Chapter 5 – Operational Amplifiers

1. Chapter 5 EGR 260 – Circuit Analysis 1 Reading Assignment: Chapter 5 in Electric Circuits, 9th Edition by Nilsson Chapter 5 – Operational Amplifiers Note: We are temporarily skipping the remaining sections of Chapter 4. We will cover them after completing Chapter 5. Operational Amplifier - An operational amplifier (op amp) is a high gain differential amplifier with nearly ideal external characteristics. Internally the op amp is constructed using many transistors. Terminology: V+ = non-inverting input voltage V- = inverting input voltage Vo = output voltage Io = output current I+ = non-inverting input current I- = inverting input current VDC = positive and negative DC supply voltages used to power the op amp (typically 5V to 30V) V = V+ - V- = difference voltage Note: Sometimes the supply voltage connections are not shown

2. Chapter 5 EGR 260 – Circuit Analysis 2 Operational Amplifiers 8-pin package pinout 8-pin package (3D view) uA741 symbol in PSPICE

3. Chapter 5 EGR 260 – Circuit Analysis 3 Typical Operational Amplifier Schematic

4. Chapter 5 EGR 260 – Circuit Analysis 4 Open-loop versus closed-loop operation: • Open-loop: • Relatively rare • Op amp specifications may be important • Closed-loop: • Most commonly used • Some sort of feedback from output to input exists • The input voltage, Vin, is defined according to the application

5. Chapter 5 EGR 260 – Circuit Analysis 5 • An op amp circuit can be easily analyzed using the following ideal • assumptions. • Ideal op-amp assumptions: • Assume that V = 0, so V+ = V- • Assume the input resistance is infinite, so I+ = I- = 0 • Realize the all voltages defined above are node voltages w.r.t. a common ground (as illustrated below) Illustration: Draw an op amp showing a common negative terminal for all node voltages.

6. Chapter 5 EGR 260 – Circuit Analysis 6 Example: Determine an expression for Vo in the inverting amplifier shown below. Illustrate the results using both DC and AC inputs.

7. Chapter 5 EGR 260 – Circuit Analysis 7 Load Connections: Vo is typically determined independent of the load (the circuit connected to the output). Once Vo has been determined, it essentially acts like a voltage source to the load. Load Current: Io is the output current for an op amp. It can be found using KCL. Example: Find Vo , I1, V2, and Io below.

8. Chapter 5 EGR 260 – Circuit Analysis 8 Example: Determine an expression for Vo in the non-inverting amplifier shown below.

9. Chapter 5 EGR 260 – Circuit Analysis 9 Example: Determine Vo in the inverting summing amplifier shown below. RF V _ 1 R 1 V o V + 2 R 2 V 3 R 3

10. Chapter 5 EGR 260 – Circuit Analysis 10 RF _ R V o + V 1 R 1 V 2 R 2 V 3 R 3 Example: Determine Vo in the non-inverting summing amplifier shown below.

11. Chapter 5 EGR 260 – Circuit Analysis 11 Example: Determine VL in the circuit shown below.

12. Chapter 5 EGR 260 – Circuit Analysis 12 Example: Determine Vo in the circuit shown below.

13. Chapter 5 EGR 260 – Circuit Analysis 13 Practical Limitations in op amps Operational amplifier circuits are generally easy to analyze, design, and construct and their behavior is fairly ideal. There are, however, some limitations to op amps which the engineer should recognize. There are three primary limitations as well as some minor limitations that will be discussed in later courses. The three primary limitations in op amps are: 1) Limited voltage - In general, the output voltage is limited by the DC supply voltages. 2) Limited current - The output current has a maximum limit set by the manufacturer (check the data sheet). 3) Frequency limitations - Op amp performance may deteriorate significantly as frequency increases (studied in later courses)

