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Phys 4051: Op-Amp Homework Due 10/15/04 In class you were shown the output of a photo multiplier tube when a particle passes through a detector. So far none of the circuits discussed allowed for both, an adjustable hysteresis setting and an adjustable threshold level. The circuit on the right fulfils this requirement. Figure out how it works. Specifically, on the graph below, draw Vout vs. Vin, V+ vs. Vin and ∆Vin vs. Vin for this circuit. (∆Vin ≡ V+ - V-). Find the numerical values for the lower and upper thresholds voltages. Use your own approach to solve this problem but if you get stuck, suggestions will be posted on our website. The link to the suggestions can be found on the announcement page at :http://mxp.physics.umn.edu/f04/Announcements.htm Use the following information to solve this problem: • V++ = -V-- = 12 V. • R1 = 1k • R2 = 3k • V- = -3V Assume that it is an ideal op-amp with infinite (open) loop gain and infinite input impedance and zero output impedance. Hints & Answers 1. 2. 3. 4. What type of feedback does this circuit use? • Positive Feedback which means the Golden Rules of Op-Amps definitely do NOT apply here; instead we need to use Vout = A ∆Vin where A is infinity. Based on 1) what output voltage(s) would you expect to measure? • Since it has positive feedback, the circuit is bistable, i.e., Vout can only be V++ or V--. Since Vout = A ∆Vin where A is infinity, if ∆Vin > 0, Vout = V++ and if ∆Vin < 0, Vout = V--. Find V+ as a function of R1, R2 and Vout. • It’s voltage divider; knowing that the current through R1 must be identical to the current through R2 we get: (Vin – V+)/R1 = (V+-Vout)/R2. Solving this for V+ yields: V+ = (Vout R1 + Vin R2)/(R1 + R2) The output changes when ∆Vin changes sign, i.e., at ∆Vin = 0. From the information above and remembering that ∆Vin ≡ V+ - V-, determine at what input voltage, Vin, this happens. Note, based on your answer in step 2, you should obtain two (different) threshold values. Indicate these values along the x-axis of the plot! • ∆Vin ≡ V+ - V- = (Vout R1 + Vin R2)/(R1 + R2) – (- 3V) = 0 (0 Volt is when the output switches!) Solving this for Vin yields: Vin = -3V (R1 + R2)/R2 – Vout R1 /R2 Plugging in the values for R1, R2 and Vout we get: Vin( Vout = V++ = 12V) = -8V Vin( Vout = V-- = -12V) = 0V. These, -8V and 0V, are two input voltages that make the output switch! In other words, they are the upper and lower input threshold voltages that are responsible for the hysteresis. The only thing left is to determine a complete hysteresis cycle. (You could “guess” this just by looking at the circuit and noting that it is a non-inverting comparator; one could draw the behavior of such a noninverting comparator directly with the above thresholds. The following steps though may give you a better feel for the details involved.) Make the following “Gedanken Experiment” to figure out how and when the output switches: 5. 6. 7. 8. 9. Start out at an extreme input voltage, for example, Vin = V++. At this point Vout is uniquely determined. What is it? What is V+? • At this point you have no choice but to make an educated guess and then check your answer: Let’s say your first guess was Vout = 12V. Then from your results in part 3 or by inspection, you can see that V+ = 12 and ∆Vin > 0, i.e., Vout = 12 V is consistent with this guess. If, on the other hand, your guess would have been Vout = –12V, then V+ (from your results in part 3) would yield +6V; checking ∆Vin shows that ∆Vin > 0, i.e., Vout cannot be –12V and +12V is the only consistent answer! Reduce your input voltage down to V-- and assume (for now) that the output voltage does not change. Temporarily plot V+ vs. Vin. • The equation for V+ has already been obtained in step 3 and since we know now that Vout = +12V, V+ = (Vout R1 + Vin R2)/(R1 + R2) = (12 + 3 Vin)/4 Use your expression for ∆Vin and (temporarily) plot it vs. Vin. • ∆Vin ≡ V+ - V- = (12 + 3Vin)/4 –(-3V) = 6 + (¾) Vin Vout will switch its state when ∆Vin crosses 0 V. From your graph from step 7 you should be able to determine this point; draw the Vout change to the new Vout level. Indicate on your graph with arrows the direction of the change. Also, does the switching point agree with the threshold determined in step 4? • From the previous equation, ∆Vin = 0 when Vin reaches -8V, i.e., the output now switches to its negative output voltage. To summarize, if Vout was 12V then the input voltage has to go all the way down to -8V to switch the output voltage –12 V. How does the circuit behave when the input voltage returns from V- - to V++? • Briefly, for Vin = V--, Vout = V--. (See step 5.) Now increasing the input voltage yields: V+ = (Vout R1 + Vin R2)/(R1 + R2) = (-12 + 3 Vin)/4 (Step 6.) Determining ∆Vin ≡ V+ - V- = (-12 + 3Vin)/4 –(-3V) = 0 + (¾) Vin. (Step 6.) In other words, Vout will switch when ∆Vin = 0 which is reached when Vin = 0V. At this point the output will go all the way positive. 15 10 5 Vout V+ V- = Vthr 0 -12 -9 -6 -3 0 -5 -10 -15 Vin 3 6 9 12 V+ - VVout