EE 321 Analog Electronics, Fall 2011Homework #2 sol utionSS 2.22. The circuit in Fig. P2.22 is frequently used to provide an output voltagevoproportional to an input signal current ii. Derive expressions for the tran-sresistance Rm≡ vo/iiand the input resistance Ri≡ vi/iifor the followingcases:(a) A is infinite.(b) A is finite.(a) We havevo= vi− iiRf= −iiRfand thus the transresistance isRm=voii= −RfFor the input resistance we have vi= 0 for all values of ii, and thusRi=viii=0ii= 0(b) We have the same expressionvo= vi− iiRfBut in this case we also have vo= −Avi(negative because the polarity of viis opp ositethe polarity on the inputs of the op-amp), and thusvo= −voA− iiRfvo1 +1A= −iiRf(1)1andRm=voii= −Rf1 +1AFor the input resistance we haveRi=viii= −voAiiFrom equation 1 we can find voand insert itRi=iiRf11+1AAii= Rf1A + 1(notice that bot h expressions go to the expressions found in (a) as we let A → ∞).SS 2.82. Measurements made on the internally compensated amplifiers listedbelow provide the DC gain and the frequency at which the gain has dropped by20 dB. For each, what are the 3 dB and unity-gain frequencies?(a) 3 × 105and 6 × 102Hz.(b) 50 × 105and 10 Hz(c) 1500 and 0.1 MHz(d) 100 and 0.1 GHz(e) 25 V/mV and 25 kHzThe 3 dB frequency, fc, is a tenth of the 20 dB frequency. The unity gain frequency, f1, isthe 3 dB frequency multiplied by the DC gain.(a) fc= 60 Hz, f1= 60 × 3 × 105= 18 MHz(b) fc= 1 Hz, f1= 1 × 50 × 105= 50 × 105= 5 MHz(c) fc= 10 kHz, f1= 10 × 1 03× 1500 = 1 5 MHz(d) fc= 10 MHz, f1= 1 GHz(e) fc= 2.5 kHz, f1= 2.5 × 103× 25 × 1 03= 62.5 MHz2SS 2.102. A noninverting amplifier with a closed-loop gain of 1000 is designedusing an op amp having an input offset voltage of 3 mV and output saturationlevels of ±13 V. What is the maximum amplitude of the sine wave that can beapplied at the input without the output clipping? If the ampliifier is capacitivelycoupled in the manner indicated in Fig. 2.36, what would the maximum possibleamplitude be?If the maximum output is 13 V, the the maximum input is 13 mV. O f that 3 mV is the offset,leaving a maximum signal amplitude of 10 mV.If the amplifier is capcitively coupled on the input we can use superposition for t heanalysis, adding the DC offset vo ltage source and the AC input source. Since for the DCsource the gain path has a open circuit t here will be no current flowing in it, and the outputvoltage f r om the offset is equal to the offset voltage, 3 mV. The result is that the AC sourcecan supply up to 13 V − 3 mV, effectively 13 V on the output, which corresponds to 13 mVon the input. This works when the input source is an AC signal only.SS 2.110. An op amp is connected in a c losed loop with gain of +100 ut ilizing afeedback resistor of 1 MΩ.(a) If the input bias curr ent is 100 nA, what output voltage results with theinput grounded?(b) If the input offset voltage is ±1 mV and the inptut bias c urrent as in (a),what is the largest possible output that can be observed with the inputgrounded?(c) If the bias-current compensation is used, what is the value of the requiredresistor? If the offset current is no more than one-tenth the bias current,what is the resulting output offset voltage (due to offset current alone)?(d) With the bias-current compensation as in (c) in place what is the largestDC voltage at the output due to the combined effect of offset voltage andoffset current?This is a non-inverting op-amp. Since the feedback resistor is R2= 1 MΩ, the resistor toground must be R1= 1 MΩ/99 = 10.1 kΩ.(a) If the input is grounded then the inverting input is also at ground and thus all the biascurrent flows through the feedback resistor, resulting in a positive output voltagevo= IBR2= 100 × 10−9× 1 × 106= 0.1 V(b) With an input offset voltage of ±1 V we expect an output vo= 100×(±1 mV) = ±0.1 V.Thus the maximum output is 0.2 V.(c) The bias compensating resistor, RB, should be equal to the ground resistor, R1, thusRB= 10.1 kΩ. This gives rise to a voltage drop which causes exactly the bias current toflow through R1, no current thro ugh R2, and zero output voltage. If the offset current3is one tenth of the bias current, then the resulting output voltage is one tenth of theoutput voltage when considering only bias current without compensation, except thatit can be either positive or negative,vo= ±0.01 V(d) The value of the output due to offset voltage is 0.1 V with either sign. The value ofoutput due to offset current is 0.01 V with either sign. The largest value of the combinedoutput is 0 .1 1 V, either positive or