EE 321 Analog Electronics, Fall 2013 Homework #3 solution

Transcription

1 EE 32 Analog Electronics, Fall 203 Homework #3 solution (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N R ] f v Nn R N R N2 R [ Nn + R ][ f RP v P + v P R ] P v Pn R N 2 n where R N = R N R N2... R Nn and = 2... n 0. (b) Design a circuit to obtain 2v N +v P +2v P2 The smallest resistor used should be 0 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 i N = v N R N. All that current flows out of the feedback resistor because the voltage across the other input resistors is zero. Thus, for only v N on, we get R f R N v N Similarly, by superposition, we can see that for only v Nx inputs turned on, then output is

2 R f n i= v Ni R Ni 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 and the parallel combination of all the other resistors. v + = = n i=0 i n i= i + v P v P + = ( + = + v P = v P )v P Next apply superposition and we get for all positive inputs on and all negative inputs off, v + = n i= v P Next, this signal is amplified by the non-inverting amplifier whose gain is composed from R f and the parallel resistance of all negative input resistors, R N, so [ + R ] n f v P R N R i= P 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 R f R N = 2 ( + R ) f RP = R N ( + R ) f RP = 2 R N 2 Note that R N = R N, so that + R f R N = 3, and 2

3 R f R N = 2 = 3 2 = 2 3 We can now select R N = 0kΩ, so that R f = 20kΩ. Next, Next, we have 2 = 2 = = 2 = = = = 0 This means that 0 =. Then we can choose = 0kΩ and 2 = 20kΩ For the circuit shown in Fig. P2.62, express v O as a function of v and v 2. What is the input resistance seen by v alone? By v 2 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 v O. First as a function of v, we have v. Next, as a function of v 2, R ( + R ) v 2 = v 2 R+R R Combining we get v 2 v. The input resistance on input is found as The relationship between the two is R i = dv di v = v 2 2 +i R 3

4 Plugging this into the expression above we get R i = R On input 2 it is clear that R i2 = 2R. For a source connected between the two inputs we have v 2 v = v, and a current through it, i, and we want the differential input resistance, R id = dv di We need a relationship between v and i. Note that v = v O. To have the same current in the two arms we must have that v v = v 2. Combining that with the input output reltionship, v = v 2 v, we can eliminate v or v v 2 = v = v 2 v v = v 2 We can also see that v 2 = 2Ri. Inserting we get v = 2Ri and see that R ic = 2R. 2.. An op-amp intended for operation with a closed-loop gain of 00 V/V uses feedback resistors of 0 kω and MΩ with a bias-current-compensation resistor R 3. What should the value of R 3 be? With input grounded, the output offset voltage is found to be +0.2 V. Estimate the input offset current assuming zero input offset voltage. If the input offset voltage can be as large as mv of unknown polarity, what range of offset current is possible? What current injected into, or extracted from, the nongrounded end of R 3 would reduce the op-amp output voltage to zero? For available ±5 V supplies, what resistor and supply voltage would you use? The bias-current compensating resistor R 3 = R R 2 = 9.9kΩ. With the bias compensated and zero offset voltage the output is equal to the offset current flowing through the feedback resistor, R 2, I OS = v O R 2 = = 20nA If the input offset voltage is up to ±V, the output offset voltage can be as large as ±0.V. In that case, the output offset due to bias current can range between = 0.V to = 0.3V. That results in a minimum offset current of I OS,min = 0nA and a maximum offset current of I OS,max = 30nA. In order to reduce the output to zero we need to add a voltage on the positive input which cancels the output of 0.2V. That means we need to add a negative voltage on the positive input which is that output divided by the non-inverting gain, 0. The voltage we 4

5 need to add is V compensation = 0.2V/0 = 2.08mV. That voltage is produced by a current through R 3. That current should be I compensation = V compensation = 2.08mV R 3 9.9kΩ = 20nA To get that current from the negative supply we need a resistor, R compensation = V supply I compensation = 5 = 7.4MΩ 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 equals that at the bottom end, so V = 3V. Then, I = 6V 0kΩ = 0.6mA. (b) The diode is not conducting, so I = 0, and V = 3V. (c) The diode is conducting so I = 0.6mA as before, but now V = 3V. (d) The diode is not conducting, so I = 0 and V = 3V In each of the ideal-diode circuits shown in Fig. P3.4, v I is a khz, 0V peak sine wave. Sketch the waveform resulting at v O. What are its positive and negative peak values? 5

6 (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 v O Minimum value: 0V. Maximum value: 0V 0 v I 0 v I v I > 0 (b) Piecewise expression for v O v I v I 0 Minimum value: 0V. Maximum value: 0V. (c) Current never flows, so 0 always. (d) This case is identical to (a) 0 v I > 0 (e) Current always flows so v I. Minimum values: 0V. Maximum value: 0V. (f) This case is identical to (a) (g) Piecwise expression for v O 6

7 v I v I 0 Minimum value: 0V. Maximum value: 0V. 0 v I > 0 (h) The output is always connected to ground, 0 always. (i) When v I < 0 this acts as a voltage divider. Otherwise they two are equal. Piecewise expression is vi 2 v I 0 v I v I > 0 Minimum value: 5V. Maximum value: 0V. (j) When v I > 0 the output is shorted to the input. When v I < 0 the output is obtained from the voltage division. The expression for v O is identical to (i). (k) The current source causes V voltage drop across the resistor, such that v O is always one volt higher than the voltage at the bottom of the resistor. When v I > 0 the voltage at the bottom of the resistor is ground. When v I < 0 the voltage at the bottom of the resistor is v I. The piecewise expression for v O is v I +V v I 0 V v I > 0 7