Chapter 11 Differential Amplifier Circuits _____________________________________________ 11.0 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. 11.1. 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. Figure 11.1: Differential amplifier shows differential inputs and common-mode inputs - 293 - 11 Differential Amplifier Circuits 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 V in(d) as V in(d) = V in1 – V in2 and + Vin 1 Vin 2 common-mode input voltage V in(c) = . From these equations, input 2 voltage one and two are respectively equal to Vin )d( Vin1 = V + (11.1) in )c( 2 and Vin )d( Vin2 = V − (11.2) in )c( 2 The input voltage represented by common-mode voltage and differential voltage is shown in Fig. 11.2. Figure 11.2: Small differential and common-mode inputs of a differential amplifier Let V out1 be the output voltage due to input voltage V in1 and V out2 be the output voltage due to V in2 . The differential-mode output voltage V out(d) be defined as + Vout 1 Vout 2 Vout(d) = V out1 – V out2 and common-mode output is defined V out(c) = . 2 Combining these equations yield V out1 as V out2 respectively as equal to Vout )d( Vout1 = V + (11.3) out )c( 2 - 294 - 11 Differential Amplifier Circuits and Vout )d( Vout2 = V − (11.4) out )c( 2 Let A V1 = V out1 /V in1 be the gain of differential amplifier due to input V in1 only and A V2 V out2/V in2 due to input V in2 only. Then from superposition theorem, the output voltage V out is equal to V out = A V1 Vin1 + A V2 Vin2 . After substituting V in1 and V in2 from equation (11.1) and (11.2), the output voltage V out is equal to V V + in )d( + − in )d( Vout = A V1 Vin )c( A V2 Vin )c( (11.5) 2 2 Equation (11.5) is also equal to V out = A V(dm) Vin(d) +A V(cm) Vin(c) , where the differential voltage gain is A V(dm) = (A V1 – A V2 )/2 and common-mode voltage gain is A V(cm) = (A V1 + A V2 ). The ability of a differential amplifier to reject common-mode signal depends on its common-mode rejection ratio CMRR, which is defined as A CMRR = V dm( ) (11.6) A V cm( ) From V out = A V(dm) Vin(d) +A V(cm) Vin(c) , output voltage V out is equal to 1 Vout = A V + V (11.7) V dm( ) in )d( CMRR in )c( Equation (11.7) clearly indicates that for large CMRR value, the effect of common-mode input is not significant to the output voltage. Example 11.1 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 µV 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 µV rms each is applied to the inputs. The common-mode interference is the same as in part A. - 295 - 11 Differential Amplifier Circuits 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 V cm = A V(dm) /CMRR= 2,500/30,000 = 0.083. 2. CMRR = 30,000. Also 20log(30,000) = 89.5dB. 3. The difference input for part A is 500µV - 0V = 500 µV. Thus, the rms output is A V(d) x 500 µV = 2,500 x 500 µV = 1.25Vrms The difference input for part B is 500 µV - (-500 µV) = 1mV Thus, the rms output is A V(d) x 1mV = 2,500 x 1mV = 2.5Vrms. 4. Since the common-mode gain A cm is 0.083 (from answer 1), then output voltage of interface from 1V 50Hz ac pick-up is A cm x 1V = 0.083V. 11.1 Bipolar Junction Transistor Differential Amplifier Consider an emitter coupled bipolar junction transistor differential amplifier shown in Fig. 11.3. Assuming that the physical parameters of transistor Q 1 and Q2 are closed to identical. With the modern fabrication technique and fabricating the transistor Q 1 and Q 2 in close approximity in the same wafer slide, close to identical physical parameters for both transistors are achievable. - 296 - 11 Differential Amplifier Circuits Figure 11.3: A bipolar junction transistor differential amplifier 11.1.1 dc Characteristics Using Kirchhoff’s voltage law, the voltage at emitter V E1 and V E2 , of the amplifier is V in1 - V BE1 = V in2 - V BE2 . From the theory of semiconductor physics, =[ − ] the collector current I C of a bipolar transistor is equal to IIVVC Sexp( BE / T ) 1 , where I S is the reverse saturation current, which is design dependent. V T is the thermal voltage, which has value approximately equal to 25.0mV at temperature 300K. Under normal operating conditions the term exp(V BE /V T) >> 1, thus, the I = C base-to-emitter voltage V BE is equal to VVBE1 T ln . The differential input IS voltage V in(d) = (V in1 - V in2 ) shall then be equal to I I C1 ⋅ S2 Vin(d) = VT ln (11.8) IS1 I C2 For identical transistor pair reverse saturation current is I S1 = I S2 and V in(d) = I C1 VT ln . The ratio of collector current of transistor Q 1 and transistor Q 2 is IC2 equal to - 297 - 11 Differential Amplifier Circuits I C1 Vin() d = exp (11.9) I C2 VT + IIC1 C2 The emitter current is I E = I E1 + I E2 , which is also equal to I E = α . Using this equation and equation (11.9), the collector current I C1 and I C2 of the transistor are separately derived shown in equation (11.10) and (11.11). αI I = E (11.10) C1 V 1+ exp − in() d VT αI I = E (11.11) C2 V 1+ exp in() d VT The current transfer characteristic curve showing the plot of collector current of transitor Q 1 and Q 2 versus the differential input voltage V in(d) is shown in Fig. 11.4. Figure 11.4: The current transfer characteristic curve of a bipolar junction transistor differential amplifier From the characteristic curve, once can notice that for several V T values such as Vin(d) > 4V T, either I C1 >> I C2 or I C1 << I C2 shall be obtained. For V in(d) < 2V T, the collector current is almost linear. - 298 - 11 Differential Amplifier Circuits At the output side, the output voltage are V out1 = V CC - I C1 RC and V out2 = VCC - I C2 RC respectively. The differential output voltage V out(d) shall be V out(d) = RC(I C2 - I C1 ). The differential output voltage V out(d) also equal to 1 1 = α − VIRout() d E C (11.12) Vin() d Vin() d 1+ exp 1+exp − VT VT − V = α in() d This equation is also equal to VIRout() d E C tanh since I C2 = 2VT ( +) =( −) ( − + ) 1/ 1 exp(VVVVVVVVindT()()()() / exp ind / 2 T /exp( ind / 2 T )exp( ind / 2 T and I C1 ( + −) =( ) ( + − ) =1/ 1 exp(VVVVVVVVind()()()() / T exp ind / 2 T /exp( ind / 2 T )exp( ind / 2 T . The transfer characteristic of the output shall be as shown in Fig. 11.5. Figure 11.5: 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. This is becasue emitter current of each transistor will be double instead of half. This configuration will also improve the bandwidth of the amplifier. 11.1.2 Differential Mode The differential input circuit of the amplifier is shown in Fig. 11.6. - 299 - 11 Differential Amplifier Circuits Figure 11.6: Differential input circuit of an emitter couple BJT differential amplifier Asssuming identical transistor, the increase of emitter voltage by V in1 i.e V in(d) /2 is compensated by the decrease of same value of emitter voltage by V in2 i.e. – Vin(d) /2. Thus, the voltage at emitter E 1 and E 2 remain unchange. Thus, the emitter current I e is approximately zero. As the result the potential at emitter is regards as same potential as ground level and R E is treated as short.