Common Drain Amplifier or Source Follower
Figure 1(a) shows the source follower with ideal current source load. Figure 1(b) shows the ideal current source implemented by NMOS with constant gate to source voltage.
Vi Vo VDD M1 (a) VG VDD Vo Vi M1 M2 (b) VT0+∆V ∆V VTN+∆V +
-Figure 1. Common drain or source follower implementation.
1. Low Frequency Small Signal Equivalent Circuit
)
0
or Z
(
Y
),
r
or Z
(
g
Y
V
-V
V
-V
V
,
V
-
V
,
V
V
,
V
V
S S ds2 L ds2 L 2 1 o i gs1 o bs1 o 2 i 1=
∞
=
=
=
=
=
=
=
=
From Figure 2(e),
2 ds1 mb1 m1 1 m1 2 ds1 mb1 2 1 m1 o ds1 mb1 gs1 m1 2 1
)V
g
g
g
(
V
-g
)V
g
g
(
)
V
-
(V
g
)V
g
g
(
v
g
I
0
I
+
+
+
=
+
+
−
=
+
+
−
=
=
gds1 gds2 D2 S2 gm1vgs1 D1 S1 G1vgs1 gmb1 + -Vo + -Vi gds1 gds2 D2 S2 gm1vgs1 Vo Vi D1 S1 G1 vgs1 gmb1vbs1 + -+ -Y + V1 + V2 YL Zi Zo I1 I2 gds1 gds2 D2 S2 gm1vgs1 Vo Vi D1 S1 G1 vgs1 gmb1vo + -+ -gds1 gds2 D2 S2 gm1vgs1 Vo Vi D1 S1 G1 vgs1 (c) + -+ -gmb1 gds1 gds2 D2 S2 gm1vgs1 D1 S1 G1 vgs1 + gmb1 Vi + - -Vo (a) (b) (d) (e) (f)
The corresponding Y-parameter matrix is,
0
detY
g
g
g
g
0
0
Y
ds1 mb1 m1 m1=
+
+
−
=
The Source Follower Properties:
∞
=
+
+
+
+
=
+
+
=
ds2 ds2 ds1 mb1 m1 L 11 L 22 i(0)g
0
g
)
g
g
g
(
Y
y
detY
Y
y
Z
ds1 mb1 m1 22 S 22 S 11 og
g
g
1
y
1
Y
y
detY
Y
y
Z
+
+
=
=
+
+
=
1
g
g
g
g
g
g
)
g
g
(g
)
g
(
Y
y
y
A
ds2 ds1 mb1 m1 m1 ds2 ds1 mb1 m1 m1 L 22 21 V0+
+
+
=
+
+
+
≈
−
−
=
+
−
=
Or)
Z
//
Z
(
g
A
V=
m1 o L∞
=
+
−
=
+
−
=
ds2 ds2 m1 L 11 L 21 I0)g
(
0
g
g
Y
y
detY
Y
y
A
2. High Frequency Small Signal Equivalent Circuit
VG
VDD
Vo
Vi
M2
M1
CL
Cgd2
Cgs1
Cdb2
Csb1
Cdb1
Cgd1
Figure 3. Source follower parasitic capacitances.
