ECE 3410 – Homework 2 Problem 1. The non-inverting amplifier shown below uses a non-ideal op amp with a finite openloop gain. R2 − Vo R1 + vIN − + (A) Calculate the ideal gain if the op amp has A → ∞, R1 = 1kΩ and R2 = 9kΩ. Solution R2 R1 = 10 V/ V G? = 1 + (B) Calculate the actual gain for the values in (A) if the gain is A = 100 V/ V. Solution For the non-inverting configuration: ? G=G A A + G? 100 = (10 V/ V) 110 = 9.0909 V/ V (C) Suppose the op amp has a finite open-loop gain of A = 1000 V/ V, and it is desired to have an actual circuit gain of G = 10V/ V. What is the ratio R2 /R1 that will achieve the desired gain? Utah State University 1 ECE 3410 – Homework 2 Solution G=G ? A A + G? ⇒ GA + GG? = AG? GA ⇒ G? = A−G Since G? = 1 + R2 /R1 , the solution is R2 GA = −1 R1 A−G = 9.10101 (D) Lastly, if the open loop gain varies from 100 to 1000 V/ V, and R1 is held constant at 1kΩ, what is the range of adjustments needed in R2 to maintain an actual gain of G = 10 V/ V? Solution Using the solution from (C), we may simply adjust the value of A to find the range of R2 : GA R2 = R1 −1 A−G = 10.1111kΩ(high end) = 9.10101kΩ(low end) Utah State University 2 ECE 3410 – Homework 2 Problem 2. The transresistance amplifier shown below is to be used in a sensor interface. The resistor Rs is an unknown resistance, which will be measured by the following method: (i) Apply a known potential Vref to the op amp’s non-inverting input terminal. Due to the virtual short effect, the same potential (or nearly the same) should then appear at the inverting input terminal. (ii) Measure the potential Vo appearing at the op amp’s output terminal. (iii) Based on Vo , estimate the value of the current flowing through Rs . The resistance is then Rs ≈ Vs /Is . Transresistance Amplifier Unknown Resistor RF Vs Rs − Vo Is + Vref − + (A) If the op amp is ideal, obtain expressions for these items: Vs , Is , and Vo . For each item, the left-hand side of your expression should only include Rs , Vref and/or RF . Finally, obtain an expression for Rs . Solution Vs = Vref Vref Is = Rs RF Vo = Vref 1 + Rs (B) Repeat (A) for the case where the op amp has finite gain equal to A V/V. This time, the left-hand side of each expression should only include Rs , Vref , RF and/or A. Utah State University 3 ECE 3410 – Homework 2 Solution The previous equations must be modified as follows: Vs = Vref − Vo A Vs Rs Vo = Vs + Is RF Is = After some algebra, the final solutions are " # A Vs = Vref A + 1 + RRFs " # A Vref Is = Rs A + 1 + RRF s " RF A Vo = Vref 1 + Rs A+1+ # RF Rs (C) Suppose the op amp is ideal, Rs = 1kΩ, RF = 2kΩ, and Vref = 1V. What are the values of Vs , Is and Vo ? Solution Vs = 1V Is = 1mA Vo = 3V (D) Repeat (C) for an op amp with finite gain A = 100 V/V Solution If the op amp has finite gain, then the corresponding error coefficient is " # A 100 = = 0.97087. RF 103 A + 1 + Rs We may obtain the solution for the finite gain case by multiplying this error Utah State University 4 ECE 3410 – Homework 2 Solution (cont.) coefficient into the previous solutions from (C): 100 Vs = 1V = 0.97087V 103 100 Is = 1mA = 0.97087mA 103 100 = 2.91262V Vo = 3V 103 . (E) For the circuit described in (C) and (D), suppose Rs can vary between 500Ω and 5kΩ. What is the corresponding range of variation that may be seen in Vo ? Solution Any variation in Rs affects the error coefficient. For this range of variation, the error coefficients vary as A (case1) = 0.82644628 RF A + 1 + 500Ω A (case2) = 