In these circuits, I1 = I2 and the voltage across D1 is made equal to the voltage across the
series combination of R and D2 to create the pseudo-PTAT current,
IPTAT’ =
VD1 - VD2
R
=
kT
Rq
ln
A2
A1
where VGS1 = VGS2 for the MOSFET only version. Psuedo-PTAT current generator using
only MOSFETs and pn junctions.
V
DD
I
1
V
DD
I
2
+
-
V
GS1
D
1
D
2
A
1
A
2
M1 M2
M3
M4
+
-
V
GS2
RI
PTAT’
V
DD
M5
I
PTAT’
Psuedo-PTAT current generator using
MOSFETs, an op amp and pn junctions.
V
DD
I
1
V
DD
I
2
D
1
D
2
A
1
A
2
M1
M2
M3
RI
PTAT’
V
DD
I
PTAT’
100326-04
+-
and where A and B are temperature independent constants, is the temperature
coefficient for Is ( ≈ 3), is the temperature coefficient for iD ( =1 for PTAT), and VBG
is the bandgap voltage of silicon (1.205V at 27°C).
The diode voltage as a function of temperature is,
Note that the term Vt(-)ln(T) is not linear with temperature and cannot completely
cancel the perfectly linear PTAT voltage.
v
D
=V
t
ln
i
D
I
s
æ
è
ç
ö
ø
÷=V
t
ln(i
D
)-V
t
ln(I
s
) I
s
=AT
g
exp
-V
BG
V
t
æ
è
ç
ö
ø
÷ and i
D
=BT
a v
D
(T)=V
CTAT
=V
t
ln(BT
a
)-V
t
lnAT
g
exp-
V
BG
V
t
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú=V
BG
-V
t
(g-a)ln(T)-V
t
ln(A/B)
The negative feedback loop shown causes the current designated as ICTAT’ to be,
ICTAT’ =
VBE
R
=
VD
R
†
I.M. Gunawan, G.C.M. Jeijer, J. Fonderie, and J.H. Huijsing, “A Curvature-Corrected Low-Voltage Bandgap Reference, IEEE J. Solid-state Circuits, vol. SC-28, No. 6, June
1993, pp. 677-670. V
DD
I
PTAT
’
+
-
V
BE
R
Q1
M2
M3
M4
M5
I
CTAT
’I
CTAT
’
+
-
V
D
R
M2
M3
M4
M5
M6
M1
V
DD
I
PTAT
’
I
CTAT
’I
CTAT
’
Generation of a pseudo CTAT current
using a bipolar transistor.
Generation of a pseudo CTAT current
using a diode.
120326-01
Generation of a pseudo CTAT current using
MOSFETs, an op amp and pn junctions.
V
DD
I
2
D
2
A
2
M1
M2
R
I
CTAT’
I
CTAT’
+-
I
PTAT
’
Differentiating with respect to T and setting T = T0 gives,
Equating the derivative to zero gives,
Substituting back gives,
At T = T0, assuming = 3.2 and = 1, the reference voltage is
VREF = VBG + Vt(-) = 1.205V + 0.057V = 1.262V (T0 = 27°C)
Because VREF ≈ VBG, this voltage reference is called the “bandgap reference”.
