ppppppp

1. Sheet
1 of 16
Bandgap reference
The schematic diagram of the bandgap voltage reference is shown in Figure 1.
The bandgap reference has largely independent of temperature and the supply rails and is
therefore used as the controlling current source for mirroring throughout a circuit eg an op-
amp.
As shown in the diagram there are two voltage sources, one generated across a diode
junction ie VBE (eg the base-emitter junction on a bipolar transistor) and the other a thermal
voltage Vt.
VBE = -2.2mV/˚C and
Vt = +0.085mV/˚C
Thus if multiply Vt by a constant K and combine with VBE it is possible to cancel the
temperature effects of each voltage source to leave a stable reference voltage VREF ie
VREF = VBE + K.Vt
VDD
VBE
VDD
Vt
K
Sum
Vref
K.Vt
Figure 1 Schematic diagram of a bandgap reference. Vref = K.Vt +VBE.

2. Sheet
2 of 16
Vbe
BJT_NPN
BJT2
Temp=Tc
Area=8
Model=BJTM1
VAR
VAR1
Tc=23
Eqn
Var
BJT_Model
BJTM1
Vje=0.7
NPN=yes
DC
DC1
Step=1
Stop=155
Start=-55
SweepVar="Tc"
DC
I_DC
SRC1
Idc=100 uA
Figure 2 ADS simulation setup to determine the temperature dependence of Vbe. The
temperature variable of the BJT model Tc is sweep by the DC simulation from –55 to
155 degrees C.
m1
Tc=-55.000
Vbe=795.6mV
m2
Tc=155.000
Vbe=418.3mV
m1
Tc=-55.000
Vbe=795.6mV
m2
Tc=155.000
Vbe=418.3mV
-40 -20 0 20 40 60 80 100 120 140-60 160
500
600
700
400
800
Tc
Vbe, mV
m1
m2
Eqn Tcoeff=1E3*(Vbe[210]-Vbe[0])/(Tc[210]-Tc[0])
Tcoeff
-1.7966 in mV
Figure 3 Temperature dependance of Vbe as simulated from the ADS simulation of
Figure 2. Temperature coefficient of VBE ~ -1.79mV/˚C.

3. Sheet
3 of 16
Temperature dependance of VBE
We can simulate the temperature effect of VBE by using the ADS simulation shown in Figure
2. In this simulation the temperature parameter of the generic spice BJT model (TC) is sweep
by the DC simulation from –55 to 155 degrees C.
The resulting plot of Vbe vs temperature is shown in Figure 3. An equation has been added to
calculate the temperature coefficient by taking the first and last data points [0] and [210] to
calculate the slope of the graph.
siliconfor1.12eVvoltagebandgapEgWhere(3)-
K.T
Eg-
.expTn
(2)-
2
3
-mWhere.Tµµ
variableskeyofdependanceeTemperatur
carriersminorityofMobilityµ
)cm(1.5x10SiliconinionconcentratcarrierIntrinsicn
Where
(1)-µ.K.T.nI
1-J.K10x1.3807constantBoltzmannsK
C10x1.602electrononchargeq
q
kT
Vtby WheregivenVoltageThermalVt
Where
Vt
V
.expIIc
-:bygiveniscurrentcollectorBipolar
32
i
m
O
3-102
i
2
iS
23-
19-
BE
s
==⎥
⎦
⎤
⎢
⎣
⎡
∝
≈∝
=
=
∝
==
==
==
=
⎟⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎝
⎛

4. Sheet
4 of 16
T
Is
.
I
V
Is
Ic
.Ln
T
V
T
V
x
y
x
1
Ln.axAs
T
Is
.
I
1
.V-
T
Is
Ln.IV-Ln.IcVgettooutexpand
T
Is
findTo
Is
Ic
.Ln
T
V
Is
Ic
.LnVof
T
V
Is
Ic
.LnVV
.KµincludesAconstantWhere
K.T
Eg-
.expA.T
K.T
Eg-
.exp.K.T.T.TµI
1into2&3equationsSub
S
TTBE
S
TSTT
T
T
BE
TBE
O
m43m
OS
∂
∂
−⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=
∂
∂
∂
∂
=
∂
∂
⇒
∂
∂
∴
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
=
⎥
⎦
⎤
⎢
⎣
⎡
=⎥
⎦
⎤
⎢
⎣
⎡
∝ +
( )
( )
( )
⎥
⎦
⎤
⎢
⎣
⎡
+
+
=
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡+
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡+
+
⎥
⎦
⎤
⎢
⎣
⎡+
⎥
⎦
⎤
⎢
⎣
⎡++
=
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡++⎥
⎦
⎤
⎢
⎣
⎡++=
∂
∂
−=⎥
⎦
⎤
⎢
⎣
⎡+=
2T
T
S
T
2TT
S
T
2
n
S
K.T
Eg-
V
T
m4V
T
Is
.
I
V
K.T
Eg-
.expm4A.T
K.T
Eg-
.
K.T
Eg-
.expm4A.TV
K.T
Eg-
.expm4A.T
K.T
Eg-
.expm3A.Tm4V
T
Is
.
I
V
K.T
Eg-
.
K.T
Eg-
.expm4A.T
K.T
Eg-
.expm3A..Tm4
T
Is
dx
dy
.1nanxaxUse
K.T
Eg-
.expm4A.TI

