Small–signal analysis
1. Introduction
Let us consider the circuit shown in Fig. 1, where the nonlinear resistor is described by the
equation i = g (v ) having graphical representation shown in Fig. 2.
i
iG
Is
is (t )
G
i(t)
v(t)
0
Fig. 1
v
Fig. 2
I s is a DC current sources whereas is (t ) is a time varying current source. We assume that
absolute value of is (t ) is at all times much smaller than the DC source.
Finding the operating point
We set to zero is (t ) and consider the DC circuit driven by I s acting alone. The DC voltage
and currents in this circuit are denoted using capital letters. Then, it holds
I s − GV = I ,
(1)
I = g (V ) .
(2)
We solve the set of equation (1)-(2) graphically, as shown in Fig. 3, finding the operating
point (VQ , I Q ) .
Now we consider the circuit of Fig. 1 including is (t ) . It is described by equations
i (t ) = I s + is (t ) − Gv(t ) ,
(3)
i (t ) = g (v(t )) .
(4)
For each t the point (v(t ) , i (t )) that satisfies equation (3) lies on a straight line parallel to the
line l shown in Fig. 3. When is (t ) = Acosω t , these straight lines are bounded by the lines k
and m depicted in Fig. 4. The points of intersection with the characteristic i = g (v ) determine
the bounded values of v(t ) and i (t ) . Thus, for all t the points (v(t ) , i (t )) lie on the arc AB of
the characteristic. This arc can be approximated by a linear segment.
1
I
Is
IQ
0
VQ
Is
V
V
l
Fig. 3
i
is(t)
t
A
IQ
B
0
m
VQ
k
v
l
Fig. 4
We present v(t ) and i (t ) as sums of two terms
v(t ) = VQ + v~ (t )
(5)
~
i (t ) = I Q + i (t )
(6)
2
~
where v~ (t ) and i (t ) can be considered as small displacements from the operating point.
Substituting (5) and (6) into (4) yield
~
I Q + i (t ) = g (VQ + v~ (t )) .
(7)
Let us expand the right hand side of equation (7) into the Taylor series and neglect the higher
order terms
dg
~
~
I Q + i (t ) = g (VQ ) +
v (t ) .
dv v=V
Q
(8)
Since I Q = g (VQ ), the equation (8) reduces to
~
i (t ) = G0 ~
v (t ) ,
(9)
dg
.
dv v =V
Q
(10)
where
G0 =
Next we substitute (5) and (6) into equation (3)
~
I Q + i (t ) = I s + is (t ) − G (VQ + v~ (t )) .
(11)
Since I Q = I s − G (VQ ) , we find
~
i (t ) = is (t ) − G~
v (t ) .
(12)
Equations (9) and (12) describe the small signal equivalent circuit shown in Fig. 5.
~
i (t )
is (t )
G
G0
v~ (t )
Fig. 5
~
This circuit is linear and enables us to find the small signals v~ (t ) and i (t ) :
i (t )
~
v (t ) = s
,
G + G0
3
(13)
G0
~
.
i (t ) = is (t )
G + G0
(14)
Example
Consider the circuit shown in Fig. 6, driven by a DC voltage source VS and a small voltage
vs(t). To find the small signal model of this circuit we determine first the DC operating point.
For this purpose we analyse the circuit shown in Fig. 7.
R=1k
i=10 -12(e38v-1)
i(t)
I
R=1k
V S=6V
v(t)
V
V S=6V
vS (t)
Fig. 6
Fig. 7
The circuit shown in Fig. 7 is described by the equation
10 −12 (e 38V − 1) + 0.001V − 0.006 = 0 ,
which can be rewritten in the from
e 38V + 10 9 V − 6 ⋅ 10 9 = 0 .