0.9708737864 RF A + 1 + 5kΩ Then, multiplying these into the ideal results gives: 10kΩ Vo (case1) = (1V) 1 + (0.82644628) 500Ω = 17.35537V 10kΩ Vo (case2) = (1V) 1 + (0.9708737864) 5kΩ = 2.912621359V (F) Estimate the value of Rs if Vo = 1.5V, RF = 10kΩ, Vref = 1V and A → ∞. Utah State University 5 ECE 3410 – Homework 2 Solution Since this is the ideal case, we can obtain the solution immediately: Rs = Vs Is Vref (Vo − Vref ) /RF Vref = RF Vo − Vref = 30kΩ. = Problem 3. The difference amplifier shown below uses an op amp, but suffers from mismatch in its resistor values. By design, R2 = R4 = 10kΩ, and R1 = R3 = 1kΩ. In reality, all resistors vary within a ±5% tolerance. R2 − − vIN R1 vOUT + + R3 R4 For this circuit, recall that there are two signal paths, called the inverting path and the non-inverting path. The goal is to balance the gains along these signal paths: R2 inverting path : − Gi = − R 1 R2 R4 non − inverting path : Gni = 1 + R1 R3 + R4 Notice here that Gi is defined as a positive number. Based on these definitions, you can show that the differential gain and the common-mode gain are: 1 (Gi + Gni ) 2 = |Gi − Gni | Ad = ACM Utah State University 6 ECE 3410 – Homework 2 (A) If all the resistors were perfectly matched to their specified values, what should be the circuit’s differential gain, common-mode gain and CMRR? Solution With perfect matching, we have: R2 = 10 V/ V R1 Acm = |Gi − Gni | = 0 Ad =∞ CMRR = 20 log10 Acm Ad = (B) Suppose all resistors are mismatched from their specified values as follows: R1 : 5% too large R3 : 5% too small R2 : 5% too small R4 : 5% too large In this case, calculate the resistor values, the differential gain, the common-mode gain and the CMRR in dB. Solution The values are: R1 = 1050Ω R2 = 9500Ω R3 = 950Ω R4 = 10.5kΩ With these values, the gains are R2 = 9.0476 R 1 R2 R4 = 1+ = 9.2140 R1 R3 + R4 Gi = Gni Since these are mismatched, we expect to see some common-mode gain: Ad = 9.1308 Acm = 0.16635 CMRR = 20 log10 Ad Acm Utah State University = 34.789dB 7 ECE 3410 – Homework 2 Problem 4. An inverting op amp configuration is shown below. Also shown are current and voltage sources inserted to model the op amp’s non-ideal bias current and offset voltage. R2 vin R1 − v− Ibias vout + − + Vofs Ibias (A) For this problem, suppose that Vofs = 0, Ibias = 20µA, and R1 = 10kΩ. The op amp’s supply rails are at ±5V, and the op amp has infinite open-loop gain. If the input signal is a sinusoid with zero DC offset and a peak-to-peak amplitude 10mV, what is the maximum gain (i.e. the maximum value for R2 ) that can be allowed without saturating the op amp? Solution Saturation will happen when the maximum output signal voltage equals the rail. Accounting for the effect of Ibias , we have: R2 max vin − + R2 Ibias = VR R1 The maximum value occurs where vin = −5mV, so the equation can be simplified to R2 (5mV) + R2 Ibias = VR R1 Solving for R2 gives R2 = 5mV R1 VR + Ibias Filling in all the known values, we find that R2 can be no greater than 243.9kΩ, corresponding to a maximum gain of 24.39 V/ V. (B) For this problem, suppose that Ibias ≈ 0, R2 = 5kΩ, R1 = 1kΩ and the op amp is otherwise ideal. If Vofs varies in the range of ±10mV, what is the range of DC offset voltages that may appear at VOUT ? Utah State University 8 ECE 3410 – Homework 2 Solution The offset voltage is multiplied by the op amp’s non-inverting gain, 1 + R2 /R1 = 6. So the range of possible output offset voltages is ±60mV. Utah State University 9