Chemistry 498A/C, Fall 1997

## Operational Amplifier Simulations

### Open loop configuration

1. Open the folder "Operational Amplifiers" on your simulations diskette and open the Hypercard stack "OpenLoop".

2. In this simulation, a operational amplifier is used "open loop", that is, without the usual external resistors. Because Vout is applied directly to the op amp's inverting input (-) and the non-inverting input (+) is grounded, the output voltage is given by Vout = -A*Vin, where A is the "open loop gain".

3. Vary the input voltage using the left-hand slider. Because A is typically very large, Vout is saturated unless Vin is very small. (In this simulation, the op amp is operated from ±15 volt power supplies, so the saturation output voltage is slightly less than 15 volts. If ±12 volt power supplies had been used instead, the output would have saturated at about ±11 volts).

4. Operational amplifiers are almost never operated in an open loop configuration such as this, for the reason that the open loop gain is much larger than usually needed and is also rather variable - that is, it varies a lot between different op amps, although it's always very large. It is standard practice to use a "closed loop" configuration in which the gain of the circuit is determined by external resistors rather than by the closed loop gain. The other stacks in this set will illustrate various closed loop configurations.

### Non-inverting amplifier

1. Open the Hypercard stack "Noninverting amp". This is a very common and useful amplifier circuit. Note that the input signal Vin is applied to the non-inverting (+) input. There are two external resistors that form a voltage divider that takes a fraction of the output voltage Vout and applies it to the inverting input (-). The voltage gain of this circuit (Vout/Vin) is determined by these two resistors.

2. Make the two resistors R1=10K and R2=90K (R1 is the one of the left). Vary the input voltage using the left-hand slider and observe the output voltage Vout. What is the voltage gain of the circuit? ________. Try other values of the resistors. Is there a pattern?

3. Observe the differential input voltage, Vdiff, the voltage between the + and - inputs, as you vary the input voltage Vin. Notice that the magnitude of Vdiff is always much smaller than either Vin or Vout, as long as the output is not saturated. In fact, Vdiff is so small that it is actually a useful approximation to treat it as zero (it's not really zero, of course, but it's just a useful mathematical simplification). Ohm's Law says that the output of the voltage divider is Vout(R1+R2)/R1. This is the voltage at the inverting (-) input. But this is almost exactly equal to the input voltage Vin, because the magnitude of Vdiff is so small. Therefore Vin = Vout((R1+R2)/R1) and the voltage gain of the circuit is Vout/Vin is very nearly (R1+R2)/R1. This is called the "nominal" closed-loop gain. Note how close it is to the actual closed-loop gain. Usually the % difference between these two is smaller than the accuracy of the resistors anyway, so there is no practical disadvantage to using this simplification. What effect does the open-loop gain (right slider) have on the gain?

4. One thing that is notable about the non-inverting amplifier is its very high input resistance. (Input resistance is defined as the ratio of the input voltage to the input current. The input voltage is of course Vin and the input current is just the differential input current, because nothing else is connected to the input). The value of this input resistance is displayed just below the circuit. It is much greater even than the differential input resistance (the resistance between the inputs) of the op amp. This makes the non-inverting amplifier a good choice for measuring voltages of very high internal resistance (such as pH electrodes).

### Inverting amplifier

1. Open the Hypercard stack "Inverting amp". This is another very common and useful amplifier circuit. Note that the input signal Vin is applied through a resistor to the non-inverting (-) input and the non-inverting (+) input is grounded . As in the previous circuit, there is a feedback resistor from the output to the non-inverting (-) input.

2. Make the two resistors R1=10K and R2=100K (left to right). Vary the input voltage using the left-hand slider and observe the output voltage Vout. What is the voltage gain of the circuit? ________. Why is this called an inverting amplifier?

Try other values of the resistors. How are the resistor values related to the gain of this circuit?

What effect does the open-loop gain (right slider) have on the gain of this circuit?

3. Note the magnitude of the differential input voltage, Vdiff. How does it compare to Vin and Vout? In inverting configurations such as this, the inverting (-) input is often called "virtual ground". Why?