5. Sheet
5 of 16
( )
( ) ( )
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
T
Eg-
q
1
T
m4V
T
V TBE
=⎥
⎦
⎤
⎢
⎣
⎡
+
+
+=
∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
=⎥
⎦
⎤
⎢
⎣
⎡
+
+
+⎟
⎠
⎞
⎜
⎝
⎛
=
∂
∂
−⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=
∂
∂
K.T
Eg-
q
kT
T
m4V
T
V
T
V
q
kT
Vt&
Is
Ic
.LnVV
K.T
Eg-
V
T
m4V
Is
Ic
.Ln
T
V
T
Is
.
I
V
Is
Ic
.Ln
T
V
T
V
2
TBEBE
TBE2T
TT
S
TTBE
Evaluation of dVBE/dT
area.Emitter-BaseA
)cm(1.5x10SiliconinionconcentratcarrierIntrinsicn
).s13cm(Typically
base.theinconstantdiffusionelectrontheforvalueeffectiveaverageTheDn
width.baseWdevice);type-pfor3x10NAprocess0.8um(for
emitter.theofareaunitperbasetheinatomsdopingofnumbertheis.NWQB
Where
A10to10arevaluesTypical
Q
Dn.q.A.n
IWhere
I
I
Vt.LnV
C10x1.602electrononchargeq
1-J.K10x1.3807constantBoltzmannsK
and
2
3
-m&1.12eV)silicon(forvoltageBandgapEqWhere
25.8mV
10x1.602
300.10x1.3807
q
KT
V
3-102
i
1-2
B
16
AB
16-14-
B
2
i
S
S
E
BE
19-
23-
19-
-23
T
=
=
=
=
==
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
==
==
===
===

6. Sheet
6 of 16
( )
0.743V
1.562x10
50x10
.Ln108.52V
A1.562x10
3x10
.0.0131.5x10..1x101.602x10
Is
3x10.3x101x10.NWQ
Q
Dn.q.A.n
IWhere
I
I
Vt.LnV
then
1umA50uA,IEassumeVbeandVbeforvaluetypicalafindTo
17-
6-
3
BE
17-
10
2106-19-
10166-
ABB
B
2
i
S
S
E
BE
2
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
=≈
===
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
==∆
−
x
( )
K1.44mV/-
300
11.1108.52
2
3
-4-0.743
T
V
2
3
-m&1.12eV)silicon(forvoltageBandgapEqWhere
0.743VVBE&25.8mVV
T
q
Eq
Vm4-V
T
V
3
BE
T
TBE
BE
o
=
−⎟
⎠
⎞
⎜
⎝
⎛
=
∂
∂
===
==
−+
=
∂
∂
−
x

7. Sheet
7 of 16
VPTAT generation
The PTAT term is realised by determining the voltage difference between two forward-biased
diodes (eg VBE). MOS transistors operating in the weak inversion region can also be used to
form the diodes.
VBE
VDD
VR
Area = nA
Area = A
V01
V02
R
Q1
Q2
Figure 4 Generation of VPTAT voltage.
We can simulate the variation VPTAT over temperature using the ADS simulation shown in
Figure 5.
If we run the same simulation again but this time on the results graph we calculate Vref given
that we know vbe1 and (vbe1-vbe2). Various values of K were tried until the temperature
compensation was achieved as shown in the graph of Figure 7. This now forms the basis of
the band-gap reference in a practical circuit the voltage summing and setting of K is achieved
using a resistive network and an op-amp.