Figure 3 shows all the parasitic capacitances of source follower circuit. Figure 4 shows the high frequency small signal equivalent circuit. From Figure 4, one obtains,
L gd2 db2 sb1 O ds2 ds1 mb1 O b 1 a 2 b 2 b 1 gs1 o b 2 b 1 i a 2 S a 1
C
C
C
C
C
,
g
g
g
G
-I
I
,
V
-V
v
,
V
V
,
V
V
V
,
V
V
+
+
+
=
+
+
=
=
=
=
=
=
=
The network a current equation is:
)
V
-V
)(
C
G
(
I
)
V
-V
)(
C
G
(
I
a 1 a 2 S S a 2 a 2 a 1 S S a 1s
s
+
=
+
=
GS VS Cgs1 Cgd1 gds1 Co G1 S1 D1 D2 S2 Co=Csb1+Cdb2+Cgd2+CL gm1 v gs1 gmb1 + Vi -(c) + Vo -+ vgs1 -CS gds2 Ya Yb Zib VS I1a V1a V2a = V1b + + V2b I2a I1b I2b
(d) gds1 D2 S2 GS Cgd1 Cgs1 Csb1 Cdb2 CL gm1 v gs1 G1 D1 S1 (a) Cgd2 gmb 1 Vo Vo + -vgs1 gds2 (a) CS IS GS Cgs1 Cgd1 gds1 Co G1 S1 D1 D2 S2 Co=Csb1+Cdb2+Cgd2+CL gm1 v gs 1 gmb1 + Vi -(b) + Vo -+ vgs1 -CS gds2 IS + +
The corresponding Y-parameter matrix is :
+
+
−
+
−
+
=
)
C
G
(
)
C
G
(
)
C
G
(
)
C
G
(
Y
S S S S S S S S as
s
s
s
The network b current equation is:
b 2 o gs1 o m1 b 1 gs1 m1 b 2 o o b 1 b 2 gs1 b 2 b 1 m1 b 2 o o b 1 b 2 gs1 gs1 m1 b 2 b 2 gs1 b 1 gs1 gd1 b 2 b 1 gs1 b 1 gd1 b 1
)]V
C
C
(
G
g
[
V
)
C
-(g
)V
C
(G
)
V
-(V
C
)
V
-(V
g
)V
C
(G
)
V
-(V
C
V
g
I
V
C
)V
C
(C
)
V
-(V
C
V
C
I
+
+
+
+
+
=
+
+
+
−
=
+
+
+
−
=
−
+
=
+
=
s
s
s
s
s
s
s
s
s
s
The corresponding Y-parameter matrix is:
)
C
(G
C
)]
C
(C
G
[g
C
)
C
(g
C
-)]
C
C
(
G
)[g
C
(C
detY
)
C
C
(
G
g
)
C
(g
-C
-)
C
(C
Y
o o gs1 o gs1 o m1 gd1 gs1 m1 gs1 o gs1 o m1 gs1 gd1 b o gs1 o m1 gs1 m1 gs1 gs1 gd1 bs
s
s
s
s
s
s
s
s
s
s
s
+
+
+
+
+
=
+
+
+
+
+
=
+
+
+
+
+
=
The gain of network b , the unloaded last stage, is:
0
Y
;
)
C
(C
G
g
C
g
y
y
Y
y
y
V
V
A
b L o gs1 o m1 gs1 m1 b 22 b 21 b L b 22 b 21 b 1 b 2 b V+
+
+
=
+
=
−
=
+
−
=
=
s
s
The input impedance of network b (or the load of network a) is given by:
)
C
(G
C
)]
C
(C
G
g
[
C
)
C
(C
G
g
detY
y
Y
y
detY
Y
y
Z
o o gs1 o gs1 o m1 gd1 o gs1 o m1 b b 22 b L b 11 b b L b 22 b is
s
s
s
s
+
+
+
+
+
+
+
+
=
=
+
+
=
The voltage gain of network a is given by:
)
C
(G
C
)]
C
(C
G
g
[
C
)]
C
(C
G
g
)[
C
(G
)]
C
(C
G
g
)[
C
(G
)
C
(C
G
g
)
C
(G
C
)]
C
(C
G
g
[
C
C
G
C
G
Z
1
Y
;
Y
y
y
V
V
A
o o gs1 o gs1 o m1 gd1 o gs1 o m1 S S o gs1 o m1 S S o gs1 o m1 o o gs1 o gs1 o m1 gd1 S S S S b i a L a L a 22 a 21 a 1 a 2 a Vs
s
s
s
s
s
s
s
s
s
s
s
s
s
s
+
+
+
+
+
+
+
+
+
+
+
+
+
+
=
+
+
+
+
+
+
+
+
+
+
+
=
=
+
−
=
=
The overall gain is given by:
]
C
C
)
C
(C
)
C
C
[(
]
G
C
)
C
(C
G
)
G
g
)(
C
C
[(
)
G
g
(
G
)
C
(G
)
C
g
(
)
C
(G
C
)]
C
(C
G
g
)][
C
C
(
G
[
)
C
(G
)
C
g
(
)
C
(G
C
)]
