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High Input Resistance Circuits: DARLINGTON PAIR

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High Input Resistance Circuits: DARLINGTON PAIR HRckts.doc 1 High Input Resistance circuits: The ideal voltage amplifier should have infinite input impedance and zero output impedance. The CC and CE with Re basic amplifiers have these properties. The Input impedance of these amplifiers is Ri = hie+(1+hfe)Re using the simplified model (assuming that hfeRe << 0.1) As Re ® ¥ this equation suggests that Ri ® ¥ . However, as Re ® ¥ the assumption hfeRe << 0.1 is no longer valid. And the more accurate equation is Ri = hie + (1+hfe)Re / (1+hoeRe) which ® hfe/hoe This is a theoretical limitation on Ri. There are other practical limitations also. 1. As Re increases the bias current causes a larger voltage drop across it. For middle of operating range VCE = VRe = VCC/2. We thus require larger impractical power supply voltage 2. In integrated circuits Re occupies chip area. Larger the value, greater is the chip area occupied, leaving less for other components. 3. Bias resistance appear in parallel with the Ri and with typical values of a few 100K the parallel combination RI is now decided by the bias resistance, and is hence lower. Solutions: The problems (1) and (2) are due to the fact that we are thinking of Re as Ohmic physical resistance. If Re is an equivalent resistance, created by a relatively smaller physical resistance in another CC circuit, this problem can be solved. (The DARLINGTON PAIR circuit). The bias resistance problem (3) may be solved by the BOOTSTRAPPING technique. DARLINGTON PAIR: The Darlington pair is a cascade of two common collector circuits as shown. With input resistance of Q2 acting as Re for Q1.Q1 thus sees a large equivalent Re, but the physical resistance producing this effect is R2 which is much smaller. Emitter of Q1 is at VR2 +VBE2 and is reasonably low voltage, not requiring a high voltage power supply. E.g. If IC1 = 0.1 mA, IC2 = 1 mA, R2 = 5 K , hie = 1K and hfe = 100 then Re for first stage is 1K + (1+100)5K = 506K, and VE1 is just 5K´1mA + 0.7V=5.7V. Had we used a physical resistance of 500K as Re for the first stage, and omitted the second stage, theVE1 would have been 500K´0.1mA = 50V, requiring a power supply of 100 V if Vce =Vre = Vcc/2. PDF created with FinePrint pdfFactory trial version http://www.fineprint.com HRckts.doc 2 For the second stage, AI2 = 1 + hfe and Ri2 @ (1+hfe)R2 For the first stage load is Ri2 and hoeRi2 is likely to be > 0.1 So AI1 = (1+hfe) / (1+hoeRi2) = (1+hfe) / (1+hoe(1+hfe)R2) @ (1+hfe) /(1+hoehfeR2) Overall current gain is 2 AI1AI2 =(1+hfe) /(1+hoehfeR2) And 2 RI = hie + AI2Ri2 =(1+hfe) R2/(1+hoehfeR2) 2 -6 The RI for the given parameters is (1+100) ´ 5K / (1+ 25´10 ´100´5K) @ 50M The Darlington pair has a drawback that emitter current of first stage is amplified by the current gain of the second stage, and so first stage drift gets amplified, making the circuit drift- prone. In general the second stage has a higher collector current (hfe2 ´ Ie1) and so the h parameters of the two transistors cannot be considered as the same.(h parameters depend on bias point) Cancelling the effect of the bias resistance by Bootstrapping: Effect of bias resistance can be minimised by bootstrapping the CC circuit as shown. Here R1 and R2 are the bias resistances. giving base bias through R3. C1 short-circuits the output to point Y at the junction of the bias resistance. Thus signal voltage at Y = Vout At point X the signal voltage is Vin. Let the signal current in R3 be Ieff Then Ieff = (Vin - Vout ) / R3 So effective resistance of the combination seen by the source is Reff = Vin /Ieff = Vin R3/(Vin - Vout ) = R3 / (1 – Av) where Av = voltage gain Vout/Vin for the CC and Darlington circuits, Av @ 1 thus Reff = R3/ e where e = 1 – Av and is a very small number. This means effective resistance of the bias combination is increased greatly. e.g. If Av is 0.95 and R3 is 50K , R1 = 100K and R2 = 100K then without bootstrapping, bias combination would give 50K + (100K parallel with 100K) = 50 + 50 = 100 K resistance. With Bootstrapping, the resistance is 50K / (1 – 0.95) = 50K ´ 20 = 1000K = 1M. Current Sources: The requirement for high effective (equivalent) resistance can also be met by the CURRENT SOURCE. The ideal current source has an infinite impedance, and so it may be used in place