4. The operation of this circuit can be understood by opening the stack "Current Flow", which shows the same inverting amplifier configuration, with arrows added to show the direction and magnitude of signal current flow. Vary the input voltage using the left-hand slider and observe the current flow. Note that because Vdiff << Vin, the voltage across R1 is very nearly Vin and the current through R1 is very nearly Vin/R1, by Ohm's Law. Almost all of this current flows through R2 rather than into the op amp inputs, as you can see by the fact that the differential input current is so small compared to the resistor current. This current (Vin/R1) flowing through R2 generates an Ohm' s Law voltage drop of (Vin/R1)*R2 = Vin(R2/R1). Again, because Vdiff << Vin, his voltage across R2 is very nearly Vout. Thus the output voltage is Vout = -Vin(R2/R1). The minus sign arises in the following way. Current flowing from left to right through R2 makes the left end positive with respect to the right end (and vice versa). The left end of R2 is connected to "virtual ground" and the right end is connected to Vout. Thus, Vout has the opposite sign of Vin.

What happens if R2 is greater than the differential input resistance (normally 1 MOhm in this simulation)? Is the differential input current still small compared to the current through R2 in that case? Does the current follow the "path of least resistance"? Explain

5. An important difference between the inverting and non-inverting amplifier configurations is the input resistance. Which of these circuits has the higher input resistance? Why is the input resistance of the inverting amplifier simply equal to R1?

### Photodiode Photometer

1. Open the Hypercard stack "Photodiode Photometer". This is a "current-to-voltage" converter that is used to measure the output of current-source transducers such as the silicon photodiode (to measure light intensity). Note that this is basically an inverting amplifier in which the input resistor has been replaced by a photodiode and in which the input voltage is fixed (-15 vols in this case).

2. Vary the light intensity on the photodiode with the slider and observe the photodiode current (the "photocurrent"). Vary the feedback resistor and determine the relationship between the photocurrent, the feedback resistor and the output voltage Vout.

3. What value of feedback resistor would you wanted to make the maximum on the light intensity slider generate a Vout of exactly 10 volts?

4. Is there any limit to how high the feedback resistor can be?

1. Open the Hypercard stack "Adder". Compare this to the inverting amplifier. What additional circuit component has been added?

2. Vary the two input voltages V1 and V2. How is the output voltage Vout related to these two input voltages?

3. How does this circuit perform its addition function? Hint: what happens to the currents through the two input resistors when they meet at the inverting input?

4. What resistor values would you use if you wanted to make the output equal to ten times the sum of V1 and V2?

### Subtractor

1. Open the Hypercard stack "Subtractor". Compare this to the inverting amplifier. What additional circuit components have been added?

2. Vary the two input voltages V1 and V2. How is the output voltage Vout related to these two input voltages?

3. What resistor values would you use if you wanted to make the output equal to ten times the difference between V1 and V2?

### Integrator

1. Open the Hypercard stack "Integrator". Compare this to the inverting amplifier. Note that R2 (the feedback resistor) has been replaced with a capacitor. Two switches have been added to control the integrator.

2. To see how this works, set the input voltage to any non-zero value and observe the output voltage as you click and release the mouse button on the run/hold switch. Click on the reset switch to reset the output voltage back to zero. Vary the input voltage and observe the effect.

3. One application of the integrator circuit is as a "ramp" generator, to generate a linearly increasing or decreasing voltage. This is used for example in polarography to scan the potential of the working electrode. In this application the input voltage Vin is held constant; the rate of change (ramp rate) of Vout is controlled by Vin, R, and C. If you wanted to create slower ramp rates, should you increase or decrease Vin, R, and C?

4. Another application of the integrator circuit is to measure the "area under the curve" of peak-shaped signals, such as those obtained in gas or liquid chromatography instruments. In this case the output signal of the chromatograph is applied to the input of the integrator. For each peak that occurs, the output of the integrator will show a step whose height is proportional to the area under the curve of that peak. This works best if the peaks start and end at zero volts. You can try this on the simulator. Start with Vin at zero, click the reset switch, then make a "peak" with the slider that goes from zero to some peak value and then back to zero. Record Vout. Now repeat this slowly, creating a peak that is wider but reaches the same peak voltage. Compare Vout to the previous.

### Differentiator

1. Open the Hypercard stack "Differentiator". Compare this to the inverting amplifier. Note that R1 (the input resistor) has been replaced with a capacitor.

2. This circuit can be used to detect and measure the changes in signals - i.e. to determine how fast a voltage is changing. How is the Vout related to the Vin in this circuit?