EE 321 Analog Electronics, Fall 2013 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by Rf Rf Rf vO = − vN 1 + vN 2 + . . . + vN n RN 1 RN 2 RN n RP RP RP Rf vP 1 + vP 2 + . . . + vP n + 1+ RN RP 1 RP 2 RP n where RN = RN 1 ||RN 2 || . . . ||RN n and RP = RP 1 ||RP 2 || . . . ||RP n ||RP 0 . (b) Design a circuit to obtain vO = −2vN 1 + vP 1 + 2vP 2 The smallest resistor used should be 10 kΩ. (a) First, for the negative inputs, let’s work with the first input via superposition. The negative input is at virtual ground, so the input current is iN 1 = RvN1 . All that current N1 flows out of the feedback resistor because the voltage across the other input resistors is zero. Thus, for only vN 1 on, we get vO = − Rf vN 1 RN 1 Similarly, by superposition, we can see that for only vN x inputs turned on, then output is 1 n X vN i vO = −Rf RN i i=1 which is identical to the expression we need to show. Next, for the positive inputs, let’s look at the first input. The voltage at the positive input is a voltage division between RP 1 and the parallel combination of all the other resistors. 1 Pn 1 1 i=0 RP i − RP 1 v+ = 1 Pn 1 1 i=1 RP i − RP 1 1 − R1 1 RP = 1 RP P1 1 − R1 + RP 1 vP 1 vP 1 P1 = 1 + RP 1 1 = + RP 1 RP 1 RP 1 1 RP 1+ −1 RP = vP 1 RP 1 − 1 RP 1 vP 1 vP 1 Next apply superposition and we get for all positive inputs on and all negative inputs off, n X vP 1 v+ = RP RP 1 i=1 Next, this signal is amplified by the non-inverting amplifier whose gain is composed from Rf and the parallel resistance of all negative input resistors, RN , so n X vP 1 Rf RP vO = 1 + RN RP 1 i=1 which is also identical with what we had to show. Combining the output from negative inputs and positive inputs we get the result we were asked to show. (b) We have the equations Rf =2 RN 1 RP Rf 1+ =1 RN RP 1 Note that RN = RN 1 , so that 1 + Rf RN = 3, and 2 RP Rf 1+ =2 RN RP 2 Rf =2 RN 1 RP 1 = RP 1 3 2 RP = RP 2 3 We can now select RN 1 = 10 kΩ, so that Rf = 20 kΩ. Next, RP RP 1 2 3 RP 1 = = × =2 RP 2 RP 2 RP 3 1 Next, we have 1 1 1 1 = + + RP RP 0 RP 1 RP 2 1= RP RP RP + + RP 0 RP 1 RP 2 1 2 RP =1− − =0 RP 0 3 3 This means that RP 0 = ∞. Then we can choose RP 1 = 10 kΩ and RP 2 = 20 kΩ. 2.62. For the circuit shown in Fig. P2.62, express vO as a function of v1 and v2 . What is the input resistance seen by v1 alone? By v2 alone? By a source connected between the two input terminals? By a source connected to both input terminals simultaneously? We can use superposition to get vO . First as a function of v1 , we have vO = −v1 . Next, as a function of v2 , R R 1+ v2 = v2 vO = R+R R Combining we get vO = v2 − v1 . The input resistance on input 1 is found as Ri1 = dv1 di1 The relationship between the two is v1 = v2 + i1 R 2 3 Plugging this into the expression above we get Ri1 = R On input 2 it is clear that Ri2 = 2R. For a source connected between the two inputs we have v2 − v1 = v, and a current through it, i, and we want the differential input resistance, dv di We need a relationship between v and i. Note that v = vO . To have the same current in the two arms we must have that v − v1 = v2 . Combining that with the input output reltionship, v = v2 − v1 , we can eliminate v1 Rid = v − v2 = v1 = v2 − v or v = v2 We can also see that v2 = 2Ri. Inserting we get v = 2Ri and see that Ric = 2R. 2.111. An op-amp intended for operation with a closed-loop gain of −100 V/V uses feedback resistors of 10 kΩ and 1 MΩ with a bias-current-compensation resistor R3 . What should the value of R3 be? With input grounded, the output offset voltage is found to be +0.21 V. Estimate the input offset current assuming