding the diode current).
The approximate equivalent circuit of the Series-Parallel Diode-Resistor Circuit of Figure-1
(with E
S
= 5V) can be analyzed using Kirchhoff's Voltage Law (KVL), Kirchhoff's Current Law (KCL), Ohm's Law and the appropriate diode approximations. The approximate equivalent circuit of the circuit originally shown in Figure-1 is drawn below with assigned current and voltage reference labels. The circuit will be analyzed by directly applying KVL, KCL, Ohm's Law and the diode approximation V
D
≈ 0.7 V for the Forward-Biased 1N400X Silicon Diode.
Es
5 V R1
560 Ω R2
1kΩ R3
220 Ω 1N400X
+
_
Iw
Ix
Iy
+
_
V1
+
_
V2
+
_
V3
+
_
VD
Approximate Equivalent of Series-Parallel Diode Resistor Circuit with assigned Current and Voltage Reference Labels
The KVL for the Mesh (Loop) on the left
side of the approximate equivalent circuit is:
+ E
S
– V
1
– V
2
= 0
The KVL for the Mesh (Loop) on the right
side of the approximate equivalent circuit is:
+ V
2
– V
3
– V
D
= 0
The KCL for the Node at the top of the approximate equivalent circuit is:
I
W
= I
X
+ I
Y
The Ohm's Law equations for each of the resistors are:
V
1
= I
W
×
R
1
= (560 Ω)
I
W
V
2
= I
Y
×
R
2
= (1 kΩ)
I
Y
V
3
= I
X
×
R
3
= (220 Ω)
I
X
The equation for the 1N400X Silicon Diode, which is Forward-Biased (assuming operation above
the knee) is V
D
≈ 0.7 V
The Ohm's Law equations, along with the value of the source voltage and diode (approximate) voltage can be substituted into the KVLs to obtain the equations (where currents I
W
,
I
X
, and I
Y
are shown in bold face):
KVL (left side Mesh):
+ 5 V – (560 Ω)
I
W
– (1 kΩ)
I
Y
= 0
KVL (right side Mesh):
+ (1 kΩ)
I
Y
– (220 Ω)
I
X
– 0.7 V = 0
The KCL ( I
W
= I
X
+ I
Y
) can be solved for I
Y
to obtain:
I
Y
= I
W
– I
X
Substituting the equation for I
Y
into the KVL (Mesh) equations results in:
+ 5 V – (560 Ω)
I
W
– (1 kΩ)(
I
W
– I
X
) = 0 for the left
side Mesh
+ (1 kΩ)(
I
W
– I
X
)
– (220 Ω)
I
X
– 0.7 V = 0 for the right
side Mesh
The last two equations represent two linear equations in the two unknowns I
W
and I
X
. The equations will be solved below using the substitution method.
We will use the Distributive Law and then combine like terms in each equation:
+ 5 V – (560 Ω)
I
W
– (1 kΩ)
I
W
+ (1 kΩ)
I
X
= 0 for the left
side Mesh (by the Distributive Law)
+ 5 V – (1.56 kΩ)
I
W
+ (1 kΩ)
I
X
= 0 for the left
side Mesh (by combining like terms)
+ (1 kΩ)
I
W
– (1 kΩ)
I
X
– (220 Ω)
I
X
– 0.7 V = 0 for the right
side Mesh (by the Distributive Law)
+ (1 kΩ)
I
W
– (1.22 kΩ)
I
X
– 0.7 V = 0 for the right
side Mesh (by combining like terms)
Therefore, we have the following two equations in the two unknowns I
W
and I
X
(obtained by using the Distributive Law and combining like terms in each equation as shown above):
+ 5 V – (1.56 kΩ)
I
W
+ (1 kΩ)
I
X
= 0 for the left
side Mesh
+ (1 kΩ)
I
W
– (1.22 kΩ)
I
X
– 0.7 V = 0 for the right
side Mesh We can solve the equation + (1 kΩ)
I
W
– (1.22 kΩ)
I
X
– 0.7 V = 0 for I
W
as follows:
+ (1 kΩ)
I
W
– (1.22 kΩ)
I
X
– 0.7 V =
0
+ (1.22 kΩ)
I
X
+ 0.7 V + (1.22 kΩ)
I
X
+ 0.7 V (
add
these terms to both sides
)
The result is:
+ (1 kΩ)
I
W
= + (1.22 kΩ)
I
X
+ 0.7 V
Solving for I
W
by dividing both sides by 1 kΩ gives:
I
W
= + (1.22 kΩ)
I
X
/
(1 kΩ) + (0.7 V)
/(
1 kΩ)
Note the cancellation of kΩ in the first term and (0.7 V)
/(
1 kΩ) = 0.7 mA in the second term.
Therefore, we have:
I
W
= 1.22
I
X
+ 0.7 mA
We can substitute I
W
= 1.22
I
X
+ 0.7 mA into the equation + 5 V – (1.56 kΩ)
I
W
+ (1 kΩ)
I
X
= 0
to obtain:
+ 5 V – (1.56 kΩ)
(1.22
I
X
+ 0.7 mA
) + (1 kΩ)
I
X
= 0
We will use the Distributive Law and then combine like terms:
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+ 5 V – (1.56 kΩ)
(1.22
I
X
) – (1.56 kΩ)
(0.7 mA
) + (1 kΩ)
I
X
= 0 (by the Distributive Law)
+ 5 V – (1.9032 kΩ)
I
X
– 1.092 V + (1 kΩ)
I
X
= 0 (
by multiplying appropriate quantities)
+ 3.908 V – (0.9032 kΩ)
I
X
= 0 (by combining like terms)
+ 3.908 V – (0.9032 kΩ)
I
X
= 0
Solving for I
X
:
I
X
= (3.908 V)
/
(0.9032 kΩ)
I
X
= 4.3268 mA
(current through 1N400X Diode)
Substitute I
X
= 4.3268 mA into the equation I
W
= 1.22
I
X
+ 0.7 mA to obtain:
I
W
= (1.22)
(4.3268 mA)
+ 0.7 mA
I
W
= 5.9787 mA
Substitute I
X
= 4.3268 mA and I
W
= 5.9787 mA into the equation I
Y
= I
W
– I
X
to obtain:
I
Y
= 5.9787 mA – 4.3268 mA I
Y
= 1.6519 mA
Substitute the currents into the Ohm's Law equations for each of the resistors:
V
1
= I
W
×
R
1
= (560 Ω)
I
W
= (560 Ω)
(5.9787 mA) = 3.3481 V
V
2
= I
Y
×
R
2
= (1 kΩ)
I
Y
= (1 kΩ)
(1.6519 mA) = 1.6519 V
V
3
= I
X
×
R
3
= (220 Ω)
I
X
= (220 Ω)
(4.3268 mA) = 0.95190 V
Since the 1N400X diode current is I
X
= 4.3268 mA, our assumption that the 1N400X diode is operating above the Forward Knee is correct. Therefore, the approximation that we used previously of V
D
≈ 0.7 V for the Forward-Biased 1N400X Silicon diode is valid.
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