EE 321 Analog Electronics, Fall 2011Homework #3 sol ution2.73.(a) Consider the instrumentation amplifier circuit of Fig. 2.20(a). If the opamps are ideal except that their outputs saturate at ±14 V in t he mannershown in Fig. 1.13, find the maximum allowed input common-mode signalfor the case R1= 1 kΩ and R2= 100 kΩ.(b) Repeat (a) for the circuit in Fig. 2.20(b), and comment on the differencebetween the two circuits.1(a) The first stage of the amplifier has a gain of 101 fo r both common mode and differentialmode. The maximum common-mode input is thus the saturation levels divided by thegain,vic,max=14101= 0.139 V(b) In this case the common mode gain is unity, whereas the differential mode g ain is2101. The maximum common-mode input is again the satraution levels divided by thecommon-mode gain,vic,max=141= 14 VThe second version o f the amplifier, which is the one named the instrumentation am-plifier, is the better o ne to use as it greatly increases the allowable common-mode rangeon the input.2.76. Design the instrumentation amplifier circuit of Fig. 2.20(b) to realize adifferential gain, variable in t he range 1 to 100, utilizing a 100 kΩ pot as variableresistor. (Hint: Design the second stage for a gain of 0.5)We can design the second stage for a gain of 0.5 by choosing R4= R3/2, for exampleR4= 10 kΩ and R3= 20 kΩ.For the first stage, we insert the pot in the 2R1spot. However if we have just the potthen the gain will be infinite when we turn the pot down t o zero resistance. So we must adda series resistor, so that 2R1= RA+ RP, where RPis the pot. The first stage gain is nowG =2R22R1+ 1 =2R2RA+ RP+ 1We have to solve for RAand R2. Notice that the gain is smallest when the pot is at thelargest resistance and the gain is largest when the pot is at the zero resistance. We want afirst-stage gain of 2 to 200. ThusG =(2 RP= 100 kΩ200 RP= 0 kΩwhich can be written as2 =2R2RA+ RP+ 1 200 =2R2RA+ 1From the second equation we get 2R2= 199 RA. Insert that in the first equation and we get2 =199 RARA+ Rp+ 1RA+ Rp= 198 RARA=1198Rp=100198= 505.1 ΩThen we g etR2=1992RA= 50.3
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