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Notational Conventions
DC Quantity — Upper case-letter, upper-case subscript VBE , ID Small-Signal Quantity — Lower case-letter, lower-case subscript vbe, id Total Quantity — Lower case-letter, upper-case subscript vBE = VBE + vbe, iD = ID + id Phasor Quantity — Upper case-letter, lower-case subscript Vbe, Id
Independent Sources
Figure 1(a) shows the diagram of an independent voltage source. The voltage v is independent of the current i that flows through the source. Fig. 1(b) shows the diagram of an independent current source. The current i is independent of the voltage v across the source.
Figure 1: (a) Independent voltage source. (b) Independent current source.
Dependent Sources
VCVS — Voltage Controlled Voltage Source
Figure 2(a) shows the diagram of a voltage controlled voltage source. The output voltage is given by a voltage gain Av multiplied by an input voltage v 1. Such a source in SPICE is called an E source.
Figure 2: (a) Voltage controlled voltage source. (b) Voltage controlled current source. (c) Current controlled voltage source. (d) Current controlled current source.
VCCS — Voltage Controlled Current Source
Figure 2(b) shows the diagram of a voltage controlled current source. The output current is given by a transconductance Gm multiplied by an input voltage v 1. Such a source in SPICE is called a G source.
CCVS — Current Controlled Voltage Source
Figure 2(c) shows the diagram of a current controlled voltage source. The output voltage is given by a transresistance Rm multiplied by an input current i 1. Such a source in SPICE is called an F source.
CCCS — Current Controlled Current Source
Figure 2(d) shows the diagram of a current controlled current source. The output current is given by a current gain Ai multiplied by an input current i 1. Such a source in SPICE is called an H source.
Passive Elements
Resistor
Figure 3(a) shows the diagram of a resistor. The voltage across it is given by
v = iR
This relation is known as Ohm’s law.
Figure 3: (a) Resistor. (b) Inductor. (c) Capacitor.
Inductor
Fig. 3(b) shows an inductor. The voltage across it is given by
v = L di dt
In the analysis of circuits having sinusoidal excitations, phasor analysis is usually used. In this case, the voltage across the inductor is given by V = LsI
where V and I are phasors, s = jω, and ω is the radian frequency of the excitation. In the phasor domain, a multiplication by s is equivalent to a time derivative in the time domain. This is because the time domain excitation is assumed to be of the form exp (st).
Capacitor
Fig. 3(c) shows a capacitor. The current through it is given by
i = C dv dt
For phasor representation of the signals, the current through the capacitor is given by
I = CsV
Voltage Division and Current Division
Voltage Division
Figure 3(a) shows a two-resistor voltage divider. The voltages v 1 and v 2 are given by
v 1 = iS R 1 =
μ vS R 1 + R 2
R 1 = vS
R 1
R 1 + R 2
v 2 = iS R 2 =
μ vS R 1 + R 2
R 2 = vS
R 2
R 1 + R 2
We substitute the solution for i 1 into the expression for vOC to obtain
vOC = vS
R 2
R 1 + R 2
vS R 1 + (1 + Ai) R 2
= vS
R 2
R 1 + R 2
R 2 +
AiR 1 R 2 R 1 + (1 + Ai) R 2
= vS 20 kΩ 1 kΩ + 20 kΩ
1 kΩ + 50 × 20 kΩ × 1 kΩ 20 kΩ + (1 + 50) 1 kΩ
vS 21
μ 1 +
50 × 20
= 0. 718 vS
The Thévenin equivalent circuit is shown in Fig. 5(b). By superposition, the short-circuit output current is given by
iSC =
vS R 1
where i 1 is given by
i 1 =
vS R 1
We substitute the expression for i 1 into the expression for iSC to obtain
iSC =
vS R 1
vS R 1 = vS
1 + Ai R 1 = vS
1 kΩ = 2. 55 × 10 −^3 vS
The output resistance is given by
Rout =
vOC iSC
- 718 vS
- 55 × 10 −^3 vS
2. 55 × 10 −^3
The Norton equivalent circuit is shown in Fig. 5(c).
Example 2 For the circuit in Fig. 6(a), it is given that R 1 = 3 kΩ, R 2 = 2 kΩ, and Gm = 0.1 S. Solve for the input resistance to the circuit.
Figure 6: (a) Circuit for Example 2. (b) Equivalent input circuit.
Solution. The input resistance is given by the ratio of the source voltage to the source current. By superposition, we can write vS = iS (R 1 + R 2 ) + Gmv 1 R 2
where v 1 is given by v 1 = iS R 1
Substitute the expression for v 1 into the expression for vS to obtain
vS = iS (R 1 + R 2 ) + GmiS R 1 R 2 = iS (R 1 + R 2 + GmR 1 R 2 )
Thus the input resistance is given by
Rin =
vS iS
= R 1 + R 2 + GmR 1 R 2 = 3 kΩ + 2 kΩ + 0. 1 × 3 kΩ × 2 kΩ = 605 kΩ
The equivalent circuit seen looking into the input is shown in Fig. 6(b).
Example 3 Solve for the open circuit output voltage and the short circuit output current for the circuit in Fig. 7(a).
Figure 7: Circuit for Example 3
Solution. The circuit contains a floating current source. To make superposition simpler to apply, this source can be broken into two series sources as shown in Fig. 7(b). The node between the sources is shown connected to ground. Although no current flows from this node to ground when both sources are active, a current does flow when either is zeroed. However, by superposition, the sum of these currents must be zero. Because the currents flow into the ground node, the voltages or current in the circuit are not affected. By superposition, the open circuit output voltage is given by
vOC = (iS + gmv 1 )
R 1
R 1 + R 2 + R 3
R 3 − gmv 1 (R 1 + R 2 ) kR 3
= iS
R 1 R 3
R 1 + R 2 + R 3
− gmv 1
R 2 R 3
R 1 + R 2 + R 3
The voltage v 1 is given by
v 1 = (iS + gmv 1 ) R 1 k (R 2 + R 3 ) − gmv 1
R 3
R 1 + R 2 + R 3
R 1
= iS R 1 k (R 2 + R 3 ) + gmv 1
R 1 R 2
R 1 + R 2 + R 3
Solution for v 1 yields
v 1 = iS
R 1 k (R 2 + R 3 )
1 − gm
R 1 R 2
R 1 + R 2 + R 3
When this is used in the equation for vOC , we obtain
vOC = iS
R 3
R 1 + R 2 + R 3
R^1 −^ gmR^2
R 1 k (R 2 + R 3 )
1 − gm
R 1 R 2
R 1 + R 2 + R 3
By superposition, the short circuit output current is given by
iSC = (iS + gmv 1 )
R 1
R 1 + R 2
− gmv 1 = iS
R 1
R 1 + R 2
− gmv 1
R 2
R 1 + R 2
The voltage v 1 is given by v 1 = (iS + gmv 1 ) R 1 kR 2
Amplifier Representations
Figure 9 shows the diagram of an amplifier. The source is represented by a Thévenin equivalent circuit. In general, the input resistance Rin is a function of the load resistance RL and the output resistance Rout is a function of the source resistance RS.
Figure 9: Amplifier model.
The output circuit of the amplifier can be represented by either a Thévenin equivalent circuit or a Norton equivalent circuit using one of the four dependent sources described above. The four equivalent circuits are summarized below.
VCVS Model
Figure 10(a) shows the amplifier model with the output represented by a voltage-controlled voltage source. The output voltage is given by
vO = Av vI
RL
Rout + RL
= Av
μ vS
Rin RS + Rin
RL
Rout + RL
Figure 10: (a) Voltage controlled voltage source amplifier. (b) Voltage controlled current source amplifier.
VCCS Model
Figure 10(b) shows the amplifier model with the output represented by a voltage-controlled current source. The output voltage is given by
vO = GmvI (RoutkRL) = Gm
μ vS
Rin RS + Rin
(RoutkRL)
CCVS Model
Figure 11(a) shows the amplifier model with the output represented by a current-controlled voltage source. The output voltage is given by
vO = RmiS
RL
Rout + RL
= Rm
μ vS RS + Rin
RL
Rout + RL
Figure 11: (a) Current controlled voltage source amplifier. (b) Current controlled current amplifier.
CCCS Model
Figure 11(b) shows the amplifier model with the output represented by a current-controlled current source. The output voltage is given by
vO = AiiS
RL
Rout + RL
= Ai
μ vS RS + Rin
RL
Rout + RL
