Practice questions for BIOEN 316 Quiz 4 Solutions for questions from 2011 and 2012 are posted with their respective quizzes.
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1 Practice questions for BIOEN 316 Quiz 4 Solutions for questions from 2011 and 2012 are posted with their respective quizzes. 1. [2011] When we talk about an ideal op-amp we usually make two assumptions. a) What are these assumptions? 1. There is no current into the two (inverting and non-inverting) inputs. 2. The voltages at these inputs are equal. b) When is each of these assumptions valid for a real amplifier circuit? The zero-current assumption is always good because the input impedance of op amps is very high. The equal-voltage assumption is good when the op-amp is operating in its linear range, i.e. when the output is not saturated. 2. [2011, 2012] (a) Explain what op-amp saturation is, with a few words and a graph of output vs. input voltage. (b) State which of the ideal op-amp assumptions it ruins. (a) Saturation is the condition when the output has reached the highest (or lowest) level permitted by the power supplies. (b) It ruins the assumption that the inverting and non-inverting input voltages are equal. 3. [2013] Suppose you build and test a circuit, and find that your output increases as you increase the positive power supply voltage but not as you increase the input signal. (a) What might be causing this strange behavior? Draw a figure to explain it. (b) What other unexpected or non-ideal measurement should you expect to make as you probe the voltages in your circuit? (a) This would be caused by saturation, because saturation is the condition in which the output has reached the limit(s) imposed by the power supplies. (b) I would expect to find the inverting and non-inverting input voltages to be unequal. 4. [2012] (a) What component would you use if you wanted very high input resistance and a gain of around 1000? (b) If you were building an electrophysiology amplifier (EEG, EKG, EMG, etc.) that isolates the patient from the 120-V power source, state where you would you put this component to provide the most benefit, and why it is good there. (a) I would use an instrumentation amplifier (INA). Just like an op-amp, its input resistance at both inputs is very high, but an instrumentation amp does not need require extra resistors that shunt current around it. (b) I would put it immediately adjacent to the patient, with just wires between the electrodes and the INA. This configuration allows only a
2 negligible amount of current to be drawn from the body. It would be on the patient side of the isolation amplifier so the signal can be amplified before it is passed through the isolation amp. 5. [2013] Difference amplifiers (made with an op-amp) and instrumentation amplifiers have similar functions, but instrumentation amps cost several times more. Name two advantages that justify the use of the instrumentation amp over the op-amp. 1. The in-amp s input impedance is higher, because both of the in-amp inputs go directly into the amplifier, which is similar to going directly into the inverting or non-inverting op-amp input. The difference amplifier has resistors that provide a lower-impedance path for current to go around the op-amp. 2. The in-amp (or INA) is simpler to use because it requires only one resistor to set the gain, compared to four for the difference amplifier. 3. In addition, the INA is optimized for measuring voltage differences, with high-precision resistors built into the integrated circuit. The resistors in your difference amp probably have a larger manufacturing error, giving the circuit a worse common-mode rejection ratio (CMRR), which is the ability of a difference-type amplifier to ignore the average of the two input voltages. 6. [2011] It is possible to make a good instrumentation amplifier out of two or three operational amplifiers. a. If you have only one op-amp, which of the amplifier circuits that we have analyzed could we use to provide the same basic function as an instrumentation amplifier? A difference amplifier. b. What is one advantage of using an instrumentation amplifier instead of an op-amp circuit? The advantages you could list are ease of use (need only one resistor to set gain), higher input impedance, and accuracy, since the integrated circuit is balanced internally and does not rely on the value of resistors with manufacturing errors. 7. [2013] Stages in an electrophysiology amplifier (EEG, EKG, EMG, etc.) might include filtering, amplification, and isolating the patient from the 120-V power source. Draw a figure or flow chart that shows the basic components that would achieve those functions. State why you have arranged the components as you did (1 sentence each). Sensor (convert variable of interest to electronic signal) -> INA (amplify signal from sensor without modifying the sensor s output) - > ISO amplifier (isolate patient from potentially dangerous voltages/currents on the display/doctor/power side of the device) -> LPF (reduce ambiguity caused by aliasing; may include other filter types) -> signal analysis & display. You might also include the
3 isolation power supply to power the INA and the patient side of the ISO amp. 8. [2012, 2013] Which of the op-amp circuits that we have studied has the highest input impedance and also includes resistors? (a) Draw the circuit and give its name. (b) Derive the gain from v in (t) to v out (t) in terms of resistances. (c) Determine as many resistances as necessary to achieve a gain of +81. (d) This circuit has great input impedance why not use it to amplify the output from a Wheatstone bridge? (a) The voltage follower has very high input impedance but it does not contain resistors. A non-inverting amplifier connects the input signal directly to the opamp s non-inverting input, so its Zin or Rin is the same as for the follower. (b) This is derived in the text to find vout(t) = (Rf/R1 + 1) vin(t). (c) A good rule of thumb is to keep your resistances in the 1 kω 1 MΩ range. Therefore, a good pair might be R1 = 1 kω and Rf = 80 kω = 33 kω + 47 kω; these last two values are part of the standard sequence of 5% resistor values: 10, 15, 22, 33, 47, 68, 82, 100. (d) A Wheatstone bridge produces a differential voltage, whereas the non-inverting amplifier has only one input. [2012] For the following questions, assume that you have the following components: Operational amplifiers, 2 μf capacitors, and resistors with values of 10 kω, 20 kω, 22 kω, 33 kω, 47 kω, 51 kω, and 1 MΩ. You should need only one of each for all of the circuits. 9. Draw a circuit with a gain of 100 at all frequencies. Calculate its input impedance. This is an inverting op-amp circuit. Our assumption that v pos = v neg means that one end of R1 is at Vin and the other end is at 0. Therefore, the ratio between Vin and input current (which is the input impedance) is R Draw a circuit with a gain of +5 at all frequencies. Calculate its input impedance. This would be a non-inverting op-amp circuit, which is also the answer to question 3. Its input impedance or resistance is nearly infinite. The gain for a non-inverting amp is (R2/R1 + 1). We can get G=5 with R2 = = 80 kω, R1 = 20 kω.
