Y Transformation 2'7 Circuits with Dependent Sources 2'8

1
?-Y Transformation (2.7)Circuits with Dependent
Sources (2.8)

• Prof. Phillips
• February 3, 2003

2
?-Y Transformation

• A particular configuration of resistors (or
  impedances) that does not lend itself to the
  using series and parallel combination techniques
  is that of a delta (?) connection
• In such cases the delta (?) connection is
  converted to a wye (Y) configuration
• The reverse transformation can also be performed

3
?-Y Transformation
a
a
Ra
R1
R2
Rb
Rc
c
b
R3
b
c
4
?-Y Transformation

• To compute the new Y resistance values
• For the balanced case (RY Ra Rb Rc)
• R? 3 RY

5
Class Example
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.

7
Example Inverting Amplifier

• The following circuit is a (simplified) model for
  an inverting amplifier created from an
  operational amplifier (op-amp).
• It is an example of negative feedback.

8
Inverting Amplifier
1kW
4kW
10kW
I



Vf
Vs100Vf
10V

• Apply KVL around loop
• -10V 1kW I 4kW I 10kW I 100 Vf 0

9
Inverting Amplifier

• Applying KVL yielded
• -10V 1kW I 4kW I 10kW I 100 Vf 0
• Get Vf in terms of I
• Vf 10kW I 100Vf 0
• Vf -(10kW/101) I

10
Inverting Amplifier

• Solve for I
• I 1.961 mA
• Solve for Vf
• Vf -0.194 V
• Solve for source voltage
• Vs -19.4 V

11
Amplifier Gain

• Repeat the previous example for a gain of 1000
• Answer Vs -19.94V

12
Transistor Amplifier

• A small-signal linear equivalent circuit for a
  transistor amplifier is the following
• Find VX

5?10-4VX
3kW
6kW
VX
5mA

13
Apply KCL at the Top Node

• 5mA VX/6kW 5?10-4VX VX/3kW
• 5mA 1.67?10-4VX 5?10-4VX 3.33?10-4VX
• VX5mA/(1.67?10-4 5?10-4 3.33?10-4)
• VX5V

14
Class Examples