VDD
IPTAT’ =
VPTAT
R1
R2
+
-
VD =
VCTAT
+
-
VREF
131010-01 V
REF=I
PTAT'R
2+V
D=(R
2/R
1)V
PTAT+V
CTAT=KV
PTAT+V
CTAT V
REF=KV
tln(A
2/A
1)+V
BG-V
t(g-a)ln(T)-V
tln(A/B) dV
REF
dT
(T=T
0
)=Kln(A
2
/A
1
)
V
t0
T
0
-
V
t0
T
0
ln(A/B)-
V
t0
T
0
(g-a)ln(T
0
)-(g-a)
V
t0
T
0 Kln(A
2/A
1)-ln(A/B)=(g-a)[1+ln(T
0)] V
REF
=(g-a)V
t
[1+ln(T
0
)]+V
BG
-V
t
(g-a)ln(T)=V
BG
+(g-a)V
t
1+ln
T
0
T
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
Approximating the ln(T/To) term with a Taylor’s series expansion gives,
Substituting this into the above gives
v
D(T)=V
CTAT=V
BG-V
t(g-a)ln(T)-V
tln(A/B) v
D(T
o)=V
BG-V
to(g-a)ln(T
o)-V
toln(A/B) v
D
(T)=V
BG
-
T
T
o
V
BG
-v
D
(T
o
)[ ]-(g-a)V
t
ln
T
T
o
æ
è
ç
ö
ø
÷ ln
T
T
o
æ
è
ç
ö
ø
÷»
T-T
o
T
æ
è
ç
ö
ø
÷+
1
2
T-T
o
T
æ
è
ç
ö
ø
÷
2
+
1
3
T-T
o
T
æ
è
ç
ö
ø
÷
3
+ v
D
(T)=V
BG
-
T
T
o
V
BG
-v
D
(T
o
)[ ]-(g-a)V
t
T-T
o
T
o
æ
è
ç
ö
ø
÷-
(g-a)V
t
2
T-T
o
T
o
æ
è
ç
ö
ø
÷
2
-
(g-a)V
t
3
T-T
o
T
o
æ
è
ç
ö
ø
÷
3
+
Differentiating VREF with respect to temperature and setting T = To gives,
Solving for K gives,
K=
V
BG
-v
D
(T
o
)-(g-a)V
to
V
to
ln
A
2
A
1
æ
è
ç
ç
ö
ø
÷
÷
Þ R
2
=
V
BG
-v
D
(T
o
)-(g-a)V
to
I
PTAT
v
D
(T)=V
CTAT
»V
BG
-
T
T
o
V
BG
-v
D
(T
o
)[ ]-(g-a)V
t
T-T
o
T
o
æ
è
ç
ö
ø
÷ V
REF
=K×V
PTAT
+V
CTAT
»KV
t
ln
A
2
A
1
æ
è
ç
ö
ø
÷+V
BG
-
T
T
o
V
BG
-v
D
(T
o
)[ ]-(g-a)V
t
T-T
o
T
o
æ
è
ç
ö
ø
÷ dV
REF
dT
(T=T
o
)=
KV
to
T
o
ln
A
2
A
1
æ
è
ç
ö
ø
÷-
V
BG-v
D(T
o)
T
o
-
(g-a)V
to
T
o
=0
Circuit:
The voltage across R2 is,
VR2 = VPTAT =Vt ln(N)
creating the PTAT current flowing through Q2.
The bandgap voltage, VBG, is,
V
BG
=
2R
1V
tln(N)
R
2
+V
BE1
The resistor divider R4-R5 “gains up” this voltage to
V
REF
=
R
4
+R
5
R
5
æ
è
ç
ö
ø
÷
2R
1
V
t
ln(N)
R
2
+V
BE1
é
ë
ê
ù
û
ú
The resistor R3 eliminates the impact of the base currents of Q1 and Q2
V
REF
'
=V
REF
+R
4
(I
B1
+I
B2
)-I
B2
R
3
2(R
4+R
5)R
1
R
2R
5
® R
3
=
R
2R
4R
5
R
1(R
4+R
5)
= V
t ln
I
2A
E1
I
1A
E2
= V
t ln
R
2A
E1
R
3A
E2
The op amp forces the relationship I
1R
2 = I
2R
3
V
REF =V
EB2+I
2R
3=V
EB2+V
R1
R
2
R
1
= V
EB2+
R
2
R
1
V
tln
R
2A
E1
R
3A
E2
= V
CTAT+
R
2
R
1
ln
R
2A
E1
R
3A
E2
V
t
Differentiating the above with respect to temperature and setting the result to zero, gives
†
K.E. Kujik, “A Precision Reference Voltage Source,” IEEE Journal of Solid-State Circuits, Vol. SC-8, No. 3 (June 1973) pp. 222-226.