8. Sheet
8 of 16
Vbe1 Vbe2
BJT_NPN
BJT3
Temp=Tc
Area=2
Model=BJTM1R
R1
R=1 kOhmDC
DC1
Step=1
Stop=155
Start=-55
SweepVar="Tc"
DC
BJT_Model
BJTM1
Vje=0.7
NPN=yes
VAR
VAR1
Tc=23
Eqn
Var
BJT_NPN
BJT2
Temp=Tc
Area=1
Model=BJTM1
I_DC
SRC2
Idc=100 uA
I_DC
SRC1
Idc=100 uA
Figure 5 ADS simulation setup to analyse the variation of VPTAT over temperature. As
for the previous examples the temperature is swept in the DC simulation box. The
resulting plot is shown in Figure 6.

9. Sheet
9 of 16
EqnVTAT=Vbe1-Vbe2
m1
Tc=155.000
VTAT=0.026
m2
Tc=-55.000
VTAT=0.013
-40 -20 0 20 40 60 80 100 120 140-60 160
0.014
0.016
0.018
0.020
0.022
0.024
0.012
0.026
Tc
VTAT
m1
m2
Eqn VTAT_Temp=1e3*(VTAT[210]-VTAT[0])/(Tc[210]-Tc[0])
VTAT_Temp
0.060 mV
Figure 6 Resulting simulation of VPTAT vs temperature after running the simulation
shown in Figure 5.
.

10. Sheet
10 of 16
Figure 4 shows how the VPTAT voltage can be realised. If we force V01 and V02 to be the
same then the voltage across the resistor R will be the difference of the two VBE’s.
E2E1
2
1
2
1
E1
E2vt
VBE2-VBE1
vt
VBE2-VBE1
S1
S2
E1
E2
E2E1
E
kT
q.VBE
S
kT
q.VBE
SE
IIIf
A
A
A
A
I
I
e
thendeviceeachforsamethearevtandJsthatsuch
process,samethefromarestransistorAssume
q
kT
vtWheree
.JA
.JA
I
I
thenIIthatsuchconfigurediscircuitAssume
0IWheneA.J1eA.JI
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
=
>
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
≈
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
2
1
2
1vt
VBE2-VBE1
A
A
vt.LnVBE2-VBE1givetoRearrange
A
A
e
( ) ( )
dx
dy
.1nanxaxUsingn.Ln
q
K
T
Vbe
n.Ln
q
KT
Vbe
q
KT
Vand
A
A
nLet
)I(I.R1IVbe
A
A
Vt.LnVBE1-VBE2thenAAIf
n
T
2
1
E2E1E1
2
1
21
−==
∂
∆∂
∴=∆∴
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
==∆=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=>
Where
K = Boltzmanns constant = 1.3807 x 10-23
J.K-1
q = charge on electron = 1.602 x 10-19
C
A = Area of base-emitter junction um2

11. Sheet
11 of 16
EqnVTAT=Vbe1-Vbe2
EqnVref=Vbe1+(K*VTAT)
EqnK=27
-40 -20 0 20 40 60 80 100 120 140-60 160
1.186
1.187
1.188
1.189
1.190
1.185
1.191
Tc
Vref
Figure 7 Graph of the simulation shown in Figure 5. In this case we have calculated
VTAT ie vbe1-vbe2 and evaluated Vref from Vbe1+(K*VPTAT), where K = 27.
BandGap reference voltage
( ) K1.44mV/-
T
Eg-
q
1
T
m4V
T
V
T
V
J
J
Ln
T
V
T
∆Vbe
thereforeand
A
A
Ln
q
KT
J
J
Ln
q
KT
V-VVbefoundwePreviously
TBEBE
C2
C1T
2
1
C2
C1
BE2BE1
o
=⎥
⎦
⎤
⎢
⎣
⎡
+
+
+=
∂
∂
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
∂
∂
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==∆

12. Sheet
12 of 16
The band-gap reference voltage is given by:-
VREF = VBE + K.Vt
The temperature stable value of VREF at 300˚K is 1.262V. Therefore the value of K required is:
20.11
25.8x10
743.0262.1
K
25.8mV
10x1.602
300.10x1.3807
q
KT
V
earlier)d(Calculate0.743VWith
V
VV
K
KgettoRearrangeK.VVV
3-
19-
23-
T
BE
T
BEREF
TBEREF
=
−
=
===
=
−
=
+=
With reference to Figure 8.
The cascode current mirror makes I1 = I2 = I3
( )
( )
( )
( ) BE3OUT
BE3OUT
BE3OUT
2
1
2
1
BE1BE2
V.KNVt.LnV
V.K.RN.Ln
R
Vt
V
N.Ln
R
Vt
I3insubI2I3AsV.K.RI3V
I3I1N.Ln
R
Vt
I2soand
A
A
NLet
A
A
Vt.LnV-VVbeRacrossvoltageThe
+=∴
+=
==+=
===
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
==∆=