C
(C
G
g
[
C
)]
C
(C
G
g
)[
C
(G
)]
C
(C
G
g
)[
C
(G
)
C
(C
G
g
C
g
V
V
since
;
A
A
V
V
V
V
V
V
V
V
A
o gs1 o gs1 gd1 S 2 o gs1 o gs1 S o m1 gd1 S o m1 S S S gs1 m1 o o gs1 o gs1 o m1 gd1 S S S S gs1 m1 o o gs1 o gs1 o m1 gd1 o gs1 o m1 S S o gs1 o m1 S S o gs1 o m1 gs1 m1 a 2 b 1 a V b V a 1 a 2 b 1 b 2 a 1 b 2 S o V+
+
+
+
+
+
+
+
+
+
+
+
+
=
+
+
+
+
+
+
+
+
+
=
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
=
=
=
=
=
=
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
The trans-impedance of the overall network is:
]
C
C
)
C
(C
)
C
C
[(
]
G
C
)
C
(C
G
)
G
g
)(
C
C
[(
)
G
g
(
G
C
g
)
C
(G
A
)
C
(G
V
V
I
V
A
o gs1 o gs1 gd1 S 2 o gs1 o gs1 S o m1 gd1 S o m1 S gs1 m1 S S V S S S o S o Z+
+
+
+
+
+
+
+
+
+
+
+
=
+
=
+
=
=
s
s
s
s
s
The DC gain is obtained by setting s=0.
)
G
(g
G
g
A
o m1 S m1 Z0=
+
The transimpedance can be re-written as follows:
)
G
(g
G
C
C
)
C
(C
)
C
C
(
a
)
G
(g
G
G
C
)
C
(C
G
)
G
g
)(
C
C
(
b
a
b
1
)
g
C
1
(
A
A
O m1 S o gs1 o gs1 gd1 S O m1 S o gs1 o gs1 S o m1 gd1 S 2 m1 gs1 Z0 Z+
+
+
+
=
+
+
+
+
+
+
=
+
+
+
=
s
s
s
If the poles are far apart, their values can be estimated as follows : gs1 m1 O gs1 O gs1 gd1 S O gs1 O gs1 S O m1 gd1 S 2 O gs1 O gs1 S O m1 gd1 S O m1 S 1
C
g
z
C
C
)
C
)(C
C
C
(
G
C
)
C
(C
G
)
G
)(g
C
(C
a
b
p
G
C
)
C
(C
G
)
G
)(g
C
(C
)
G
(g
G
b
1
p
−
=
+
+
+
+
+
+
+
+
−
=
−
=
+
+
+
+
+
+
−
=
−
=
The bandwidth and gain bandwidth product are defined by the dominant pole p1. That is,
π
π
2
w
f
w
A
w
2
w
f
p
w
GBW GBW BW Z0 GBW BW BW 1 BW=
=
=
=
The phase margin PM for non-inverting amplifier is the distance between the phase angle at the unity gain bandwidth frequency with respect to –180. For the source follower transfer function this is calculated as follows:
−
+
=
+
−
=
−
−
+
=
+
∠
−
∠
−
+
∠
+
∠
=
∠
>>
+
+
≈
+
+
+
=
+
+
+
=
2 GBW 1 -GBW 1 -2 GBW 1 -GBW 1 -2 GBW 1 GBW GBW Z0 GBW Z 1 GBW 2 GBW 1 GBW GBW Z0 2 GBW 1 GBW GBW Z0 GBW Z 2 1 Z0 Zp
w
tan
z
w
tan
90
PM
PM
180
p
w
tan
90
z
w
tan
0
)
p
jw
1
(
)
p
jw
(
)
z
jw
1
(
A
)
jw
(
A
p
w
since
;
)
p
jw
1
)(
p
jw
(
)
z
jw
1
(
A
)
p
jw
1
)(
p
jw
(1
)
z
jw
1
(
A
)
jw
(
A
)
p
1
)(
p
(1
)
z
1
(
A
)
(
A
s
s
s
s
3. Source Follower as DC Level Shifter
We assume the source follower is operating at saturation mode. That is the drain current is given by:
)
V
V
(
V
;
)
V
V
)(
2
/
(
I
2 DS GS T T GS DS=
β
−
>
−
where: β =K(W / L) Therefore,β
DS T GSI
2
V
V
−
=
orβ
DS T GSI
2
V
V
=
+
β
DS T o in GSI
2
V
V
V
V
=
−
=
+
If VBS=0, this is not possible for NMOS transistor in Nwell process.