4 11. (a) Draw a difference amplifier that has a gain of 2.3. (b) Determine its input impedance. (a) This amplifier has two pairs of resistors, each of which should have a ratio of 2.3. Two of the many options of R2/R1 are 51/22 and 47/20. (b) Input impedance for v1 is R1 when v2 is constant; for v2 it is R1+R2 when v1 is constant. I calculate that Rinput is 2/(R1+R2) when v1 and v2 are raised equally. 12. [2012] Determine the DC gain (i.e. when ω=0) and cutoff frequency for one of the two circuits that we have studied that has a cutoff frequency. Extra credit: Draw the circuit. These are meant to be op-amp circuits, such as the low-pass filter shown here or a differentiator with C in series with R1. You will get credit if you calculate G(0)=1 and ω C =1/R 2 C from a passive RC circuit, but you need to draw an op-amp circuit if you want full extra credit (3 pts). A simple integrator (no R2) has a DC gain of and does not have a cutoff frequency; its slope is always -20 db/decade. Similarly, a straight differentiator has a DC gain of 0 and does not have a cutoff frequency; it is always +20 db/decade.
5 13. [2013] Suppose that we wish to convert an arterial pulse signal into a square wave to use as a clock signal for a digital counter. Let s consider three signals of increasing complexity. (a) Let v p1 (t) = 4 sin(2πt) V. Draw and name the simplest op-amp circuit that will convert this signal into a square wave. Draw the input and output signals on the same time axis. This would be a comparator: just an op-amp with the inverting input grounded. The average of v p1 (t) is 0 so we don t need a non-zero reference voltage. (b) Let v p2 (t) = 4 sin(2πt) + 5 V. Draw the simplest op-amp circuit that will convert this signal into a square wave. Include resistor value(s). This would need a reference voltage around 5 V, connected to the inverting input. Use a voltage divider connected to the power supplies (whose values I would have to give you). (c) Let v p3 (t) = 4 sin(2πt) sin(120πt) + 5 V. Draw and name an op-amp circuit that will convert this signal into a square wave. Show on the plot what parameters you would need to choose in order to select resistor values for this circuit. Draw the input and output signals on the same time axis. Write two equations that would, after much algebra, allow you to calculate these resistor values (you do not need to solve for the resistances). Briefly explain the operation of your proposed circuit. This would be a Schmitt trigger, which would produce two thresholds on V p3 for the transitions between positive and negative saturation.
6 14. [2011] Suppose that we wish to convert a smooth pulse signal into a square wave to use as a clock signal for a digital counter. The pulse signal is modeled as the sum of three sinusoids: v p (t) = sin(2πt) sin(4πt) sin(120πt) a) What are the period and fundamental frequency of this signal? b) Propose a circuit, including component values, that will produce one cycle of a square wave per fundamental period of the signal. c) Briefly explain the operation of your proposed circuit. 2 pulse signal, volts time, seconds a) The period is 1 second and the fundamental frequency is 1 Hz or 2π rad/sec. The signal also contains the first and 59 th harmonic frequencies, where the 1 st harmonic is 2x the fundamental and the 59 th is 60x the fundamental. The figure can t really show the 59 th due to wait for it Aliasing! b) A Schmitt trigger will be an effective circuit. (show figure) A standard comparator will not be effective because the signal crosses v in = 0 more than once per period. It is possible to improve things by low-pass filtering to remove the 60-Hz noise, but the secondary peak will still cause two pulses per period. c) The Schmitt trigger provides hysteresis that causes the output voltage to stay low (or high) until the input voltage has passed a particular positive (or negative) threshold as the input voltage rises (or falls). The threshold voltages are +Vsat(R1/R2) when input voltage is rising and Vsat(R1/R2) when input voltage is falling. R1 and R2 must be chosen to provide thresholds that are above (or below) the secondary peak (or trough).
7 15. [2011] In the following circuit, the downward-pointing triangles indicate ground. Each of the op amps is supplied +9 and 9 V as power. Two positive voltages are applied at terminals 1 and 2. Determine the function v(t) at each of the circuit locations, a-e, as a function of v 1 and v 2. Va = (V2 V1)*51/17. Vb = Va * 6.8/1.7 Vc = 470^2 * 10 6 * dva/dt Vd = 1/(600*300x10 6 *integral(va dt) Ve = 1/1*Vb + 1/2*Vc + 1/3*Vd 16. [2011] The previous problem six shows five operational amplifiers, which may be identified as a through e by the circled letters at their outputs. Each op-amp is combined with 2 resistors, or 1 resistor and 1 capacitor to make a sub-circuit. a. Which op-amp sub-circuit is most like a high-pass filter? a b c d e b. The sub-circuit from question 4.a is a poor filter because its gain just keeps increasing as frequency increases. Draw the sub-circuit and add one component to it to make it a better high-pass filter, i.e. one that has a finite cutoff frequency. We would add a resistor in series with the capacitor.
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