zero input offset voltage. If the input offset voltage can be as large as 1 mV of unknown polarity, what range of offset current is possible? What current injected into, or extracted from, the nongrounded end of R3 would reduce the op-amp output voltage to zero? For available ±15 V supplies, what resistor and supply voltage would you use? The bias-current compensating resistor R3 = R1 ||R2 = 9.9 kΩ. With the bias compensated and zero offset voltage the output is equal to the offset current flowing through the feedback resistor, R2 , vO 0.21 = = 210 nA R2 106 If the input offset voltage is up to ±1 V, the output offset voltage can be as large as ±0.1 V. In that case, the output offset due to bias current can range between 0.21 − 0.1 = 0.11 V to 0.21 + 0.1 = 0.31 V. That results in a minimum offset current of IOS,min = 110 nA and a maximum offset current of IOS,max = 310 nA. In order to reduce the output to zero we need to add a voltage on the positive input which cancels the output of 0.21 V. That means we need to add a negative voltage on the positive input which is that output divided by the non-inverting gain, 101. The voltage we IOS = 4 need to add is Vcompensation = −0.21 V/101 = −2.08 mV. That voltage is produced by a current through R3 . That current should be −2.08 mV Vcompensation = = 210 nA R3 9.9 kΩ To get that current from the negative supply we need a resistor, Icompensation = − Rcompensation = Vsupply Icompensation = 15 = 71.4 MΩ 210 × 10−9 3.2. For the circuits shown in Fig. P3.2 using ideal diodes, find the values of the voltages and currents indicated. (a) In this case the diode is conducting, such that the voltage at the top end of the diode V equals that at the bottom end, so V = −3 V. Then, I = 106 kΩ = 0.6 mA. (b) The diode is not conducting, so I = 0, and V = 3 V. (c) The diode is conducting so I = 0.6 mA as before, but now V = 3 V. (d) The diode is not conducting, so I = 0 and V = −3 V. 3.4. In each of the ideal-diode circuits shown in Fig. P3.4, vI is a 1 kHz, 10 V peak sine wave. Sketch the waveform resulting at vO . What are its positive and negative peak values? 5 (My plotting program is temporarily out of commission so you get a description in words and equations instead. Note that this procedure is probably the best way to go about plotting the waveform anyway; getting an expression and then plottinng that expression.) (a) Piecewise expression for vO ( 0 vO = vI vI ≤ 0 vI > 0 Minimum value: 0 V. Maximum value: 10 V (b) Piecewise expression for vO ( vI vO = 0 vI ≤ 0 vI > 0 Minimum value: −10 V. Maximum value: 0 V. (c) Current never flows, so vO = 0 always. (d) This case is identical to (a) (e) Current always flows so vO = vI . Minimum values: −10 V. Maximum value: 10 V. (f) This case is identical to (a) (g) Piecwise expression for vO 6 ( vI vO = 0 vI ≤ 0 vI > 0 Minimum value: −10 V. Maximum value: 0 V. (h) The output is always connected to ground, vO = 0 always. (i) When vI < 0 this acts as a voltage divider. Otherwise they two are equal. Piecewise expression is vO = ( vI 2 vI vI ≤ 0 vI > 0 Minimum value: −5 V. Maximum value: 10 V. (j) When vI > 0 the output is shorted to the input. When vI < 0 the output is obtained from the voltage division. The expression for vO is identical to (i). (k) The current source causes 1 V voltage drop across the resistor, such that vO is always one volt higher than the voltage at the bottom of the resistor. When vI > 0 the voltage at the bottom of the resistor is ground. When vI < 0 the voltage at the bottom of the resistor is vI . The piecewise expression for vO is ( vI + 1 V vI ≤ 0 vO = 1V vI > 0 7