β
DS TO o in GSI
2
V
V
V
V
=
−
=
+
This difference between the voltage level of the input and output can be specified by controlling W/L ratio for a given K and IDS.
In Figure 1(b), VBS=-Vo. Hence,
β
φ
φ
γ
DS o TO o in GSI
2
)
V
(
V
V
V
V
=
−
=
+
+
−
+
Note Vo appears on both sides of the equation, need iteration to determine the solution.
Common Drain Amplifier or Source Follower Experiments
4. Source Follower as DC Level Shifter
Source follower is a voltage follower, its gain is less than 1. The DC transfer characteristic has a slope of less than 1. First let us determine the maximum output voltage. For source follower this occurs
when the input voltage Vin is at maximum or equal to VDD.
0
26
.
3
V
6
.
0
V
)]
6
.
0
V
6
.
0
(
1
1
[
5
V
)]
V
(
[V
-V
V
-V
V
O(max) O(max) O(max) O(max) O(max) T0 DD T DD O(max)=
−
+
+
−
+
+
−
=
−
+
+
=
=
γ
φ
φ
This a non-linear equation, its solution requires iteration. A MATLAB m file is created to solve this problem. A call to MATLAB fzero function is invoke to obtain the solution as shown below: *MATLAB m file “srcf.m” stored in \MATLAB directory
function y=srcf(vo) y=vo+sqrt(0.6+vo)-3.26
*MATLAB “fzero” function invocation vo=fzero(‘srcf’,1)
vo=1.7327
Pspice Vo(max)=1.7595
One can estimate the required bias of M1, using the biasing principle. The transistor M2 is a current sink biased to generate 100uA with a fixed gate to source voltage of VGS=VT0+∆V=1+1.5=2.5. That is the
output voltage Vo=∆V=1.5. Since M1 and M2 are connected in series, means their current are the same. With both transistor have the same W/L means that the gate to source must be equal to VT+∆V. But VT for
transistor M1 is equal to VTN accounting for the non-zero bulk bias.
4.6745
1.5
1.6745
1.5
V
V
V
V
6745
.
1
)
6
.
0
)
5
.
1
(
6
.
0
(
1
1
)
V
(
V
V
-1.5
V
V
TN O bias BS T0 TN O BS=
+
+
=
∆
+
+
=
=
−
−
−
+
=
−
−
+
=
=
−
=
φ
φ
γ
Due to gradual slope of the voltage DC transfer characteristic, it is difficult to obtain the proper operating point from Pspice simulation. To help on the bias determination the current DC transfer characteristic is also plotted for transistor M1. We know the bias current was designed for 100 uA. From Pspice simulation DC transfer characteristic, the bias voltage is 4.75V at IDSQ=101.205uA. This is very closed to the
theorically calculated value.
Pspice netlist at this operating is simulated to obtain the small signal characteristics. The theoretical small signal parameters are determined and compared with Pspice simulation results.
73
.
0
6
-E
)
9642
.
1
9642
.
1
18
.
45
95
.
130
(
6
-E
95
.
130
g
g
g
g
g
A
Z
K
6
.
5
6
-E
)
9642
.
1
18
.
45
95
.
130
(
1
g
g
g
1
Z
6
-E
9642
.
1
r
1
g
M
509
.
6)
-E
21
.
98
)(
02
(.
1
I
1
r
6
-E
9642
.
1
r
1
g
M
509
.
6)
-E
21
.
98
)(
02
(.
1
I
1
r
45.18umho
6)
-E
95
.
130
(
)
5
.
