egr285 book chp 5, 2026Feb18

5 Operation Amplifiers Study Objectives • Know the 2 operating modes of the op amp and when they apply. • Know the given op amp models when in the linear mode. • Know how to solve an op amp circuit using various models. • Be able to recognize common op amp circuits such as adders, multipliers. History In World War II op amps were invented to perform mathematical operations like addition, multiplication, integration, and differentiation. At the time the war department needed to solve differential equations in order to determine the path of bomb projectiles. The op amps were used in analog computers which solved the differential equations. Prior to that people would do these calculations by hand and those people were called computers. Between WW II and the 1980s analog computers were important machines for solving differential equations. In resent years the digital computer has become powerful enough to do this job. 18. Feb. 2026 5 Operation Amplifiers 1.

5.1 Theory Op amps are used for many other applications such as in radios, sensors, amplifiers, etc. and are still very important in electronics. They can be made very small and very cheaply. Unlike a resistor the op amp needs a power source to activate it. For this reason it is called an active device, resistors are passive. We will learn how to analyze an op amp in a circuit and how to use the op amp to condition or scale a signal, compare two signals, add signals, integrate signals, etc. It can also be used to make a digital to analog converter. The op amp has 2 main modes (linear and comparator). The circuit is designed so that the op amp is in one of these modes. This book focuses on the linear mode applications. Diagram showing relationship between Output and Input difference for Op Amp. If the gain, A, is 100,000 and V+ subtract V- then Vout = 0 0 .00002 2 .00010 10 .00015 12 .01000 12 18. Feb. 2026 5 Operation Amplifiers 2.

Comparator Mode. This mode is not covered in this chapter. Is often used in making a comparison of 2 voltages and the output will either be +Vcc or -Vcc. In this mode V- and V+ will differ by an amount (about .00012 volts or more) such that the output is at a maximum. Thus the output is at the rails which is either +Vcc or -Vcc. Remember Vout = A*(v+ subtract v-). For example, one may want to know when the voltage from a temperature sensor is above 1 volt. To do this connect V- to 1 volt, connect the sensor to V+ and monitor Vo to see when it is +Vcc. Linear Mode. In this mode the op amp operates such that the output, Vo, does not reach the rails. The rails are +Vcc and –Vcc as shown by the graph. In this mode the op amp can be modeled as an amplifier. The output equation is: Vo = A*Vin. = Gain * Vin = 100,000 * (v+ subtract v-) This equation is a linear straight line relationship. This line is very steep. On the order of 100,000 or more for the slope. With this value of A, when Vo reaches Vcc, which is typically 12v, then Vin would be .00012 volts. This is practically zero. Thus in this mode we may say Vin is nearly 0. That means V- essentially equals V+ within .00012 volts. Again, this means V+ and V- are very nearly equal in the linear mode. If v+ increases compared to v- then Vo increases. If there is a feedback resistor going from the output back to v-, then the increased voltage at Vo causes v- to increase and match v+. Said another way, the feedback resistor provides a means for v- to change and to track changes in v+. Thus they stay equal to each other, nearly. As an example, this mode can be used to amplify the small voltage from a LM35 temperature sensor before going on to an A/D converter. 18. Feb. 2026 5 Operation Amplifiers 3.

LM741 This is a very common op amp package. The cost is a few dimes if that. The op amp is contained in an 8 pin chip as shown. Power must be applied. Pin 7 (Vcc) is typically 12 to 18 volts, Pin 4 (-Vcc) is typically -12 to -18. The inverting input, V-, is on pin 2. The non-inverting input, V+, is on pin 3. The output, Vo, is on pin 6. We often just draw the three connections V-, V+ and Vo while leaving out the power supply connections in order to simplify the drawing. But keep in mind that power must be applied and that current flows through it. Models for the Op Amp Electrical symbol of the op amp. A mathematical model describes how something works. For example, Ohms law, V = IR, is a model for the resistor. There are many models representing an op amp, each with each with varying levels of accuracy and complexity. The simplest and yet fine model is the ideal model. 18. Feb. 2026 5 Operation Amplifiers 4.

5.2 Linear Mode vs. Comparator Mode. Usually one wants to operate in only one of these 2 modes. In either case, the output can never be more than the rails. The rails refer to the value of the positive power supply and negative power supply value. In the linear mode, the output should never reach the rails (See first picture in chapter). Using the equation on the first page, and a gain of 100,000 then: Vo = 100,000*(V+ subtract V-). This means V+ and V- are very nearly equal in the linear mode. For v+ and v- to be practically the same consider this next circuit. If v+ increases compared to v- then Vo increases sending more current through Rf, through Ri and thus increasing the voltage at v-. Said another way, the feedback resistor provides a means for v- to change and to track changes in v+. Thus they stay equal to each other, nearly. By contrast, in the comparator mode, V+ and V- are typically not very close. This means that the output will either be at the plus or minus rail because Vo = A*(V+ minus V-) = 100,000*(some non-zero value) which evaluates to a large number. But again the output cannot be a larger than the rails. This mode is used to determine if the output of a sensor is above or below a certain voltage. The output of the comparator Op Amp could them be sent to a microprocessor to make some decision. 18. Feb. 2026 5 Operation Amplifiers 5.

5.3 Ideal Op Amp Model. The ideal op amp model is a very simple model of how the op amp operates. It is the easiest model to use and it is fairly accurate in representing how an op amp works. In an ideal world the op amp would have ideal characteristics. Ideally we would like our op amps to work this way. Here are the rules of an ideal op amp: Operating Conditions for an Ideal Op Amp: 1. No current goes into v-, inverting node 2. No current goes into v+, non-inverting node 3. The voltage at the inputs are equal, v- = v+ These 3 rules will help us solve and understand how the op amp works in the circuit. Typically one applies a KCL at one of the op amp input nodes when solving these circuits. Optional: The ideal op amp has the following qualities:  infinite gain  infinite frequency response  input terminals draws no current  infinite input resistance  output resistance is zero  power supply has no limits 18. Feb. 2026 5 Operation Amplifiers 6.

5.4 Finite Gain Model This model is a little more complicated. It says:  The input V- may be different than the input V+ by a small amount.  Current into the op amp inputs = zero.  Vout = A (V+ subtract V-) where A is the open loop gain.  Open Loop Gain, A, is typically 105 to 107 Note. You will not know the open loop gain, A, and should not design a circuit by relying on its value. 5.5 Finite Gain Model with input and/or output resistance This model shown next is yet a little more complicated. In addition to the finite gain attributes it has:  an input resistance, Rin, typically very high  an output resistance, Ro, typically very low  an open loop gain of A, which might be 1E5 to 1E7. 5.6 Real Op Amps An op amp in real life has the following qualities. When working with op amp circuits it is important to keep these in mind.  The output current has a limit. Usually < 10 mA.  The output voltage has a limit. Typically within 1 volt of the power supply.  Finite gain. Open loop gain, A, is about 105 to 107. Vout = A(V+ - V-)  Inputs will draw a small current. On the order of micro amps.  Minimum value of R load is generally about 2K 18. Feb. 2026 5 Operation Amplifiers 7.

5.7 Op Amp Examples For most examples we will be using the ideal op amp model. Example 1 - Inverting Amplifier. For this circuit find vo/vi using the ideal model. The rules of the ideal op amp say that the input nodes (v- and v+) of the op amp have the same voltage. In other words v- = v+. Note that the v+ is the non-inverting input while v- is the inverting input. Thus here both inputs of the op amp have zero volts because v+ is connected to ground. Next do a KCL at v-. There are 3 wires to consider. Remember that: current = voltage across / resistance. KCL at V-, 0 = (Vi-0)/Ri + (Vo-0)/Rf + 0 Solving: Vo = - Rf/Ri Vi This means that the output is –Rf/Ri times the input Vi. This circuit is called an inverting amplifier. You can use this circuit to scale a signal. If Ri=2k and Rf= 6k then what is the output for the following values: Vi = 0, 2, -3, -8? Answer: Vo = 0, -6,9, 12. Check out that last value. The output cannot be greater than the power supply value and that is why we get 12 instead of 24. Two inverting amplifiers in series would yield a non-inverting amplifier. 18. Feb. 2026 5 Operation Amplifiers 8.

Example 2 - Non-Inverting Amplifier. For this circuit find vo/vi. As you can see this circuit is quite similar. In fact, we will take the same approach. The rules of the ideal op amp say that the input nodes (v- and v+) of the op amp have the same voltage. In other words v- = v+ and thus both inputs of the op amp are at Vi volts. If Vi is 10 volts then v+ = v- = 10 volts. Note that the v+ is the non-inverting input while v- is the inverting input. Next write a KCL at v-. There are 3 wires to consider at v-. Remember that current = voltage across / resistance. KCL at v-, 0 = (0-V-)/Ri + (Vo-V-)/Rf + 0 Solving: Vo = (1+Rf/Ri)Vi This means that the output is (1 +Rf/Ri) times the input. This circuit is called a non- inverting amplifier. You can use this circuit to scale a signal. If Ri = 2k and Rf = 6k then what is the output for the following values: vi = 0,2, -3, -8? Answer: Vo = 0, 8, -12, -15. Check out that last value. The output cannot be greater than the power supply value and that is why we get -15 instead of -32. As an example, this circuit can be used to amplify the small voltage from a LM35 temperature sensor before going on to an A/D converter. 18. Feb. 2026 5 Operation Amplifiers 9.

Example 3 - Voltage Follower. For this circuit find Vo. Here Vo is the voltage of Rload. Remember that for the ideal op amp V+ = V- and there is no current going into either op amp input. That means there is no current through Rth and no voltage across it. On the left of Rth the voltage is Voc and on the right of Rth is Voc (again the voltage across it is 0). Next V+ is = to Voc, thus V- is also equal to Voc and then the output voltage of the op amp is equal to Voc too. Finally Vo is also equal to Voc. Why have the op amp there if it acts like a wire? Why not just do this: In the above circuit without the op amp, if the load resistance is low then the load starts to draw lots of current and that current comes from the voltage source. If the source Voc is not powerful or is a weak communication signal, it may not have enough power to run the load. That is where the op amp comes in. It isolates the weak Voc source from the load. The op amp gets its power from a power supply and uses that energy to effectively boost the power of the weak signal Voc. This configuration of the op amp is also called a buffer/driver. It buffers the source. It can also increase the fan out so as to be able to drive several loads at the same time. These are used by a TV cable company to boost the TV signal periodically along the road going to the customers. The next circuit uses the voltage follower to increase the fan out. The original signal would not be strong enough to power all of the loads by itself. 18. Feb. 2026 5 Operation Amplifiers 10.

Example 4 - Adder or D/A Conversion This design is of an adder circuit. It can also be used in digital to analog conversion because the ratio of the resistors is 2 to 1 at each input. The circuit adds the 4 signals v0, v1, v2, v3. To solve for Vo write a KCL at V-, and V- = 0. V+ is ground and v- must be the same. The result for Vo turns out to be: Vo = -Rf (V0/8k + V1/4k + V2/2k + V3/1k). As you can see V1 has twice the weight of V0, V2 has twice the weight of V1, etc. If used as an A/D converter Vo would be the least significant bit. The weighting can be changed by changing the input resistors. Rf can be used to scale the maximum output value. These weights are set up to convert digital to analog. 18. Feb. 2026 5 Operation Amplifiers 11.

Example 5 Nothing special about this circuit other than the fact that 2 op amps are in series. The output of the first stage drives the second state, so solve for Vo1 then solve for Vo2. To solve for the final output do the following steps: 1. Write a KCL at the non-inverting (+) node of the first op amp to get the voltage V1 at the + input. 2. 1st op amp. The voltage at the (-) input = the voltage at the (+) input = V1. 3. Write a KCL at the inverting (-) node of the first op amp to get Vo1. 4. 2nd op amp. The voltage at the (-) input = the voltage at the (+) input = Vo1. 5. V2 = Vo1 6. Write a KCL at the inverting (-) node of the second op amp to get Vout2. Voltage divider: V1 = 12*60/(60+30) = 8 KCL at V1, 0 = (10-8)/40k + 0 + (Vo1-8)/80k Yields: Vo1 = 4 and V2 = 4 KCL at V2, 0 = (0-4)/25k + 0 + (Vo2-4)/50k Yields: Vo2 = 12 18. Feb. 2026 5 Operation Amplifiers 12.

More Examples 18. Feb. 2026 5 Operation Amplifiers 13.

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5.8 Homework 1a. Use the ideal op amp model. Find Vo and io. 1b. Use the ideal finite gain model. Find Vo and io in terms of Rout, Rin, Gain A. 2. Use the ideal op amp model. Find Vo in terms of the unknowns: Va, Vb and the resistors. 3. Use the ideal op amp model. Find Vo in terms of the unknowns: Va, Vb and the resistors. Hint: Find Vx first. 18. Feb. 2026 5 Operation Amplifiers 15.

4. Use the ideal op amp model. Find Vo in terms of the unknowns. Hint: Find V2 first. 5a. Use the ideal op amp model. Find Vo. 5b. Find Vo in terms of Rout, Rin, Gain A, using the finite gain model that has input and output resistance. 6. Use the ideal op amp model. Find R such that Vo = -3. 7. Use the ideal op amp model. Find Vo1 and Vo2 in terms of V1 and V2. 18. Feb. 2026 5 Operation Amplifiers 16.

Solution to HW. 1a. 1b. 2. 3. 4. 18. Feb. 2026 5 Operation Amplifiers 17.

5a. 5b. 6. 7. 18. Feb. 2026 5 Operation Amplifiers 18.