Chapter 04.ppt dependent sources for electrical

Chapter Four Dependent Sources
4.1 Linear Dependent Sources
4.2 Suppression and Dependent Sources
4.3 Power in Dependent-Sources Networks
Contents of this Chapter:

4.1 Linear Dependent Sources
• Definitions and Notation
A dependent source is a source element whose value ( voltage or
current), called output variable, depends on some other variable
(voltage or current), called controlling variable. A dependent
source has two pairs of terminals:

Input port:
If the controlling variable is a voltage, it is a open-circuit.
If the controlling variable is a current, it is a short circuit.

Output port:
The output may be voltage (dependent voltage source)

The output may be current (dependent current source).
Output port:

Therefore, there are four type linear dependent sources:
Voltage-controlled voltage source, VCVS
where m is the voltage transfer coefficient
(no dimension).

Voltage-controlled current source, VCCS
where gm is the transfer conductance
(conductance dimension, unit S).

Current-controlled voltage source, CCVS
where r is the transfer resistance
(resistance dimension, unit W).

Current-controlled current source, CCCS
where  is the current transfer coefficient
(no dimension).

Generally, we left off the input or controlling port of the dependent
source in circuit graph and draw only the output port with labeled
control variable. In this manner, the controlling variable, which
may be a voltage or current located in other part of the circuit,
should be labeled in the circuit.

Examples Using Dependent Sources
Example 1. A dependent voltage source is inserted between a
Thévenin source network and a load resistor. Let us calculate the
voltage vo across the load resistor assuming the output current io is
zero.

Writing KVL around the input loop (containing vs ) yields
1
S S S
v R i v
  
Since the v1 terminal pair is an open circuit, the current is must be
zero. Therefore, v1= vS
Consider now the output loop (containing RL ). Writing KVL for
this loop yields Av1=v0
Then we obtain v0= AvS
Thus we see that the output voltage vo is proportional to the input
voltage vs. If A is greater than unity, vo is also larger in magnitude
than vs. In this particular case, therefore, the dependent source is
functioning as a linear amplifier, increasing the magnitude of the
signal vs by a constant factor, A.

Example 2. A current-controlled current source is connected
between a Thévenin equivalent source network and a Thévenin
equivalent load network.
Let us determine the relation between the source voltage vs and the
output voltage vo assuming that the output current io is zero.

For the input loop (containing the source voltage vs)
KVL and Ohm's law yield iB=vS /RS
For the output loop, containing thr output voltage vo, write KVL
0 OC L L OC F L B
v V R i V R i

     
Therefore, we obtain the relationship between input and output
0
F L
OC S
S
R
v V v
R

  
The figure below shows this
relation, transfer characteristic.
This example indeed represents a transistor
amplifier over part of the transistor range of
operation.
Notice that the slope is negative. This means
that as vs increases, vo must decrease. Any
variations in vs, therefore, appear "upside-
down" or inverted (反相) at the output of the
network.

4.2 Suppression and Dependent Sources
•Suppression of Dependent Sources
We have already remarked that when the relation between the
dependent-source value and the control variable is a simple
proportionality, the dependent source is a linear element. Therefore, it is
possible to use super-position in networks containing linear dependent
sources, and we can use Thévenin and Norton equivalents to represent the
terminal characteristics of networks containing these elements.
Recalling that superposition involves finding the response of a network to
each independent source, one at a time, then summing results, a question
now arises about what to do with the dependent sources when one is
suppressing independent sources.
Because dependent sources are used to represent the behavior of resistive
elements such as transistors. To suppress a dependent source, therefore,
amounts to removing some elements from the network.
In the superposition problem for linear circuits,
only the independent sources can be suppressed,
and the dependent sources can never be suppressed.

An Example of How Not to Use Superposition
This circuit contains one independent source and a dependent
source. Recall the result in the previous section, the correct
answer for the output voltage is v0=AvS
Now we use the superposition on both independent and
dependent sources to solve the circuit again.

1. Suppress the dependent source, and calculate the response for
the independent source
In the left loop, because there is no source, the current is is zero
and v1=0.
In the right loop, according to the KVL v01=Av1=0.

2. Suppress the dependent source, and calculate the response
for the dependent source.
In the left loop, because of open-circuit for the input port of
dependent source, the current is is zero and v1= vs.
In the right loop, because the dependent source is suppressed,
the output is v02=0.
By the superposition, the total output voltage should be zero.
This result is obviously incorrect.