1
(
6
.
0
2
1
g
V
2
g
umho
95
.
130
6)
-6)(98.21E
-E
3
.
87
(
2
I
2
g
g
uA
21
.
98
)
1
5
.
2
(
2
6
-E
27
.
87
)
V
-V
(
2
I
I
6
-E
3
.
87
6
-1)E
-(5.4
6
-9.6E
6)
-E
40
(
L
W
K
ds2 ds1 mb1 m1 m1 V0 i ds1 mb1 m1 O ds2 ds2 DSQ P ds2 ds1 ds1 DSQ N ds1 m1 BS mb1 DSQ N mN m1 2 2 T0 GSN N 2 DSQ N N N=
+
+
+
=
+
+
+
=
∞
=
=
+
+
=
+
+
=
=
=
=
=
=
=
=
=
=
=
=
−
−
=
−
=
=
=
=
=
=
−
=
=
=
=
=
=
λ
λ
φ
γ
β
β
β
Comparing with Pspice results of: Zi=1 E+20 =∞
Zo=5.323K
AV0=.7315
*PSpice file for NMOS Inverter with PMOS Current Load *Filename="Lab3.cir" VIN 1 0 DC 4.75VOLT AC 1V VDD 3 0 DC 5VOLT VSS 4 0 DC 0VOLT VG2 5 0 DC 2.5VOLT M1 3 1 2 4 MN W=9.6U L=5.4U M2 2 5 4 4 MN W=9.6U L=5.4U .MODEL MN NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6
+ TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10
+ U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL MP PMOS VTO=-1 KP=15U
+ GAMMA=0.6 LAMBDA=0.02 PHI=0.6
+ TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10
+ U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 *Analysis
.TF V(2) VIN
.AC DEC 100 1HZ 10GHZ .PROBE
.END
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.7500 ( 2) 1.5757 ( 3) 5.0000 ( 4) 0.0000
( 5) 2.5000
**** SMALL-SIGNAL CHARACTERISTICS V(2)/VIN = 7.315E-01
INPUT RESISTANCE AT VIN = 1.000E+20 OUTPUT RESISTANCE AT V(2) = 5.323E+03
5. Current Source Driven Source Follower As Transimpedance
Amplifier
The voltage gain transfer function has two-zero at the LHPand two-pole at the LHP. That is phase plot is expected cancel each other. Hence the dynamic range of the phase angle is small, hence large phase margin. The gain plot is below 0 db. That is the determination of fBW and fGBW is not possible. Source
follower does not have a voltage gain but has a transimpedance gain or voltage to current gain. The source follower is re-simulated using a current source input. The gain is highly dependent on the input impedance Rs. The biasing current is determined to established the same bias as in the voltage source. That is the bias current and input impedance Rs must be selected to produce a bias of 4.75V. This can be achieved by a current bias of 100uA and Rs of .0475Meg. The Pspice netlist with this bias is shown below:
*PSpice file for NMOS Inverter with PMOS Current Load *Filename="Lab3a.cir" IIN 0 1 DC 100u AC 1 VDD 3 0 DC 5VOLT VSS 4 0 DC 0VOLT VG2 5 0 DC 2.5VOLT RS 1 0 .0475Meg M1 3 1 2 4 MN W=9.6U L=5.4U M2 2 5 4 4 MN W=9.6U L=5.4U .MODEL MN NMOS VTO=1 KP=40U + GAMMA=1.0 LAMBDA=0.02 PHI=0.6
+ TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10
+ U0=550 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 .MODEL MP PMOS VTO=-1 KP=15U
+ GAMMA=0.6 LAMBDA=0.02 PHI=0.6
+ TOX=0.05U LD=0.5U CJ=5E-4 CJSW=10E-10
+ U0=200 MJ=0.5 MJSW=0.5 CGSO=0.4E-9 CGDO=0.4E-9 *Analysis
*.DC VIN 0 5 0.05 .TF V(2) IIN
.AC DEC 100 1HZ 100THZ .PROBE
.END
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 4.7500 ( 2) 1.5757 ( 3) 5.0000 ( 4) 0.0000
( 5) 2.5000
**** SMALL-SIGNAL CHARACTERISTICS V(2)/IIN = 3.475E+04
INPUT RESISTANCE AT IIN = 4.750E+04 OUTPUT RESISTANCE AT V(2) = 5.323E+03
6. Current Driven Source Follower High Frequency Model
Experiments
The parasitic capacitances will be determined to check the theory against Pspice simulation results. These capacitances will be determined at the operating point. The reverse biases are first calculated, using the node voltages at the operating point from Pspice simulation.
