Chapter 11 Differential Amplifier Circuits

1 - 293 - Chapter 11 Differential Amplifier Circuits _____ Introduction Differential Amplifier or diff-amp is a multi-transistor Amplifier . It is the fundamental building block of analog circuit. It is virtually formed the Differential Amplifier of the input part of an operational Amplifier . It is used to provide high voltage gain and high common mode rejection ratio. It has other characteristics such as very high input impedance, very low offset voltage and very low input bias current. Differential Amplifier can operate in two modes namely common mode and Differential mode. Each type will have its output response illustrated in Fig. Common mode type would result zero output and Differential mode type would result high output. This shall mean the Amplifier has high common mode rejection ratio.

2 Figure : Differential Amplifier shows Differential inputs and common-mode inputs 11 Differential Amplifier Circuits - 294 - If two input voltage are equal, the Differential Amplifier gives output voltage of almost zero volt. If two input voltages are not equal, the Differential Amplifier gives a high output voltage. Let s define Differential input voltage Vin(d) as Vin(d) = Vin1 Vin2 and common-mode input voltage Vin(c) = 2VV2in1in+. From these equations, input voltage one and two are respectively equal to Vin1 = 2VV)d(in)c(in+ ( ) and Vin2 = 2VV)d(in)c(in ( ) The input voltage represented by common-mode voltage and Differential voltage is shown in Fig. Figure : Small Differential and common-mode inputs of a Differential Amplifier Let Vout1 be the output voltage due to input voltage Vin1 and Vout2 be the output voltage due to Vin2.

3 The Differential -mode output voltage Vout(d) be defined as Vout(d) = Vout1 Vout2 and common-mode output is defined Vout(c) = 2VV2out1out+. Combining these equations yield Vout1 as Vout2 respectively as equal to Vout1 = 2VV)d(out)c(out+ ( ) 11 Differential Amplifier Circuits - 295 - and Vout2 = 2VV)d(out)c(out ( ) Let AV1 = Vout1/Vin1 be the gain of Differential Amplifier due to input Vin1 only and AV2 Vout2/Vin2 due to input Vin2 only. Then from superposition theorem, the output voltage Vout is equal to Vout = AV1 Vin1 + AV2 Vin2. After substituting Vin1 and Vin2 from equation ( ) and ( ), the output voltage Vout is equal to Vout = + +2 VVA2 VVA)d(in)c(in2V)d(in)c(in1V ( ) Equation ( ) is also equal to Vout = AV(dm)Vin(d) +AV(cm)Vin(c), where the Differential voltage gain is AV(dm) = (AV1 AV2)/2 and common-mode voltage gain is AV(cm) = (AV1 + AV2).

4 The ability of a Differential Amplifier to reject common-mode signal depends on its common-mode rejection ratio CMRR, which is defined as CMRR = )cm(V)dm(VAA ( ) From Vout = AV(dm)Vin(d) +AV(cm)Vin(c), output voltage Vout is equal to Vout = +)c(in)d(in)dm(VVCMRR1VA ( ) Equation ( ) clearly indicates that for large CMRR value, the effect of common-mode input is not significant to the output voltage. Example A Differential Amplifier shown in figure below has Differential gain of 2,500 and a CMRR of 30,000. In part A of the figure, a single-ended input of signal 500 V rms is applied. At the same time a 1V, 50Hz interference signal appears on both inputs as a result of radiated pick-up from ac power system. In part B of the figure, Differential input signal of 500 V rms each is applied to the inputs. The common-mode interference is the same as in part A.

5 11 Differential Amplifier Circuits - 296 - 1. Determine the common-mode gain. 2. Express CMRR in dB. 3. Determine the rms output signal for part A and B. 4. Determine the rms interface voltage on the output. Solution 1. The common-mode gain Vcm = AV(dm)/CMRR= 2,500/30,000 = 2. CMRR = 30,000. Also 20log(30,000) = 3. The difference input for part A is 500 V - 0V = 500 V. Thus, the rms output is AV(d) x 500 V = 2,500 x 500 V = The difference input for part B is 500 V - (-500 V) = 1mV Thus, the rms output is AV(d) x 1mV = 2,500 x 1mV = 4. Since the common-mode gain Acm is (from answer 1), then output voltage of interface from 1V 50Hz ac pick-up is Acm x 1V = Bipolar Junction Transistor Differential Amplifier Consider an emitter coupled bipolar junction transistor Differential Amplifier shown in Fig. Assuming that the physical parameters of transistor Q1 and Q2 are closed to identical.

6 With the modern fabrication technique and fabricating the transistor Q1 and Q2 in close approximity in the same wafer slide, close to identical physical parameters for both transistors are achievable. 11 Differential Amplifier Circuits - 297 - Figure : A bipolar junction transistor Differential Amplifier dc Characteristics Using Kirchhoff s voltage law, the voltage at emitter VE1 and VE2, of the Amplifier is Vin1 - VBE1 = Vin2 - VBE2. From the theory of semiconductor physics, the collector current IC of a bipolar transistor is equal to []IIVVCSBET= exp(/) 1, where IS is the reverse saturation current, which is design dependent. VT is the thermal voltage, which has value approximately equal to at temperature 300K. Under normal operating conditions the term exp(VBE/VT) >> 1, thus, the base-to-emitter voltage VBE is equal to VVIIBETCS1= ln.

7 The Differential input voltage Vin(d) = (Vin1 - Vin2) shall then be equal to Vin(d) = VIIIITC1 SSCln122 ( ) For identical transistor pair reverse saturation current is IS1 = IS2 and Vin(d) = 2C1 CTIIlnV. The ratio of collector current of transistor Q1 and transistor Q2 is equal to 11 Differential Amplifier Circuits - 298 - IIVVC1 Cin dT2= exp( ) ( ) The emitter current is IE = IE1 + IE2, which is also equal to IE = IIC1C+2 . Using this equation and equation ( ), the collector current IC1 and IC2 of the transistor are separately derived shown in equation ( ) and ( ). IIVVC1 Ein dT=+ 1 exp( ) ( ) IIVVCEin dT21=+ exp( ) ( ) The current transfer characteristic curve showing the plot of collector current of transitor Q1 and Q2 versus the Differential input voltage Vin(d) is shown in Fig. Figure : The current transfer characteristic curve of a bipolar junction transistor Differential Amplifier From the characteristic curve, once can notice that for several VT values such as Vin(d) > 4VT, either IC1 >> IC2 or IC1 << IC2 shall be obtained.

8 For Vin(d) < 2VT, the collector current is almost linear. 11 Differential Amplifier Circuits - 299 - At the output side, the output voltage are Vout1 = VCC - IC1RC and Vout2 = VCC - IC2RC respectively. The Differential output voltage Vout(d) shall be Vout(d) = RC(IC2 - IC1). The Differential output voltage Vout(d) also equal to VI RVVVVout dECin dTin dT( )( )( )expexp=+ + 1111 ( ) This equation is also equal to VI RVVout dECin dT( )( )tanh= 2 since IC2 = ()()()1 1222 /exp(/exp// exp(/) exp(/( )( )( )( )+= +VVVVVVVVin dTin dTin dTin dT and IC1 =()()()1 1222 /exp(/exp// exp(/) exp(/( )( )( )( )+ =+ VVVVVVVVin dTin dTin dTin dT. The transfer characteristic of the output shall be as shown in Fig. Figure : Output transfer characteristic curve of a BJT Differential Amplifier From the analysis, one can see that to increase the range of input voltage so that it has more linear operating region, a seperate emitter resistor which is termed as emitter-degeneration resistor, can be added to each transistor instead of sharing emitter resistor.))))

9 This is becasue emitter current of each transistor will be double instead of half. This configuration will also improve the bandwidth of the Amplifier . Differential Mode The Differential input circuit of the Amplifier is shown in Fig. 11 Differential Amplifier Circuits - 300 - Figure : Differential input circuit of an emitter couple BJT Differential Amplifier Asssuming identical transistor, the increase of emitter voltage by Vin1 Vin(d)/2 is compensated by the decrease of same value of emitter voltage by Vin2 Vin(d)/2. Thus, the voltage at emitter E1 and E2 remain unchange. Thus, the emitter current Ie is approximately zero. As the result the potential at emitter is regards as same potential as ground level and RE is treated as short. Based on the analysis, the ac Differential input circuit of the Amplifier can be splitted into two half Circuits as one is shown in Fig.

10 Figure : ac Differential mode half circuit of an emitter coupled BJT Differential Amplifier 11 Differential Amplifier Circuits - 301 - The corresponding ac circuit of the half circuit Amplifier is shown in Fig. Figure : ac circuit of circuit shown in Fig. The output voltage is equal to ()2Vr||Rg2V)d(inoCm)d(out = ( ) Thus, the Differential -mode gain AV(dm) is equal to ()oCm)d(in)d(out)dm(Vr||RgVVA == ( ) The Differential input impedance Rin(d) can be obtained from equation Vin(d)/2 = ib1r . Thus, the Differential input impedance is equal to Rin(d) = 2r ( ) The Differential output impedance Rout(dm) can be obtained from equation Vout(dm)/2 = iC(ro||RC). Thus, the Differential output impedance Rout(dm) is equal to Rout(d) = 2(ro||RC) ( ) Common Mode The common input circuit of the Amplifier is shown in Fig. and its corresponding half circuit is shown in Fig.