ECE201Lect-8
Transcript of ECE201Lect-8
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ECE201 Lect-8 1
-Y Transformation (2.7);
Circuits with Dependent Sources
(2.8)
Dr. Holbert
February 13, 2006
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ECE201 Lect-8 2
-Y Transformation
A particular configuration of resistors (orimpedances) that does not lend itself to the
using series and parallel combinationtechniques is that of a delta () connection
In such cases the delta () connection isconverted to a wye (Y) configuration
The reverse transformation can also beperformed
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ECE201 Lect-8 3
-Y Transformation
a
c b
a
bc
R1 R2
R3
Ra
RbRc
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ECE201 Lect-8 4
-Y Transformation
To compute the new Y resistance values
For the balanced case (RY=Ra=Rb=Rc)
R
= 3RY
3
1
#
2
1
#
i
inode
k
nodeYbesideknode
nodeY
R
RR
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ECE201 Lect-8 5
Class Example
Learning Extension 2.17
Learning Extension 2.18
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ECE201 Lect-8 6
Circuits with Dependent Sources
Strategy:
Apply KVL and KCL, treating dependent
source(s) as independent sources.
Determine the relationship between
dependent source values and controlling
parameters.
Solve equations for unknowns.
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ECE201 Lect-8 7
Example: Inverting Amplifier
The following circuit is a (simplified) model
for an inverting amplifiercreated from an
operational amplifier(op-amp).
It is an example of negativefeedback.
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ECE201 Lect-8 8
Inverting Amplifier
1kW
+
4kW 10kW
+
+
Vf Vs=100Vf10V
I
Apply KVL around loop:
-10V + 1kWI+ 4kWI+ 10kWI+ 100 Vf= 0
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ECE201 Lect-8 9
Inverting Amplifier
Applying KVL yielded:
-10V + 1kWI+ 4kWI+ 10kWI+ 100 Vf= 0
Get Vfin terms ofI:
Vf+ 10kWI + 100Vf = 0Vf= -(10kW/101)I
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ECE201 Lect-8 10
Inverting Amplifier
Solve forI:
I = 1.961 mA
Solve for Vf:
Vf = -0.194 V
Solve for source voltage:Vs = -19.4 V
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ECE201 Lect-8 11
Amplifier Gain
Repeat the previous example for againof
1000
Answer: Vs = -19.94V
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ECE201 Lect-8 12
Another Amplifier
1kW 4kW100nF
+
Vf Vs=100Vf10V0
I
Find the output voltage Vsfor this circuit,
assuming a frequency of w=5000
+
+
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ECE201 Lect-8 13
Find Impedances
1kW 4kW-j2kW
+
Vf Vs=100Vf10V0
I
+
+
Apply KVL around loop:
-10V0+ 1kWI+ 4kWI-j2kWI+ 100 Vf= 0
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ECE201 Lect-8 14
Another Amplifier
KVL provided:
-10V0+ 1kWI+ 4kWI-j2kWI+ 100 Vf= 0
Get Vfin terms ofI:
Vf
-j2kWI+ 100 Vf
= 0
Vf= (j2kW/101)I
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ECE201 Lect-8 15
Another Amplifier
Solve for I:
I= 2mA 0.2
Solve for Vf:
Vf = 39.6mV90.2
Solve for source voltage:Vs = 3.96V90.2
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ECE201 Lect-8 16
Transistor Amplifier
A small-signal linear equivalent circuit for a
transistor amplifier is the following:
Find VX
3kW6kW
+
VX5mA
510-4VX
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ECE201 Lect-8 17
Apply KCL at the Top Node
5mA = VX/6kW+ 510-4VX+ VX/3kW
5mA = 1.6710-4VX+ 510-4VX+ 3.3310
-4VX
VX=5mA/(1.6710
-4
+ 510
-4
+ 3.3310
-4
)
VX=5V
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ECE201 Lect-8 18
Class Examples
Learning Extension E2.19
Learning Extension E2.20