For M1, VDB=V(4)-V(3)=0-5=-5 VBS=V(4)-V(2)=0-1.5757 For M2, VDB=V(4)-V(2)=0-1.5757=-1.5757 VBS=0
The MATLAB program is invoked to obtain the parasitic capacitances. For M1, [CGS,CGD,CBD,CBS]=CAP(9.6,5.4,-5,-1.5757) CGS=23.2704, CGD=3.84, CBD=15.6331, CBS=39.1607 For M2, [CGS,CGD,CBD,CBS]=cap(9.6,5.4,-1.5757,0) CGS=23.2704, CGD=3.84, CBD=25.0807, CBS=61.84 The small signal low frequency gain is
90.6db
or
K
5
.
34
6
-E
)
1084
.
49
95
.
130
(
6)
-E
E6)(130.95
0475
(.
)
G
(g
g
R
)
G
(g
G
g
A
6
-49.1084E
6
-E
)
9642
.
1
9642
.
1
18
.
45
(
g
g
g
G
m1 S m1 Z0 ds2 ds1 mb1 O=
+
=
+
=
+
=
=
+
+
=
+
+
=
Pspice simulation result is Azo =90.818 db.
In this experiment, only Cgs and Cgd will be included in the simulation. Also, Cs=CL=0.
fF
84
.
3
C
C
C
C
C
C
O=
sb2+
db2+
gd2+
L=
gd2=
90.007
89.987
-89.994
90
12.43E9
54.372E12
tan
5.63E9
54.372E12
tan
90
p
w
tan
z
w
tan
90
PM
T
65
.
8
2
w
f
54.372T
54.372E12
9)
E3)(1.576E
5
.
34
(
p
A
w
M
250
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576
.
1
2
p
f
G
896
.
0
2
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63
.
5
2
z
f
E9
63
.
5
15
-23.2704E
6
-E
95
.
130
C
g
z
12.43E9
30
-3.84)E
(23.2704)(
30
-E
)
84
.
3
2704
.
23
)(
84
.
3
0
(
21
-1142.77E
21
E
67
.
570
21
-E
42
.
691
C
C
)
C
)(C
C
C
(
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C
)
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(C
G
)
G
)(g
C
(C
p
G
576
.
1
21
-1142.77E
21
E
67
.
570
21
-E
42
.
691
6
-49.1084)E
6)(130.95
-(21.05E
G
C
)
C
(C
G
)
G
)(g
C
(C
)
G
(g
G
p
21
-1142.77E
6)
-4E
15)(49.108
-E
(23.2704
G
C
21
-570.67E
15)
-3.84)E
4
6)((23.270
-(21.05E
)
C
(C
G
21
-691.42E
6)
-E
)
1084
.
49
95
.
130
)((
15
E
)
84
.
3
0
((
)
G
)(g
C
(C
1 1 -2 GBW 1 GBW 1 -GBW GBW 1 Z0 GBW 1 BW Z gs1 m1 O gs1 O gs1 gd1 S O gs1 O gs1 S O m1 gd1 S 2 O gs1 O gs1 S O m1 gd1 S O m1 S 1 O gs1 O gs1 S O m1 gd1 S=
+
=
−
+
=
−
+
=
=
=
=
=
=
=
=
=
=
=
=
=
=
−
=
−
=
=
+
+
+
+
−
+
=
+
+
+
+
+
+
+
+
−
=
=
+
−
+
+
=
+
+
+
+
+
+
−
=
=
=
=
+
=
+
=
+
−
+
=
+
+
− −π
π
π
π
π
The Pspice simulation results are:
