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Note for Linear and Digital IC integration - ICA By JNTU Heroes

  • Linear and Digital IC Application - ICA
  • Note
  • Jawaharlal Nehru Technological University Anantapur (JNTU) College of Engineering (CEP), Pulivendula, Pulivendula, Andhra Pradesh, India - JNTUACEP
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SCAD Engineering College 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) SCAD Engineering College 2

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SCAD Engineering College 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. SCAD Engineering College 3

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SCAD Engineering College 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 SCAD Engineering College 4

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SCAD Engineering College = I f C1 1 + 1ff b g I + ffffffff C2 b ----------------(7) (Assuming b 2 = b1 = b for identical transistors) In the Widlar current source IC2<<IC1,therefore the term Thus I ref t IC1 f g IffC2 ffffff may be neglected in (7) b 1 1 + ffff b fffffb fffffffffffIref IC1= b + 1 Vfffffccffff@ ffffffffB ffffEfff ffffffV R1 t I b For >>1 C1 I ref Wilson current source: Where I ref = The Wilson current source shown in fig It provides an output current I o, which is very nearly equal to V ref and also exhibits a very high output resistance. Analysis: Since VBE1=VBE2 IC1=IC2 and IB1=IB2 =IB At node’b’ f g 2ffff + 1 IC2 -----------(1) IE3=2IB +IC2= b IE3 is equal to f g 1ffff 1+ b IE3= IC3+IB3 =IC3 -----------(2) From Eqn (1)&(2) we obtain SCAD Engineering College 5

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