Download Band Gap Regulator Analysis

Band Gap Regulator Analysis
J. R. Biard, Ph.D.
Honeywell
9/10/02
1
Silicon Band Gap

The band gap of silicon changes with
temperature
4 .7 3  10 4 T 2
E G  117
. 
T 636

Between -40oC and 200oC this can be fit with
a linear equation to within ± 1.5meV
EG  1.20585  2.745  10 4 T
2
Silicon Band Gap
1.22
1.6
1.2
EG = -2.74471E-4T + 1.205852
1.18
0.8
1.16
0.4
1.14
0.0
2
EG = 1.17-(4.73E-4*T )/(T + 636)
1.12
-0.4
1.10
-0.8
o
-40 C
1.08
o
o
200 C
25 C
1.06
0
100
200
300
 EG (meV)
Band Gap (eV)
1.20
400
-1.2
-1.6
500
o
Temperature ( K)
3
Transistor Equation
 For
an NPN bipolar transistor the
collector current is given by
dn dn nx 0
IC   A e qDn ,

dx dx W b
2
ni
 qVbe 
nx 0  no exp
 , no 
 kT 
NA
4
Transistor Equation
 Combining
these expressions gives
2
ni
 qVbe 
IC   A e qDn
exp

 kT 
W NA
2
ni 
kT
 qEG 
AT exp 
n
 , Dn 
 kT 
q
3
5
Transistor Equation
 The
intrinsic carrier density expressed
in terms of the linear fit to the band gap
is given by
n i2 
2
ni 
 q EGo  mT 
AT exp 

kT


3

 qEGo 
 qm   3
A exp k   T exp  kT 


6
Transistor Equation
-40oC and 200oC the intrinsic
carrier concentration is given by
 Between
2
ni 
 qE Go 
A' T exp 

 kT 
3
 Thus,
silicon behaves as though it had a
constant band gap of EGo. Constant
band gap is used in PSPICE but the
default value is wrong (1.11eV).
7
Transistor Equation
 Silicon
mobility follows a power law with
absolute temperature; the exponent is a
function of the doping level.
 To 
 n  o 
 T

 The
same general form also works for
P-type material
8
Transistor Equation
/sq P-type resistor is PTAT Proportional To Absolute Temperature.
 This resistor gives a constant collector
current in a band gap regulator.
 All other resistors require a temperature
dependent collector current.
 A 2000
 To 
I C  I Co  
 T

9
Transistor Equation
 Combining
all these equations gives the
form of the transistor equation needed
for the band gap regulator
  

3
 qE  V  
 
T
ICo   A e A 'kT o  
 To 
T
exp
W b NA

Go
be
kT




A e A 'k o   4       q EGo  Vbe  


ICo 
T
exp


   

kT


W
N
T
 b A o

10
Differential Equation
 Taking
the derivative of the transistor
equation with respect to temperature
gives a differential equation
dVbe
E Go  Vbe
 k

 4      
 q
dT
T
 This
nonlinear differential equation can
be solved using an integration factor.
11
Differential Equation
 Solve
the differential equation for Vbe
T
 kT   T 
Vbe  E Go  E Go  Vbeo   4       ln 
 q   To 
To
Where Vbeo is the value of Vbe @ To
 Vbe is a linear function of T only when
(4 +  - ) = 0.
12
Transistor Model
PSPICE the term (4 - ) is called XTI,
EGo is called EG and must be set to
EG = 1.2058; PSPICE uses 1.11.
 Modeling Honeywell NPN transistors for
PSPICE with EG = 1.2058 and  = 0
gives XTI = 2.0599.
  = 4 - 2.0599 = 1.9401
 In
13
Transistor Model
= 100A and To = 298oK (25oC)
linear regression of Vbe data between
-40oC and 200oC gives:
 For IC
dVbe
Vbeo  0.67005 and
  19978
.
mV / o C
dT
 The
error between measured Vbe and
the best straight line fit over
temperature is ±2.0mV
14
Transistor Model
1.20
6
1.00
4
0.80
2
0.60
0
0.40
-2
0.20
-4
0.00
-6
200
-50
-25
0
25
50
75
100
125
150
175
Vbe Deviation (mV)
Vbe (V)
Vbe = -1.9978E-3(T-25) + 0.67005
o
Temperature ( C)
15
Band Gap Regulator
 For
a PTAT resistor in series with a
diode connected transistor biased to Vref
the drop across the resistor is PTAT
regardless of the value of the resistor.
 For a zero TC resistor the resistor
current and voltage are both PTAT,

= 1, Vref is different, and the variation
with temperature is greater.
16
Band Gap Regulator
 Vref =
1.2654V
 PTAT Resistor
 Resistor voltage is PTAT
independent of resistor
value.
 @ 0.1mA, Vbe = 0.6701V
 @ 1.0mA, Vbe = 0.7292V
 (dVbe/dT) drops by
0.1984mV/oC
17
Band Gap Regulator
PTAT Resistors 5.953K & 3.0K
1.0
2.5
2.0
Vb = 1.266V
0.8
1.5
0.7
1.0
VR1 (V)
5.953K
3.0K
0.6
0.5
0.5
0.0
0.4
-0.5
0.3
-1.0
0.2
-1.5
TC1 = 3.3557E-3
TC2 = 0
0.1
Error Voltage (mV)
0.9
-2.0
0.0
-2.5
-40
10
60
110
160
o
Absolute Temperature ( K)
18
Band Gap Regulator
Implant Resistors 5.953K & 3.0K
1.0
5
Vb = 1.266V
0.9
4
0.8
3
5.953K
0.7
2
VR1 (V)
3.0K
0.6
1
0.5
0
0.4
-1
0.3
-2
TC1 = 3.660E-3
TC2 = 7.030e-7
0.2
-3
0.1
-4
0.0
-5
-40
10
60
110
160
o
Absolute Temperature ( K)
19
Band Gap Regulator
 The
voltage difference between the Vbe
values of two transistors operated at
different current density has a positive
linear temperature coefficient and Vbe
has a nearly linear negative temperature coefficient. These two characteristics can be combined to give a near
zero temperature coefficient voltage
source.
20
Differential Equation
 Taking
the derivative of the transistor
equation with respect to temperature
gives a differential equation
dVbe
E Go  Vbe
 k

