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Common Gate Amplifier
Figure 1(a) shows a common gate amplifier with ideal current source load. Figure
1(b) shows the ideal current source implemented by PMOS with constant gate to source
voltage.
VDD
VDD
VG2
M2
Vo
M1
VG1
Vo
M1
VG1
Vi
Vi
(a)
(b)
Figure 1. Common gate amplifier.
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1. Low Frequency Small Signal Equivalent Circuit
vgs1
+
Vi
-
gmb1v bs1
I1
I2
D1
g m1 v gs1
G1
g ds1
D2
g ds2
+
Vo
S2
S1
-
(a)
g mb1 Vi
I1
vgs1
+
Vi
-
I2
D1
gm1 Vi
G1
g ds1
D2
g ds2
+
Vo
S2
S1
-
(b)
I1
I2
+
+
Y
V1
Zi
(c)
V2
YL
Zo
Figure 2. Common gate amplifier low frequency small signal equivalent circuit.
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Figure 2(a) and 2(b) show the low frequency small signal equivalent circuit.
Figure 2( c) shows the two-port, its port variables assignment are as follows:
YL = g ds2 (or Z L = rds2 ); YS = ∞(or Z S = 0)
V1 = Vi = − v gs1 , V2 = Vo
From Figure 2(b), the current equations are derived to obtain thr Y parameters:
I1 = g m1V1 + g mb1V1 + g ds1 (V1 - V2 ) = (g m1 + g mb1 + g ds1 )V1 - g ds1 V2
I 2 = −(g m1 + g mb1 )V1 + g ds1 (V2 - V1 ) = −(g m1 + g mb1 + g ds1 )V1 + g ds1 V2
The corresponding Y-parameter matrix is,
 (g + g mb1 + g ds1 ) − g ds1 
Y =  m1

- (g m1 + g mb1 + g ds1 ) g ds1 
detY = 0
The input impedance of common gate amplifier is,
g ds1 + g ds2
2g ds
y 22 + YL
2
=
≈
=
detY + y11 YL 0 + (g m1 + g mb1 + g ds1 )g ds2 g m g ds g m
This input impedance is low, since gm is large. The output impedance of common gate
amplifier is,
Zi =
Zo =
y11 + YS
1
1
=
=
detY + y 22 YS y 22 g ds1
This output impedance is high, since gds1 is very small. A common gate amplifier is
primary used as impedance transformer from low to high impedance.
The voltage gain of common gate amplifier is,
A V0 =
g + g mb1 + g ds1
− y 21
= m1
= (g m1 + g mb1 + g ds1 )R out ≈ g m1R out
y 22 + YL
g ds1 + g ds2
1
g ds1 + g ds2
This voltage gain is practically the same as the common source amplifier, except for no
signal inversion. The current gain of common gate amplifier is,
where : R out = Z O //Z L =
AI =
(g m1 + g mb1 + g ds1 )g ds2
− y 21 YL
=1
=
detY + y11 YL 0 + (g m1 + g mb1 + g ds1 )g ds2
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2. High Frequency Small Signal Equivalent Circuit
VDD
M2
VG2
C db2
Cgd2
Vo
C gd1
C db1
M1
VG1
C gs1
CL
C bs1
Vi
Figure 3. Common gate amplifier parasitic capacitances.
Figure 3 shows all the parasitic capaciatnces needed for high frequency model.
Figure 4(a) shows the high frequency small signal equivalent circuit of common gate
amplifier. The two-port assignment is shown in Figure 4(b).
The network current equation is:
I1 = (g m1 + g mb1 )V1 + sC i V1 + g ds1 (V1 - V2 ) = (g m1 + g mb1 + g ds1 + sC i )V1 - g ds1 V2
I 2 = −(g m1 + g mb1 )V1 + g ds1 (V2 - V1 ) + sC o V2 = −(g m1 + g mb1 + g ds1 )V1 + (g ds1 + sC o )V
The corresponding Y-parameter matrix is:
− g ds1 
g + g mb1 + g ds1 + sC i
Y =  m1
g ds1 + sC o 
 - (g m1 + g mb1 + g ds1 )
detY = (g m1 + g mb1 + g ds1 + sC i )(g ds1 + sC o ) − g ds1 (g m1 + g mb1 + g ds1 )
= s[g ds1C i + (g m1 + g mb1 + g ds1 )C o ] + s 2 C o C i
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D1
+
VS
Vi
vgs1
Ci
D2
gm1 Vi
+
gmb 1Vi
G1
g ds1
Co
g ds2
+
Vo
S2
S1
-
Ci=C gs1+C bs1
Co=Cgd1 +Cdb1+C gd2+Cdb2+CL
(a)
I2
I1
+
=
+
V1
+
V2
Y
VS
GL
Zi
(b)
Figure 4. Common gate amplifier high frequency small signal equivalent circuit.
The input impedance is given by:
(g ds1 + sC o ) + g ds2
y 22 + YL
Zi =
=
detY + y11 YL s[g ds1C i + (g m1 + g mb1 + g ds1 )C o ] + s 2 C o C i + (g m1 + g mb1 + g ds1 + sC i )g ds2
=
(g m1 + g mb1 + g ds1 )g ds2
(g ds1 + g ds2 ) + sC o
+ s[(g ds1 + g ds2 )C i + (g m1 + g mb1 + g ds1 )C o ] + s 2 C o C i
The output impedance is given by:
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ZO =
y11 + YS
1
1
=
=
detY + y 22 YS y 22 g ds1 + sC o
The voltage gain is given by:
g + g mb1 + g ds1
g + g mb1 + g ds1
V2
- y 21
=
= m1
= m1
V1 y 22 + YL (g ds1 + sC o ) +g ds2 (g ds1 +g ds2 ) + sC o
AV =
=
(g m1 + g mb1 + g ds1 )R out
A
= V0
s
1 + sC o R out
1+
p1
The bandwidth is defined by the dominant pole p1. That is,
w BW = p1 =
1
R out C o
f BW = f -3db = f p1 =
w BW
2π
The gain bandwidth is given by:
w GBW = A V0 w BW = (g m1 + g mb1 + g ds1 )R out
f GBW =
(g + g mb1 + g ds1 )
1
= m1
R out C o
Co
w GBW
2π
The phase margin PM for the non-inverting amplifier , which the case here, is the
distance of the phase angle at the unity gain (or 0db) frequency (fGBW) to –180. That is,
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∠A(jw GBW ) = −180 + PM
A(s ) =
A V0
s
1+
p1

jw GBW
∠A(s ) = ∠A V0 − ∠1 +
p1

w
= 0 - tan -1  GBW
 p1
w
PM = 180 - tan -1  GBW
 p1

 = −180 + PM


 = −180 + PM = 180 - tan -1 (A V0 )

w

 = 180 - tan -1  GBW

 w BW

 = 180 - tan -1 (A V0 )

The transfer signal is a single pole with no zero. The PM is always greater 90°. Hence it
is stable.
Common Gate Amplifier Experiments
3. Common Gate Amplifier Biasing and Low Frequency Small
Signal Parameters Determination.
The DC or large signal transfer characteristic is difficult to obtain analytically.
The reasons are the gate to source voltage and the threshold voltage changes with the
input voltage. From Figure 1, these voltages are given by:
VGS1 = VG1 - Vi
VBS = VSS - Vi
Instead of obtaining the complete DC transfer characteristic, only the bias point or
operating point is of interest. The operating point should lie in the region when both
transistors are in saturation. The goal is to obtain the upper and lower bound of the
saturation region.
M2, the PMOS transistor in Figure 1 is in saturation when the condition is satisfied.
|VGSP|-|VTP|<|VDSP|
|VG2-VDD|-|VTP|<|VO-VDD|
|0-VDD|-|-1|<|VO-VDD|
VDD-1<VDD-VO
VO<1
M1, the NMOS transistor in Figure 1 is in saturation when:
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VGSN-VTN<VDSN
VG1-Vi –VTN < VO-Vi
VG1<VO+VTN = 1+1=2
VGSN -VTN>0.3
VG1-Vi – VTN >0.3
VG1>0.3+Vi+VTN =0.3+0+1=1.3 ; at bias point Vi=0
That is,
1.3 <VG1<2
this define the range of VG1 when VG2=0 to guarantee that both transistor are in
saturation, select say VG1=1.5. With this value selected PSpice simulation is conducted to
determine the actual operating point, the bias voltage of the input signal. Initially the
PSpice file simply enter a guess value of the bias voltage. This bias will affect the AC
response but not the DC response. The bias voltage is obtained by selecting a point at the
middle of the steepest slope. The simulation shows that this bias voltage is –1.4903. This
information is entered in the PSpice file and re-run to obtain the proper AC response. The
operating point and small signal parameters are determined as follows:
The threshold voltage is computed at the operating point as follows:
VBS = VB - VS = VSS - Vi = −2.5 − (−1.4903) = −1.0097
VTN = VT0 + γ ( φ − VBS − φ ) = 1 + 1( 0.6 − (−1.0097) − 0.6 ) = 1.4909
The operating point current is
 9.6E - 6 
W
 = 87.27 E - 6
 = 40E - 6
 L N
 (5.4 - 1)E - 6 
= I N = I1 = I 2 = I P
βN = K N 
I DSQ
β 
β 
=  N (VGSN - VTN ) 2 =  N (VG1 - Vi - VTN ) 2
 2 
 2 
 87.27 E - 6) 
2
=
(1.5 − (−1.4903) − 1.4909) = 98.21uA
2


The transconductances and resistances are computed:
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g m1 = g mN = 2β N I DSQ = 2(87.27 E - 6)(98.2E - 6) = 130.95E - 6 = 130.95umho
g mb1 =
γ
2 φ − VBS
rds1 = R ON =
g ds1 =
1
rds1
1
rds2
1
2 0.6 − (−1.0097)
(130.95E - 6) = 51.7625E - 6 = 51.7625umhoA
1
1
=
= .509E6 = .509M
λ N I DSQ (.02)((98.21E - 6)
= 1.9642E - 6
rds2 = R OP =
g ds2 =
g m1 =
1
1
=
= .509E6 = .509M
λ P I DSQ (.02)((98.21E - 6)
= 1.9642E - 6
R out = (R ON //R OP ) = .2545E6 = .2545M
Zi =
g ds1 + g ds2
(1.9642 + 1.9642)E - 6
=
= 1.08E4
(g m1 + g mb1 + g ds1 )g ds2 (130.95 + 51.7625 + 1.9642)E - 6(1.9642E - 6)
A V = (g m1 + g mb1 + g ds1 )R out = (130.95 + 51.7625 + 1.9642)E - 6(.2545E6) = 47
A Vdb = 20 log10 (A V ) = 20 log10 (47) = 33.44 db
The PSpice simulation shows that
Zi=1.032x104
Zo=2.543x105
Av=33.756db.
*PSpice file for NMOS Common Gate Amplifier with
*PMOS Current Load
*Filename="Lab3.cir"
VIN 1 0 DC -1.4903VOLT AC 1V
VDD 3 0 DC 2.5VOLT
VSS 4 0 DC -2.5VOLT
VG1 5 0 DC 1.5VOLT
VG2 6 0 DC 0VOLT
M1 2 5 1 4 MN W=9.6U L=5.4U
M2 2 6 3 3 MP W=25.8U 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
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*Analysis
.DC VIN -2.5 2.5 0.05
.TF V(2) VIN
.AC DEC 100 1HZ 10GHZ
.PROBE
.END
V(2)/VIN = 4.894E+01
INPUT RESISTANCE AT VIN = 1.032E+04
OUTPUT RESISTANCE AT V(2) = 2.543E+05
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4. Common Gate Amplifier High Frequency Model Experiments
The parasitic capacitances will be determined to check the theory against Pspice
simulation results. The capacitances are determined at the operating point. The reverse
bias are first calculated.
For PMOS,
VBD=VB-VD=VDD-VO=2.5-0.56=1.94 (entered as negative PMOS polarity are
reversed)
VBS=0
For NMOS
VBD=VB-VD=VSS-VO=-2.5-0.56=-3.06 *
VBS=VB-VS=VSS-Vi=-2.5-(-1.4903)=-1.0097
* VO=0.56 is the output voltage at operating point.
The MATLAB program is invoked to obtain the parasitic capacitances.
For NMOS,
[CGS,CGD,CBD,CBS]=cap(9.6,5.4,-3.06,-1.0097)
CGS=23.2704fF CGD=3.84fF CBD=19.3375fF CBS=43.2441fF
For PMOS,
[CGS,CGD,CBD,CBS]=cap(25.8,5.4,-1.94,0)
CGS=62.5392fF CGD=10.32fF CBD=55.4944fF CBS=152.02fF
In the first Pspice simulation, only the Cgs and Cgd are included. That is, the output
capacitance is calculated as follow:
Co=Cgd1+Cgd2=3.84fF + 10.32fF=14.16fF
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The theoretical bandwith is calculated as follows:
w BW =
f BW =
w GBW
1
1
=
= 277.49E6
R out C O (.2545E6)(14.16E - 15)
w BW 277.49E6
=
= 44E9 = 44M
2π
2π
(g + g mb1 + g ds1 ) (130.95 + 51.7625 + 1.9642)E - 6
= m1
=
= 13.042E9
CO
14.16E - 15
w GBW 13.04E9
=
= 2.077E9 = 2.077G
2π
2π
w

 13.042E9 
PM = 180 - tan -1  GBW  = 180 − tan -1 
 = 91.22
 277.49E6 
 w BW 
The Pspice simulation results are :
f GBW =
f BW = f -3db = 44.452M
f GBW = 2.1779G
PM = 91.163
To include the effect of all the parasitic capacitances. The area and perimeter of the
source and drain of each transistor are included in the PSpice netlist. The output
capacitance is re-calculated to include all the parasitic capacitances.
C O = C gd1 + C db1 + C gd2 + C db2 + C L
= (3.84 + 19.3375 + 10.32 + 55.4944)fF = 88.9919fF
w BW =
f BW =
w GBW
1
1
=
= 44E6
R out C O (.2545E6)(88.9919E - 15)
w BW 44E6
=
= 7 E9 = 7M
2π
2π
(g + g mb1 + g ds1 ) (130.95 + 51.7625 + 1.9642)E - 6
= m1
=
= 2.075E9
CO
88.9919E - 15
w GBW 2.075E9
=
= 0.3305E9 = 330.5M
2π
2π
w

 2.075E9 
PM = 180 - tan -1  GBW  = 180 − tan -1 
 = 91.215
 44E9 
 w BW 
f GBW =
The Pspice simulation results are:
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f BW = 6.6343M
f GBW = 305.995M
PM = 91.169
*PSpice file for NMOS Common Gate Amplifier with
*PMOS Current Load
*Filename="Lab31.cir"
VIN 1 0 DC -1.4903VOLT AC 1V
VDD 3 0 DC 2.5VOLT
VSS 4 0 DC -2.5VOLT
VG1 5 0 DC 1.5VOLT
VG2 6 0 DC 0VOLT
M1 2 5 1 4 MN W=9.6U L=5.4U AD=40.32P AS=40.32P PD=27.6U
PS=27.6U
M2 2 6 3 3 MP W=25.8U L=5.4U AD=108.36P AS=108.36P PD=60U
PS=60U
.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 -2.5 2.5 0.05
.TF V(2) VIN
.AC DEC 100 1HZ 10GHZ
.PROBE
.END
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