13. Sheet
13 of 16
( )
Ω=∴
===∴
=
=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
=
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
=
∆
=
=
+
==
=Ω=
−
10.45KK.R
10.45
)0.025.Ln(8
0.63-1.262
.Ln(N)V
Vbe-V
K
1.262VC23atvoltagereferencetcoefficienetemperaturzeroThe
0.0536V
1
8
(0.025).Ln
A2
A1
Vt.Ln∆ ∆VWith
C23at52uA
1x10
1
8
(0.025).Ln
R1
A2
A1
Vt.Ln
R1
Vbe
I
0.025V
10602.1
)23273.1.38x10
q
KT
vt
8N;1KRthatandC23ofetemperaturaAssume
T
REF
32
19
23-
o
o
o
x

14. Sheet
14 of 16
X.N
Vt
Vref
X
R
IE1
X.N
K.R
IE2
VEE
VDD
M1 M2
M3 M4
M5 M6
M7 M8
M10
M9
Q3
Q1
Q2
Figure 8 Circuit of the bandgap reference used in the example.

15. Sheet
15 of 16
The Bandgap circuit of Figure 8 was entered as a schematic into ADS as shown in figure and
analysed using a DC simulation block. For the simulation, the Temperature variable was
added to the active devices and resistor and the resistor initially set to 10K was varied until
the correct compensated curve resulted.
vout
vbe
R
R7
Temp=T
R=9 kOhm
VAR
VAR1
W=1.0
VDD=7.5
T=23
L=1.0
Eqn
Var LEVEL1_Model
MOSFETM2
Tox=24.7e-4
Mjsw=0.35
Cjsw=350e-12
Mj=0.5
Cj=560e-12
Cgbo=700E-12
Cgdo=220e-12
Cgso=220e-12
Lambda=0.04
Phi=0.8
Gamma=0.57
Kp=50e-6
Vto=-0.7
PMOS=yes
LEVEL1_Model
MOSFETM1
Tox=24.7e-4
Mjsw=0.38
Cjsw=380e-12
Mj=0.5
Cj=770e-12
Cgbo=700E-12
Cgdo=220e-12
Cgso=220e-12
Lambda=0.04
Phi=0.7
Gamma=0.4
Kp=110e-6
Vto=0.7
NMOS=yes
BJT_PNP
BJT1
Mode=nonlinear
Trise=
Temp=T
Region=
Area=10
Model=BJTM1
BJT_PNP
BJT3
Mode=nonlinear
Trise=
Temp=T
Region=
Area=10
Model=BJTM1
DC
DC1
Step=1
Stop=125
Start=-50
SweepVar="T"
DC
MOSFET_PMOS
MOSFET9
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
MOSFET_NMOS
MOSFET1
Temp=T
Width=W um
Length=L um
Model=MOSFETM1
MOSFET_NMOS
MOSFET3
Temp=T
Width=W um
Length=L um
Model=MOSFETM1
MOSFET_NMOS
MOSFET6
Temp=T
Width=W um
Length=L um
Model=MOSFETM1
MOSFET_NMOS
MOSFET5
Temp=T
Width=W um
Length=L um
Model=MOSFETM1
MOSFET_PMOS
MOSFET7
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
MOSFET_PMOS
MOSFET4
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
MOSFET_PMOS
MOSFET8
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
MOSFET_PMOS
MOSFET2
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
MOSFET_PMOS
MOSFET10
Temp=T
Width=2.2*W um
Length=L um
Model=MOSFETM2
BJT_PNP
BJT2
Mode=nonlinear
Trise=
Temp=T
Region=
Area=1
Model=BJTM1
BJT_Model
BJTM1
Vje=0.7
NPN=no
R
R6
R=1 kOhm
V_DC
VDD
Vdc=VDD
Figure 9 ADS schematic setup for analysing the bandgap example circuit. R7 was
initially set to 10K (as per the hand calculations) and then varied to optimise the
bandgap voltage vs temperature curve shown in Figure 10

16. Sheet
16 of 16
-40 -20 0 20 40 60 80 100 120-60 140
1.164
1.166
1.168
1.162
1.170
T
vout, V
Figure 10 Resulting plot from the simulation shown in Figure 8. For this simulation the
temperature parameter for the active devices and resistor was varied over the
temperature range –50 to 125 deg C using the parameter sweep within the DC
simulation block.
One disadvantage of the example circuit is the headroom required on the supply rails. This is
because there are 4 VSAT+VT and a Vbe, resulting in the need to raise the supply from the
nominal +5V to +7.5V. Lower rail circuits tend to use low supply differential op-amp circuits.