(15)
We solve this equation using the Newton-Raphson method
V
( j +1)
=V
( j)
−
e 38V
( j)
+ 10 9 V ( j ) − 6 ⋅ 10 9
,
( j)
38e 38V + 10 9
using the initial guess V (0 ) = 0.6 . As a result we obtain: V(1) = 0.591,
V(3) = 0.590. Thus, we assume Vo = 0.590 and find
G0 =
di
dv
v =V0
= 38 ⋅ 10 −12 e 38⋅0.590 = 0.207 S ,
or
R0 = G0−1 = 4.8 Ω .
The small signal model is shown in Fig. 8.
~
i
1000Ω
v~
vs=(t)
Fig. 8
4
4.8 Ω
(16)
V(2) = 0.590,
2. Small signal analysis of bipolar transistor circuits
Let us consider a simple amplifier circuit, containing BJT, shown in Fig. 9.
iC
iB
vCE
R1
R2
vBE
E1
E2
vS (t)
Fig. 9
E1 and E2 are DC voltage source, whereas vs (t ) is a small time varying voltage source.
Finding the operating point.
We set to zero vs (t ) and solve the DC circuit driven by the DC voltage sources E1 and E2
graphically, as illustrated in Figs. 10 and 11.
IB
IC
E2
R2
E1
R1
0
(I B )Q
(IC )Q
(IB )Q
(VBE )Q
E1
VBE
0
Fig. 10
(VCE )Q
E2
VCE
Fig. 11
The operating point is specified by (VBE )Q , (I B )Q , (VCE )Q , (I C )Q .
Generally the operating point is the solution of the set of hybrid equations describing the
transistor
VBE = v̂BE (I B ,VCE )
(17)
I C = îC (I B ,VCE )
(18)
and the equations describing the R1 , E1 and R2 , E2 branches
VBE = E1 − R1I B ,
5
(19)
VCE = E2 − R2 I C .
(20)
Now we consider the circuit shown in Fig. 9 including the small time voltage vs (t ) . In this
circuit voltages and currents are time varying: vBE (t ) , iB (t ) , vCE (t ) , iC (t ) . We present them as
follows:
vBE (t ) = (VBE )Q + v~1 (t ) ,
(21)
~
iB (t ) = (I B )Q + i1 (t ) ,
(22)
vCE (t ) = (VCE )Q + v~2 (t ) ,
(23)
~
iC (t ) = (I C )Q + i2 (t ) ,
(24)
~
~
where v~1 (t ) , i1 (t ) , ~
v2 (t ) , i2 (t ) represent the small displacements from the operating point. We
set these equations into the hybrid equations
(VBE )Q + v~1 (t ) = v̂BE ((I B )Q + ~i1 (t ) , (VCE )Q + v~2 (t )) ,
(25)
(I C )Q + ~i2 (t ) = îC ((I B )Q + ~i1 (t ) , (VCE )Q + v~2 (t )) ,
(26)
and expand the functions on the right hand sides into the Taylor series, about the operating
point, neglecting the higher order terms
(VBE )Q + v~1 (t ) = v̂BE ((I B )Q , (VCE )Q ) + ∂ v̂BE
∂ iB
∂ v̂
~
i1 (t ) + BE v~2 (t )
∂ vCE Q
Q
(I C )Q + ~i2 (t ) = îC ((I B )Q , (VCE )Q ) + ∂ îC
∂ î
~
i1 (t ) + C ~
v2 (t ) .
∂ iB Q
∂ vCE Q
(27)
(28)
Taking into account equations (17) and (18) we obtain
∂ v̂
∂ v̂
~
~
v1 (t ) = BE i1 (t ) + BE ~
v2 (t ) ,
∂ iB Q
∂ vCE Q
(29)
∂ î
∂ î
~
~
i2 (t ) = C i1 (t ) + C v~2 (t ) .
∂ iB Q
∂ vCE Q
(30)
The set of equations can be rewritten in the form
~
~
v1 (t ) = h11 i1 (t ) + h12 v~2 (t ) ,
(31)
~
~
i2 (t ) = h21 i1 (t ) + h22 v~2 (t ) ,
(32)
6
∂ v̂BE
∂ î
∂ î
∂ v̂
, h12 = BE , h21 = C , h22 = C
.
∂ iB Q
∂ iB Q
∂ vCE Q
∂ vCE Q
The circuit described by equation (31)-(32), called the small signal model of the transistor, is
shown in Fig. 12.
~
~
i2
i1
where h11 =
R = h11
~
h21 i1
v~1
G = h22
v~2
h12 v~2
Fig. 12
Typical values of the coefficients are as follows: h11 = 103 Ω , h12 = 2 ⋅ 10−4 , h21 = 50 ,
h22 = 10−5 S .
Now we consider the resistor-source branches of the circuit shown in Fig. 9 and write the
equations
vBE = E1 + vs (t ) − R1iB ,
(33)
vCE = E2 − R2iC .
Substituting (21)-(24) and taking into account (19) and (20) we obtain
(34)
~
~
v1 (t ) = vs (t ) − R1 i1 (t ) ,
(35)
~
v~2 (t ) = − R2 i2 (t ) .
(36)
Combining the equations (35)-(36) with the equations (31)-(32) describing the small signal
model of the transistor we find the small signal equivalent circuit of the amplifier circuit,
shown in Fig. 13.
~
i2 (t )
~
i1 (t )
R1
vs (t )
~
h21 i1 (t )
h11
v~1 (t )
h22
~
v2 (t )
R2
h12 ~
v2 (t )
Fig. 13
We find the voltage gain of this circuit defined as v~2 vs . For this purpose we write the
equations (29)-(30) and (35)-(36) describing the circuit and rearrange them as shown
underneath:
7
v~ (t )
~
~
i2 (t ) = − 2 = h21 i1 (t ) + h22 ~
v2 (t ) ,
R2
~
v2 =
− h21 ~
i (t ) ,
1 1
h22 +
R2
~
~
vs (t ) − R1 i1 (t ) = h11 i1 (t ) + h12
(− h21 ) ~i (t ) ,
1 1
h22 +
R2

1 
 h22 + vs (t )
R2 
~

i1 (t ) =
,

1 
(h11 + R1 ) h22 +  − h12 h21
R2 

v~2 (t )
=
vs (t )
− h21


(h11 + R1 ) h22 + 1  − h12h21
R2 

.
(37)
Substituting in the voltage gain (37): R1 = R2 = 1kΩ h11 = 1kΩ , h12 = 2 ⋅ 10−4 , h21 = 50 ,
h22 = 10−5 S we find
v~2 (t )
= −24.8 .
vs (t )
3. Small signal analysis – a general case
Let us consider a circuit containing linear and nonlinear resistors, inductors, capacitors,
linear controlled sources, driven by a DC source and a small time varying source. To form the
small signal equivalent circuit we find first the DC operating point. For this purpose we set to
zero the small time varying source and consider the circuit driven by the DC source only. As a
result we obtain a DC circuit, by short-circuiting all the inductors and open-circuit all the
capacitors. It can be solved using an arbitrary method for the analysis of resistive circuits. If
the circuit is nonlinear we apply the Newton-Raphson method. The operating point Q is
specified by branch current (I j )Q and branch voltages (V j )Q .
In the circuit driven by the DC source and the small time varying source all branch
voltages and currents depend on time. We present each branch voltage as
v(t ) = VQ + v~ (t )
and each branch current as
~
i (t ) = I Q + i (t )
8
~
v (t ) and i (t ) are small displacements from the operating point. Since both voltages
where ~
v j (t ) and (V j )Q satisfy KVL in an arbitrary loop we state that voltages ~
v j (t ) also satisfy KVL
in an arbitrary loop.
~
Similarly we prove that the currents ij (t ) satisfy KCL at all nodes.
Now we consider circuit elements and describe them in terms of small time-varying
signals.