An Example of How to Use Superposition Correctly
Above circuit contains two independent sources and a dependent
source. Recall the result obtained in previous section, the correct
answer for the output voltage is
0
F L
CC S
S
R
v V v
R

  
Now we use superposition on only independent sources to solve
the output voltage again.

1. Calculate the response to DC voltage source Vcc
Suppressing the input voltage source vs (Notice the dependent
source is reserved):
There is no source in the left loop, so iB=0. In the right loop, KVL
and Ohm's law
01 CC L L CC F L B CC
v V R i V R i V

      

2. Calculate the response to input voltage source vs
Suppressing the DC voltage source Vcc (Notice the dependent
source is reserved):
In the left loop, KVL and Ohm's law yield
For the right loop, apply KCL and Ohm's law
According to the superposition, we yield
S
B
S
v
i
R

02
F L
L L F L B S
S
R
v R i R i v
R


      
0 01 02
F L
CC S
S
R
v v v V v
R

    
Notice that the dependent source
is never suppressed. It should be
included in all of the networks
involved in the superposition
solution.

Thévenin and Norton Equivalents with Dependent Sources
Whenever superposition is valid, that is, in networks containing
independent sources, resistors, and linear dependent sources, it is
possible to represent the terminal characteristics of any network
by a Thévenin or Norton equivalent. Dependent sources introduce
several complications into the problem of finding Thévenin and
Norton equivalents, but they do not alter the validity of the
concept.

Where, iSC is the short-circuit
current, which is determined in
following circuit.
vOC is the open-circuit voltage,
which is determined.
RO is the Thévenin resistance
(internal resistance), which is
determined:

Now let's demonstrate how to determine the Thévenin
equivalent circuit.
1. Determine the open-circuit voltage across terminals a and b,
vOC

2. Determine the Thévenin equivalent resistance, RT
Suppressing all independent source inside the network
Because vs is suppressed, iB= 0 and iB= 0. Therefore, the output of
the current-control current source is equivalent to an open-circuit.
The only element connected to terminals a and b is the load
resistance RL。
Finally, we draw out the Thévenin equivalent of the original network

Feedback and Thévenin Equivalents
What is the feedback?
Feedback is the connections from a dependent source (output)
back to the control input (input).
Function of the feedback
When there is a feedback in the network, the dependent source
(such as a transistor) can have enormous, and sometimes
surprising effects on the terminal characteristics of a network.
If the feedback weaken the input, called negative feedback, the
output will become smaller.
If the feedback strengthen the input, called positive feedback,
the output will become greater and greater (finally up to
infinite).

Now we discuss the feedback in non-sources (independent)
network with dependent sources. As we know, a two-terminal
network with dependent sources but independent sources is
equivalent to a pure resistance. Therefore, we can use a test
source to determine the characteristic of the network.
Example 1. We connect a test current source to the input port
and determine the input terminal voltage.
According to KCL,
KVL gives the
characteristic of input port
Therefore, the Thévenin
equivalent resistance is

Example 2.
Because the controlling current of the
dependent source can be easily determined
by a test voltage, We connect a test voltage
source to the input port and determine the
input current.
Since resistor Rs and the test voltage source
are in parallel, the controlling current
T
B
S
v
i
R
 
Recall the KCL,
0
T B F B
i i i

  
Then we yield the characteristic
(1 )
( ) (1 ) F
T B F B F B T
S
i i i i v
R

 

      
Therefore, the Thévenin equivalent resistance is
1
S
T
T
T F
R
v
R
i 
 


4.3 Power in Dependent-Sources Networks
An Example and a Dilemma
Let us look at an example and notice the power transfer.
In the example the power received by RL must come from the
dependent source. However, the dependent sources are only
used to model the behavior of devices like transistors, there are
not independent power sources in deed. How is this dilemma to
be resolve

Modeling and Energy Conservation
The key to the dilemma lies in the word "model". A model
network is a useful network only to the extent that it can
represent or model the behavior of a network built out of actual
devices. If the model network tells us that power is being
delivered from the dependent source to the load and that this
power cannot be coming from any existing independent sources,
we must conclude that the device or combination of devices being
modeled by the dependent source must include an energy source
somewhere. Indeed, every electronic device that is modeled with a
dependent source requires some energy source, or power supply,
to operate.

Exercises of this chapter:
E4.1, EE4.2, E4.4, E4.5

Chapter 04.ppt dependent sources for electrical

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