V
REF = V
EB2 -
1 +
R
2
R
1
V
OS +
R
2
R
1
V
t ln
R
2A
E1
R
3A
E2
1 -
V
OS
I
1R
2
R
2
R
1
ln
R
2 A
E1
R3A
E2
=
VGO - VCTAT + (-)Vt0
VPTAT
=
1.205 - 0.7 + (2.2)(0.026)
0.026
= 21.62
Therefore, R2/R1 = 9.39. In order to use the equation for V
REF with V
OS ≠ 0, we must
know the approximate value of VREF and iterate if necessary because I1 is a function of
VREF. Assuming VREF to be 1.262, we obtain from
V
REF = V
EB2 -
1 +
R2
R
1
V
OS +
R2
R
1
V
t ln
R2AE1
R
1A
E2
1 -
VOS
V
REF - V
EB2 - V
OS
a new value VREF = 1.153 V. The second iteration makes little difference on the result
because VREF is in the argument of the logarithm
VREF =
R3
R1
VPTAT +
R3
R2
VCTAT
Comments:
• The BJT of the I
CTAT’ generator can be replaced with an MOSFET-diode equivalent
• Any value of VREF can be achieved
• Part (b.) of Example 17-1 showed how to design the resistors of this implementation
Comments:
• Ability of the ZTC point not to drift with temperature restricts the temperature range
• The reference voltage must be equal to the ZTC voltage
• The voltage V
REF will suffer the bandgap curvature problem which can be translated
into I
REF. 060529-09
V
DD
I
PTAT
V
DD
I
VBE
R
3
+
-
V
REF
=V
GS
(ZTC)
I
REF
Comments:
• True temperature independence is only achieved over a small range of temperatures
• References that do not correct this problem have a temperature dependence of 10
ppm°/C to 50 ppm/°C over 0°C to 70°C.
(Iconstant a quasi-temperature independent current subject to the TCF of the resistors)
where
Vt = kT/q
Ic1 and Ic2 are the collector currents of Qn1 and Qn2, respectively
Rx = a resistor used to define IPTAT
VREF =
VBE
R2
+
Vt
R3
ln
2IPTAT
INL + Iconstant
+ IPTAT R1
Temperature coefficient 3 ppm/°C with a total quiescent current of 95µA. VDD VDDVDD
IPTAT IPTAT
IPTAT
INL
IVBE
+INL
IVBE
IConstant
Qn1
x1
Qn2
x2R2
R3
VREF
3-Output Current Mirror (IVBE
+INL)
VDD
Fig. 400-02
R1
V
GO-V
CTAT(T
0) -( -)V
t ln
T
T
0
so that
V
CTAT(I
PTAT) =V
GO-
T
T
0
V
GO-V
BE(T
0)-(-1)Vt ln
T
T
0
and VCTAT(IConstant) =VGO -
T
T0
VGO -VCTAT(T0) -Vt ln
T
T0
Combining the above relationships gives,
V
REF(T) = V
PTAT + V
GO - (T/T
0)[V
GO - V
CTAT(T
0)] - [ - 3] V
t ln
(T/T
0)
If 3, then V
REF(T) ≈ V
PTAT + V
GO
1 - (T/T
0) + V
CTAT(T
0)(T/T
0)
The design procedure for the parallel, curvature corrected reference is:
1.) Pick R
1. Can be used to set the magnitude of current flow based on IPTAT.
2.) Choose R
2 to satisfy
R
2
R
1
= - T
dV
CTAT/dT
V
PTAT
3.) Pick R
3 to satisfy
R
3
R
2
=
−
4.) Finally, select R
4A = R
4B = R
4 to achieve the desired magnitude of V
REF
V
REF =
R
4
R
1
V
PTAT +
1
R
4
R
2
V
CTAT =
R
4
R
2
R
2
R
1
V
PTAT +
V
CTAT
=
R
4
R
2
-T
d
dT
V
GO -
T
T
0
V
GO +
T
T
0
V
CTAT +
V
CTAT
=
R
4
R
2
If the bus metal is 50m/sq. and is 5µm wide, the resistance of the bus in one direction
is (50m/sq.)x(1000µm/5µm) = 10 The difference in drain currents for an overdrive
of 0.1V is,
VGS1 = 1mV + VGS2 + 1mV = VGS2 + 2mV