 4      
 q
dT
T
 This
nonlinear differential equation can
be solved using an integration factor.
11
Band Gap Regulator
 When
the collector (emitter) current is
changed by a factor of 10 at 298oK
(25oC) the Vbe changes by 59.13mV and
dVbe/dT changes by 0.1984mV/oC.
 Thus, the difference in Vbe of two
transistors operated at a 10/1 current
density ratio is proportional to absolute
temperature.
21
Band Gap Regulator
 The
band gap regulator is a compact
analog computer that solves for a
constant reference voltage (~EGo).
 The +0.1984mV/oC temperature coefficient difference between two junctions
operated at 10/1 current density must
be multiplied by 10.070 to equal the
dVbe/dT of -1.9978mV/oC.
22
Band Gap Regulator
Vbe difference of 59.13mV @ 25oC
must also be multiplied by 10.070 which
gives a voltage of 0.5953V.
 Adding this voltage to Vbeo = 0.6701V
gives a reference voltage of 1.2654V.
 This reference voltage is equivalent to
Vref = EGo + 2.32(kTo/q) @ 25oC.
 The
23
Band Gap Regulator
 The
positive TC voltage that must be
added to the Vbe to get the reference
voltage is proportional to absolute
temperature.
 With a 2000/sq implanted resistor
which is proportional to absolute
temperature the collector current is
constant and  = 0.
24
Band Gap Regulator
 The
deviation of Vref from its nominal
value over the temperature range is of
the order of the ±2.0mV deviation in the
Vbe from the best straight line fit.
 For smaller temperature ranges the best
value of Vref is slightly different and the
deviation from nominal is less.
25
Band Gap Regulator
26
Band Gap Regulator
27
Band Gap Regulator
28
Band Gap Regulator
PTAT Resistors
29
Band Gap Regulator
PTAT Resistors
1.2775
1.35
1.2770
TC1 = 3.3557E-3
TC2 = 0
EG = 1.2058V
XTI = 2.0599
1.2765
1.33
1.2755
1.31
1.2750
1.2749V
1.2745
VBG (V)
VBG (V)
1.2760
1.29
1.2845V
1.2740
EG = 1.110V
XTI = 3.0
1.2735
1.27
1.2730
1.2725
1.25
-40
10
60
110
160
o
Temperature ( C)
30
Band Gap Regulator
31
Band Gap Regulator
Implant Resistors
32
Band Gap Regulator
Implant Resistors
1.2735
1.33
TC1 = 3.660E-3
TC2 = 7.030e-7
1.2730
1.2725
EG = 1.2058V
XTI = 2.0599
1.31
1.2720
1.30
1.2715
1.29
1.2715V
1.2843V
1.2710
VBG (V)
VBG (V)
1.32
1.28
EG = 1.110V
XTI = 3.0
1.2705
1.27
1.2700
1.26
1.2695
1.25
-40
10
60
110
160
Temperature ( oC)
33
Band Gap Regulator
34
Band Gap Regulator
35
Band Gap Regulator
36
Summary
 Know
your circuit
– What does it have to do?
 Know
your simulator
– When do you take exception to defaults?
 Characterize
your devices in terms of
the parameters used in your simulator
– How were the measurements done?
37
PSPICE Models

PTAT Resistor
.MODEL
Rptat RES (r=1.0 tc1=3.3557e-3 tc2=0)

IMP Resistor
.MODEL Rimp RES (r=1.0 tc1=3.6600e-3 tc2=7.0300e-6)

NPN Transistor with N+ sinker, 10m x 10m emitter
.MODEL QNPN NPN (is=8.41E-17 bf=1.928e+2 nf=1.00e+0
+ vaf=1.046e+2 ikf=7.85e-3 ise=1.056e-18 ne=1.091e+00
br=3.257e+1
+ nr=1.00e+0 var=1.052e+1 ikr=2.2775e-3 isc=4.1715e-15
+ nc=1.752e+0 nk=5.00e-1 iss=0.00e+0 ns=1.00e+0
re=1.1025e+0
+ rb=5.1475e+2 rbm=1.072e+2 rco=0.00e+0
vo=1.8519e+0
+ gamma=1.1586e-10 qco=2.3742e-14
cje=1.80e-13 vje=7.40e-1 + mje=3.40e-1 cjc=5.9584e-14 vjc=6.00e-1
mjc=4.00e-1 xcjc=1.00e+0
+ cjs=2.9648e-14 vjs=5.00e-1
mjs=3.10e-1 fc=5.00e-1 tf=2.1e-10 + xtf=0.00e+0 vtf=1.00e+2 itf=0.00e+0
ptf=0.00e+0 tr=1.00e-8
+ eg=1.20585e+0 xtb=1.905e+0
xti=2.0599e+0 kf=0.00e+0 af=1.00e+0
+ tre1=6.64e-4 tre2=-4.72e-7
trb1=4.00e-3 trb2=8.00e-6 trm1=8.24e-4
+ trm2=5.13e-6 trc1=7.02e-3
trc2=1.91e-5)
38