Dependent Sources and Operational Amplifiers
Samantha R. Summerson
18 September, 2009
1
Dependent Sources
A dependent source is similar to a regular (independent) source except that the voltage or current is a
function of other voltages or currents in the circuit. A dependent course is also drawn using a diamond
rather than a circle.
i
+
vin
±
v
−
Figure 1: Circuit with a dependent voltage source.
In Fig.1, vin may be defined in terms of a voltage or current that appears elsewhere in the circuit. For
example,
vin = 2v.
In this case, the dependent source is a voltage-controlled voltage source because it is defined in terms of a
voltage across another circuit element. The gain parameter, 2 in this example, is unit-less. Alternatively, we
could also have
vin = 3i.
This assignment would mean that the dependent source is a current-controlled voltage source since it is
defined in terms of a current through another element. The gain paramenter, 3 for the example, has units
of volts/amps; this ensures that the units for the dependent voltage source work out.
i
+
iin
↑
v
−
Figure 2: Circuit with a dependent current source.
1
In Fig.2, a dependent current source is shown. Note that we use the diamond shape for both dependent
voltage and current sources. The current iin may be defined in terms of both currents or voltages that
appear in the circuit. For example,
iin = −2i.
In this case the dependent source is a current-controlled current source. The gain parameter in this case is
unit-less. Supposing we define the input current in terms of a voltage, e.g.
iin = 3v,
the source is now a voltage-controlled current source and the gain parameter has units amps/volts.
In total, there are four types of dependent sources. These sources are useful for modeling transistors,
amplifiers, generators, and other active circuits. Circuits with dependent sources can still be solved using
KCL, KVL, and the Node Method.
2
Operational Amplifers
An operational amplifier, or op-amp, is a circuit element which is actually itself an active circuit; specifically,
it is a type of integrated circuit.
a
+
b
−
Figure 3: Operational Amplifier.
Rout
Rin
± G(ea − eb )
Figure 4: Equivalent circuit for the operational amplifier.
There are two inputs to the op-amp, labeled a and b in Fig.3. Terminal a is called the non-inverting
input terminal and terminal b is the inverting input terminal. The differential signal is the difference in
node voltages at the terminals,
Vab = ea − eb .
The output signal is a scaled version of the differential; the scalar G is the voltage gain. In Fig.4, we show an
equivalent model for the op-amp which uses a voltage-controlled voltage source. In general, Rin has a very
2
large impedance; in an ideal op-amp, the impedance is infinite. Rout is very small and, ideally, zero (short
circuit). Also, the gain parameter G is also very large. One important note, however, is that it is impossible
for electronic voltages to yield voltages higher than that provided by the power supply.
Op-amps are often used with negative feedback, meaning the output signal is returned to the inverting
input terminal. A common op-amp is the inverting amplifier.
RF
R
iin
+
vin
±
−
+
v
+
RL
−
vout
−
Figure 5: Inverting Amplifier (note the negative feedback).
To solve a circuit with an op-amp, we take the following steps:
1. Since we assumed Rin was very large, it must be that iin is very small. Set iin = 0.
2. No current flow through the resistor, Rin , means the voltages across the input terminals to the op-amp
is zero. Set v = 0.
3. Now solve the circuit using standard techniques.
Example 1. Solve for vout for the circuit in Fig.5.
We begin by setting iin = 0 and vin = 0. Also, we identify the nodes so that we can use the node method
to solve the circuit.
RF
ea
R
eb
0
+
vin
±
ec
+
−
+
RL
0
−
vout
−
Figure 6: Example 1: identifying nodes and setting values for op-amp.
Since the bottom node is the ground node, we automatically can write:
ea
= Vin ,
eb
=
ec
= Vout .
3
0,
Writing the KCL equation for node b we have
eb − ea
eb − ec
+
+ 0 = 0.
R
RF
The last zero on the left-hand side of the equation represents iin = 0. Now we plug in the node voltage
values and solve.
−Vout
−Vin
+
R
RF
Vout
⇒
RF
=
=
⇒ Vout
=
0,
Vin
,
R
RF
−
Vin .
R
We note that the gain is entirely determined by the input resistor, R, and the feedback resistor, RF . The
output voltage is independent of the load resistance, meaning the ouput acts as an ideal voltage source (i.e.
puts out the same voltage for all values of RL ).
RF
Ra
iin
Rb
Va ±
vb
+
±
−
+
v
+
RL
−
vout
−
Figure 7: Summing amplifier.
Example 2. Solve for Vout in the summing amplifier circuit.
We begin again by identify the nodes and setting iin = 0 and v = 0.
RF
Ra
ea
eb
Va
0
ec
Rb
+
ed
+
−
+
±
vb
±
0
RL
−
vout
−
Figure 8: Example 2: Summing amplifier.
Immediately we see that
ea
= Va ,
eb
= Vb ,
ec
=
4
0, ed
=
Vout .
Using KCL at node c, we have
ec − ea
ec − eb
ec − ed
+
+0+
= 0.
Ra
Rb
Rf
Now we plug in the node voltage values and solve.
−Vb
−Vout
−Va
+
+
Ra
Rb
Rf
Vout
⇒
Rf
=
=
⇒ Vout
=
0,
−Va
−Vb
+
,
Ra
Rb
Rf
Rf
−
Va −
Vb .
Ra
Rb
We see that our answer here is similar to the inverting amplifier, except we have a sum of two terms since
there was two input voltages. In light of superposition, this should be intuitive. If we take away one of the
sources, we have a normal inverting amplifier.
5