Mechanical Engineering - Answers

Homework 5 1. In the circuit below: a) Determine the gain of this amplifier. b) Is it inverting or non-inverting? Vi 3 9 Vo + 0 Ans: Treating the (-) terminal of the op-amp as a node, we see that its voltage is zero, since the (+) terminal is at ground. (Remember: V+-V-=0) Therefore, we can write the nodal equation: Vi  0 0  Vo  3 9 Remember: no current flows in or out of the (-) or (+) terminals on the op-amp. Reducing and rearranging variables: Gain  Vo 9    3 Vi 3 This gives us the gain and the minus sign identifies this as an inverting amplifier. 2. In the circuit below: a) Determine the gain of the amplifier. b) Is it inverting or non-inverting? 4 2 0 Vo Vi + Ans: We use the (-) terminal of the op-amp as a node again. The voltage at the (-) terminal is the same as the voltage on the (+) terminal, i.e. Vi. The nodal equation is: 0  Vi Vi  Vo  4 2 Rearranging and combining: Gain  Vo 2  1   1.5 Vi 4 The positive gain identifies this as a non-inverting amplifier. 3. In the circuit below: a) Determine the gain of the amplifier. b) Is it inverting or non-inverting? Vi 10 2 Vo + 0 Ans: As in Problem 1, treating the (-) terminal of the op-amp as a node, we see that its voltage is zero, since the (+) terminal is at ground. (Remember: V+-V-=0 ). Therefore we can write the nodal equation: Vi  0 0  Vo  10 2 Remember, no current flows in or out of the (-) or (+) terminals on the op-amp. Reducing and rearranging variables: Gain  Vo 2    0.2 Vi 10 This gives us the gain and the minus sign identifies this as an inverting amplifier. 4. What would R2 have to be to make the gain of this amplifier -10? Vi 10 R2 Vo + 0 Ans: As in the inverting op-amp problems above, we treat the (-) terminal of the op-amp as a node and see that its voltage is zero, since the (+) terminal is at ground. (Remember: V+-V-=0 ). Therefore we can write the nodal equation: Vi  0 0  Vo  10 R2 Remember, no current flows in or out of the (-) or (+) terminals on the op-amp. Reducing and rearranging variables to solve for R2: Gain  Vo R   2  10 Vi 10 R2  10  10  100 5. Design a non-inverting amplifier with a gain of +5. Ans: We would use the non-inverting amplifier topology shown below: R1 0 R2 Vo Vi + We use the (-) terminal of the op-amp as a node again. The voltage at the (-) terminal is the same as the voltage on the (+) terminal, i.e. Vi. The nodal equation is: 0  Vi Vi  Vo  R1 R2 Rearranging and combining: Gain  Vo R  1 2  5 Vi R1 R2 4 R1 Since, we only have one equation and two unknowns, we can choose the value of R1 or R2 and then calculate the value of the other resistor based on the relation above. So we choose R1 to be 10 and that makes R2 =40. 6. What is Vo in terms of Va , Vb and Vc? Va 2 Vb 2 4 Vc 4 Vo + 0 Ans: Since the voltage at the (+) terminal of the op-amp is 0, the voltage at the (-) terminal is also 0. We can use superposition to build up the output relation by applying each input voltage and shorting out the other inputs in turn. To start, we apply Va and short Vb and Vc inputs to ground. Since the (-) terminal is also at ground, no current flows through the resistors to the inputs shorted to ground. They can be ignored in the nodal analysis. (REMEMBER: You CANNOT always do this. ONLY when the voltage is 0 across the resistors.) Using nodal analysis as in the inverting amplifiers above, the relation for Va is then simply: 4 Vo   Va  2Va 2 Similarly, we can develop the relations for Vb and Vc. 4 Vo   Vb  2Vb 2 4 Vo   Vc  Vc 4 Superposition gives us the total relation as the sum of the responses: Vo  2Va  2Vb  Vc 7. What is Vo in terms of Va and Vb? Va 2 2 5 10 Vb + 10 Vo + 0 0 Ans: We can break the circuit into two parts, analyzing the first amplifier as an inverting amplifier with a gain of -1. This places –Va at the output of the first op-amp and the input to the second part of the circuit. Using superposition, we can carry this input to the final output, Vo, recognizing the second part of the circuit as an inverting amplifier with gain: Vo   10 (Va )  2Va 5 Continuing the superposition with Vb: Vo   10 Vb  Vb 10 So the total response is: Vo  2Va  Vb 8. Design an op-amp circuit to carry out this equation: Vo  Va  2(Vb  Vc ) Ans: There are a number of circuits that could carry this out. But the circuit in Problem 7 gives us the easiest solution: Vb 100 100 Vc 50 - 100 Va 100 100 Vo + + 0 0 The first op-amp is configured as an inverting amplifier with a gain of -1 for both Vb and Vc. The output of the first op-amp is carried through the second op-amp with a gain of -2 (100 divided by 50). So the final response to Vb and Vc is -1*-2=2. The second op-amp is also an inverting amplifier for Va with a gain of –(100/100)=-1. So the total response is: Vo=-Va+2(Vb+Vc). Note: Different valued resistors could be used, however, the ratios must be the same. 9. What is Vo in terms of Va , Vb and Vc? Va 2 Vb 2 3 Vo Vc + 3 1 0 Ans: As in the problem 1 above, the (-) terminal of the op-amp forms a node to analyze the circuit. However, the voltage at this node is NOT 0. We look at the (+) terminal and see that the 3 ohm and 1 ohm resistors form a voltage divider with Vc. The voltage at the (+) terminal is then: V  Vc 1  0.25Vc (1  3) This is also the voltage at the (-) terminal. We can now write the nodal equation: Va  0.25Vc Va  0.25Vc 0.25Vc  Vo   2 2 3 Solving for Vo in terms of the inputs gives: 1.5Va  1.5Vb  Vc  Vo 10. Try problem 9 with superposition. That is, turn on each of the inputs in turn with the others shorted to ground, analyze the response and sum the resulting outputs. Ans: Starting with Va, we short the other inputs to ground. The key thing to remember here is that the 3 ohm and 1 ohm resistor have no current through them to the (+) terminal since no current can flow in or out of the input terminals on an op-amp. This places the (+) terminal at ground and consequently the (-) terminal is also at 0 volts. We can then analyze the circuit as a simple inverting amplifier from Va. Note: since the Vb input is grounded and the (-) terminal is also at 0 volts, no current flows in the 2 ohm resistor to Vb and it does not enter into the calculation. The gain of the inverting amplifier is as we saw above: -3/2 or 1.5. So the response to Va is -1.5Va. In the same way, we can short Va and Vc and then apply Vb. The analysis is the same as for Va and the output due to Vb is -1.5Vb. Now, things can get difficult if you are not careful. Shorting Va and Vb to ground and applying Vc , we see that the 3 ohm and 1 ohm resistors form a voltage divider and the voltage at the (+) terminal is 0.25Vc. This voltage is also at the (-) terminal. The key is to remember that this voltage causes current to flow in BOTH 2 ohm resistors to the shorted inputs for Va and Vb. The total current is: I 0.25Vc 0.25Vc   0.25Vc 2 2 Nodal analysis says that this current flows from the 3 ohm resistor connected to Vo. This gives us the drop across the 3 ohm resistor and that plus the voltage at the (-) terminal gives us Vo: Vo  3I  0.25Vc  3(0.25Vc )  0.25Vc  0.75Vc  0.25Vc  Vc Together with the results for the inputs above, the total output is: 1.5Va  1.5Vb  Vc  Vo The same result as before.