Nonlinear resistor described by the equation
i (t ) = g (v(t )) .
(38)
~
We substitut i (t ) = I Q + i (t ) , v(t ) = VQ + v~ (t ) and expand the function g (v(t )) in the Taylor
series, about the operating point, neglecting the higher order terms
dg
~
~
I Q + i (t ) = g (VQ ) +
v (t ) .
d v v=V
Q
(39)
Since in the DC circuit I Q = g (VQ ) we obtain
~
i (t ) = Gv~ (t ) ,
(40)
dg
.
d v v =V
Q
(41)
where
G=
Equation (40) describes the small signal model of the nonlinear resistor, being a linear resistor
having conductance G specified by (41). If the resistor is linear its small signal model is the
same resistor.
Similarly we prove that the small signal model of any linear controlled source is the same
controlled source.
Nonlinear capacitor described by the equation
q(t ) = c(v(t )) .
(42)
We substitute v(t ) = VQ + v~ (t ) , expand the function c(v(t )) into the Taylor series about the
operating point and neglect the higher order terms
q (t ) = c(Vs ) +
Since
i (t ) =
dc
~
v (t ) .
d v v=V
Q
d q(t ) ~
= i (t )
dt
9
(43)
(44)
then
d v~ (t )
~
i (t ) = C
,
dt
(45)
dc
.
d v v =V
Q
(46)
where
C=
Equations (45) and (46) describe a linear capacitor being the small signal model of the
nonlinear capacitor.
If the capacitor is linear its small signal model is the same capacitor.
Nonlinear inductor described by the equation
φ (t ) = f (i (t )) .
(47)
Similarly as in the case of capacitor we state that the small signal model of the nonlinear
inductor is a linear inductor specified by the equation
~
d i (t )
~
v (t ) = L
,
dt
(48)
df
.
d i i=I
Q
(49)
where
L=
If the inductor is linear, its small signal model remains the same inductor.
To construct the small signal equivalent circuit of a given circuit we set to zero DC source
and replace all its elements by their small signal models.
Example
For the circuit shown in Fig. 14 construct the small signal equivalent circuit about its
operating point.
Data: q5 = v5 + 0.1v53 , q7 = 0.08v73 , φ 4 = 0.5e0.8i4 .
L2=1H
i1
i5
R1=1Ω
vs (t )
R3=2Ω
v5
i3
i4
v4
i6 = 0.5i1
i7
v7
Is = 1A
Fig. 14
Finding the operating point.
We set to zero vs (t ) , remove the capacitors and short circuit the inductor (see Fig. 15).
10
I1
I4
R3
R1
V5
E
V7
I 6 = 0.5I1
Is = 1A
Fig. 15
The circuit can be described by the node equation
E
E
+ 0.5 − 1 = 0
1
1
having the solution E =
2
V.
3
Hence, it holds:
1
E 2
I1 = = A , I 6 = 0.5 I1 = A , I 4 = I s = 1A ,
1 3
3
The small signal models of the nonlinear elements:
d v~ (t )
~
i5 (t ) = C5 5 ,
dt
C5 =
d v~ (t )
~
i7 (t ) = C7 7 ,
dt
C7 =
~
d i4 (t )
~
(
)
v4 t = L4
,
dt
d q5
d v5
V5 = E =
2
V,
3
8
V7 = E + R3 I 4 = V .
3
= 1 + 0.3V52 = 1.133 F ,
V5
d q7
d v7
= 0.24V72 = 1.707 F ,
V7
L4 = 0.4e0.8 = 0.89 H .
Using the above results we construct the small signal equivalent circuit shown in Fig. 16.
~
i1 (t )
1Ω
1H
~
i5 (t )
2Ω
1.333F
~
~
i6 (t ) = 0.5 i1 (t )
vs (t )
Fig. 16
11
0.89H
~
i4 (t )
1.707F
~
v7 (t )