14. Chapter 5 EGR 260 – Circuit Analysis 14 Voltage Limitations Vo for any op amp circuit is limited by the supply voltage. In general, If Vo attempts to exceed these limits, the output is limited to +VDC or -VDC and we say that the op amp is “saturated” or is “in saturation.” Practically, Vo is often about 2V under the supply voltage, or Example: Consider the inverting amplifier shown below (covered earlier). 1) Determine the voltage gain, ACL = Vo/Vin

15. Chapter 5 EGR 260 – Circuit Analysis 15 2) Determine Vo for various possible values of Vin (fill out the table shown below) 3) Graph Vo versus Vin . Identify the saturation and linear regions of operation. Vo Vin

16. Chapter 5 EGR 260 – Circuit Analysis 16 R2 _ Vin 2k Vo + • Example: Consider the inverting amplifier shown below (covered earlier). • If R2 = 4k, determine range of values for Vin such that the op am will operate in the linear range • If Vin = 2V, determine R2 (max) such that the op am will operate in the linear range +12V -12V

17. Chapter 5 EGR 260 – Circuit Analysis 17 4k _ Vin 2k Vo + Example: Sketch Vo (on the same graph) for the circuit shown below if Vin is a triangle wave as specified below. +10V 20V -10V Vin 10V 0V -10V -20V

18. Chapter 5 EGR 260 – Circuit Analysis 18 Current Limitations The maximum output current, Io for an op amp is specified by the manufacturer. Exceeding this limit will typically destroy the op amp. The output current can be calculated using KCL at the output node. An op amp circuit should be designed to insure that its output current does not exceed the maximum value, Io (max), specified by the manufacturer. Example: If Io (max) is specified at 25 mA by the manufacturer, determine the minimum value of RL that can safely be used in the circuit shown below.

19. Chapter 5 EGR 260 – Circuit Analysis 19 Using node equations to analyze op amp circuits As op amp circuits become more complex, simultaneous equations may be needed to analyze them. Node equations are a natural choice since many op amp voltages are expressed as node voltages. Note: Writing a KCL (node) equation at the output node is not typically helpful except to find Io since it introduced another unknown (Io). Example: Determine Vo in the circuit shown below. Note: Writing a KCL equation (node equation) at the output is only helpful for finding Io since it introduces another unknown (Io).

20. Chapter 5 EGR 260 – Circuit Analysis 20 Example: Determine Vo in the circuit shown below.

21. Chapter 5 EGR 260 – Circuit Analysis 21 Example: The circuit shown below is a current amplifier. Determine an expression for IL . Also find the current gain, AI = IL/IS.

22. Chapter 5 EGR 260 – Circuit Analysis 22 Op amp models So far we have analyzed op amps using a few ideal assumptions about op amps (such as V+ = V- and I+ = I- = 0) . How would we convey these assumptions to a circuit analysis program like PSPICE? Typically, we would construct a circuit model that acts like the op amp that we desire. A simple op amp model is shown below. Also note that PSPICE can be used to model op amps with specific part numbers (such as the uA741 in the PSPICE library EVAL.SLB). In these cases, ORCAD develops very detailed circuit models that match the characteristics of the particular op amp. The ORCAD model might look like the model shown below plus additional components to more accurately model additional features. Typical values for the op amp model shown: AOL = 100,000 Rin = 2M - 10M

23. Chapter 5 EGR 260 – Circuit Analysis 23 Example: Determine Vo in the circuit shown in two ways: 1) by making ideal op amp assumptions

24. Chapter 5 EGR 260 – Circuit Analysis 24 Example: Determine Vo in the circuit shown in two ways: 2) by using an ideal op amp model with Rin = 2M and AOL = 100,000.

25. Chapter 5 EGR 260 – Circuit Analysis 25 Analyzing operational amplifier circuits using PSPICE: There are two ways to analyze op amp circuits using PSPICE. 1) Use the general op amp model just introduced (consisting of a resistor and a voltage-controlled voltage source. 2) Use one of the op amp models from a PSPICE library (such as the uA741). Refer to two examples on the course web site: 1) Op Amp Example using a General Op Amp Model 2) Op Amp Circuit using a Library Model ( uA741) Note: End of Test #2 material here.