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JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY
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SCAD Engineering College
SCAD ENGINEERING COLLEGE
CHERANMAHADEVI
EC-6404 LINEAR INTEGRATED CIRUITS
(ANNA UNIVERSITY – REGULATION 2013)
II YEAR / IV SEMESTER ECE
SUBJECT NOTES
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UNIT -I
CURRENT MIRROR AND CURRENT SOURCES:
Constant current source(Current Mirror):
A constant current source makes use of the fact that for a transistor in the active mode of
operation, the collector current is relatively independent of the collector voltage. In the basic
circuit shown in fig 1
Transistors Q 1&Q2 are matched as the circuit is fabricated using IC technology. Base and
emitter of Q1&Q2 are tied together and thus have the same VBE . .In addition, transistor Q1 is
connected as a diode by shorting it s collector to base. The input current I ref flows through the
diode connected transistor Q1 and thus establishes a voltage across Q1.
This voltage in turn appears between the base and emitter of Q 2 .Since Q2 is identical to
Q1, the emitter current of Q 2 will be equal to emitter current of Q1 which is approximately equal to
I r ef
As long as Q2 is maintained in the active region ,its collector current IC2=Io will be
approximately equal to Iref .
Since the output current Io is a reflection or mirror of the reference current Iref, the circuit is
often referred to as a current mirror.
Analysis:
The collector current IC1 and IC2 for the transistor Q1 and Q2 can be approximately
expressed as
IC1 t a F I ES e
IC2 t a F I ES e
V
V
Bfffff
Eff1ffff
VT
Bffffffff
fE2ffff
VT
- - - - - - - - - ( 1)
------------(2)
From equation (1)&(2)
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ffff@f V
ffffBfffE1ff
f BV
fffff
Eff2ffffffffffffff
IffCfff2f
VT
=e
- - - - - - - - - - - - - - - - - ( 3)
IC1
Since VBE1=VBE2 we obtain
IC2=IC1=IC =IO
Also since both the transistors are identical , b1
KCL at the collector of Q1 gives
Iref= IC1+IB1 +IB2
f
g
IffffC1ffff If C2
2fff
f
= IC1 +
+
= IC 1 +
b1 b 2
b ----------(4)
= b2 = b
solving Eq (4).
IC may be expressed as
b
I C = ffffffffffffffffI ref ------------(5)
b+2
Where Iref from fig can be seen to be
VffffBE
VfffffCC
ffffff
fff
fffff
I ref = V CC @
≈≈
(as V BE=0.7V is small)
R1
R1
ffffb
ffffffffffff
From Eq.5 for b >>1, b + 2 is almost unity and the output
current I0 is equal to the reference
current, Iref which for a given R1 is constant. Typically Io varies by about 3% for 50 ≤ b ≤200.
It is possible to obtain current transfer ratio other than unity simple by controlling the area
of the emitter-base junction (EBJ) of the transistor Q 2 . For example, if the area of EBJ of Q2 is 4
times that of Q1,then
IO=4 I ref
The output resistance of the current source is the output resistance,r0 of Q2,
Vffffffffff
Vf A
R0=I02= A ≈= I [V is the Early voltage]
r ef
A
I
O
The circuit however operates as a constant current source as long as Q 2 remains in the active
region.
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Widlar current source:
Widlar current source which is particularly suitable for low value of currents. The circuit
differs from the basic current mirror only in the resistance R E that is included in the emitter lead of
Q2.
It can be seen that due to R E the base-emitter voltage VBE2 is les than VBE1 and consequently
current Io is smaller than IC1
The ratio of collector currents IC1&IC2 using
fffffffff
f@f V
ffffBfffE1ff
f BV
fffff
Eff2ffffffff
IffCfff2f
VT
=e
- - - - - - - - - - - - ( 1)
IC1
Taking natural lohgaritihm of both sides, we get
Iffffffff
lnjk
V BE1-VBE2=V T C1
-------(2)
IC2
Writing KVL for the emitter base loop
VBE1=VBE2+(IB2+IC2)RE ----------------(3)
or
V BE1-VBE2=(1/ b +1)I C2R E -----------(4)
From eqn (2)&(4) we obtain
f
g
Iffffffff
1ffff
+ 1 I c2 RE = V T ln C1 --------------(5)
b
IC2
Or
V
Iffffffff
R E = dfffffffffffffffffffffffffffffTffffef
lnC1
IC2 -------------1 + 1ffff IC2
(6)
b
A relation between IC1 and the reference current Iref is obtained by writing KCL at the
collector point of Q1
Iref= IC1+IB1 +IB2
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