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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