of Re , ensuring correct bias current and providing high resistance simultaneously. Current sources may also be used as loads in high gain circuits (Active Loads) PDF created with FinePrint pdfFactory trial version http://www.fineprint.com HRckts.doc 3 Two Transistor Current Mirror: Also called current repeater used frequently for IC transistor biasing Q1 and Q2 are identical transistors, Q1 is controlling and Q2 is controlled transistor. As Vbe of both are same, IC1 = IC2, and as hFE’s are equal IB1 = IB2 = IB Now IR = Ic1 + 2Ib = Ic1 + 2(Ic1 / hFE) Thus IC2/IR = IC1/IR = hFE / (2+ hFE) Also IR = (VCC – VBE )/R I Hence I = =[ h / (2+ h )][ (V – V )/R] C2 FE FE CC BE For typical values of hfe the difference between IR and IC2 is negligible so the circuit is a current mirror, reflecting value of Ic1 Ic2 IR in the controlled side.IC2. Adding more transistors (or using multi collector fabricated transistor currents in several Ib I b branches can be controlled by the Q1 transistor., with all transistors delivering same currents. Hence the circuit is a CURRENT REPEATER circuit. For N controlled transistors the value of ICn ==[ hFE / (N + 1+ hFE)][ (VCC – VBE )/R]. If hfe is low, then the IR = IC2 approximation is not valid, and the circuit will not be an accurate current mirror / repeater. In that case the Three transistor current mirror should be used.. I I I c2 c3 c4 I R R c2 Ic3 Ic4 Q1 Q1 Three Transistor current mirror The three transistor current mirror is used to obtain current mirror action when the hfe of the transistors is low. For the circuit shown As Vbe of both Q1 and Q2 are same, IC1 = IC2 = IC, and as hFE’s are equal IB1 = IB2 = IB IR = IB3 + IC2 = (IE3 / 1+hfe) + hfeIB But IE3 = 2IB Hence IR = IB([2/(1+hfe)]+hfe) And IC = hfeIB Hence 2 IC/IR = hfe(1+ hfe)/( hfe + hfe + 2) Thus even with hfe of 10 the ratio is close to 1. This is a better current mirror for low hfe transistors PDF created with FinePrint pdfFactory trial version http://www.fineprint.com HRckts.doc 4 IR is given by (VCC - 2VBE )/ R WILSON SOURCE: This is also a three-transistor circuit, IC And VBE of the Q2 and Q3 being equal, IB2 = IB3= IB and IC2 = IC3= IR R IC Now, Q1 IR = IB1+IC3 = IB1+ hfeIB = IE1/(1+hfe) + hfeIB = (2IB+ IC)/ (1+hfe) + hfeIB =(2IB+ hfeIB)/ (1+hfe) + hfeIB Q3 Q2 =IB[(2+ hfe)/ (1+hfe) + hfe] IC1 = hfe IB1 = hfe IE1/(1+hfe) = hfe (2IB+ IC)/ (1+hfe) = hfe (2IB+ hfeIB)/ (1+hfe) = IB[hfe(2+ hfe)/ (1+hfe)] Then 2 IC1/IR = hfe(2+ hfe)/( hfe + 2hfe + 2) and IR is given by (VCC - 2VBE )/ R Again this is useful when hfe of the transistors are low. CURRENT SOURCES FOR LOW CURRENT APPLICATIONS For getting low currents like 50 uA the current sources above require too large a value of R. (several 100 K) Example: In a two transistor current mirror it is required to get a output current 50 uA. Find R if hfe = 100. And VCC = 15V. Here 0.05 = 100/(100+2) ´(15-0.7)/R Hence R = 280 k. The value is too high Here the WIDLAR source is used. VBE2 =VBE1 + (IB1 + IC1) Re VBE2 - VBE1 = (IB1 + IC1) Re V V Now IC1 = aFIESe be1/VT and IC2 = aFIESe BE2/VT , also, I c1 = e(VBE1-VBE2 ) /VT Ic2 Hence VBE1-VBE2 = VTln (IC2/IC1) VT æ Ic2 ö Thus Re = lnç ÷ æ 1 ö è Ic1 ø I ç1+ ÷ c1ç ÷ è h fe ø The reference current IR1 = IC2 + IB2 + IB1 = IC2(1+(1/hfe)) + (Ic1/hfe) The last term an be neglected as IC1 << IC2 And IC2 = (VCC- VBE2)/R for hfe >>1 From this IC2 is determined, and Using desired value of IC1 , Re can be found 100 (15 - 0.7) In the given problem taking R = 14K IC2 = =1.01 mA (100 +1) 14 PDF created with FinePrint pdfFactory trial version http://www.fineprint.com HRckts.doc 5 0.025 1.01 And Re = ln = 1.49K æ 1 ö 0.05 0.05ç1+ ÷ è 100 ø Thus much smaller resistor values are required for the same current. VOLTAGE SHIFTERS The differential voltage at the input of a differential amplifier may be AC or DC. and the corresponding output too may be AC or DC. Therefore Coupling capacitors and transformer coupling cannot be used to interconnect stages. It is also required that for 0 input output should be 0. In the normal differential amplifier circuit the output is taken from the collector and this has a DC component equal to the bias VC. with 0 input this is the quiescent voltage at the collector. To make the final output voltage 0, capacitors and transformers cannot be used, so the standing VC has to be shifted down to 0V. For this VOLTAGE SHIFTER circuits are used.
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