1. Lecture 241 Circuits with Dependent Sources Strategy: Apply KVL and KCL, treating dependent source(s) as independent sources. Determine the relationship between dependent source values and controlling parameters. Solve equations for unknowns.

2. Lecture 242 Example: Inverting Amplifier The following circuit is a (simplified) model for an inverting amplifier created from an operational amplifier (op-amp). It is an example of negative feedback.

3. Lecture 243 Inverting Amplifier 1k  + - 4k  10k  + - + - VfVf V s =100V f 10V I

4. Lecture 244 Inverting Amplifier Apply KVL around loop: -10V + 1k  I + 4k  I + 10K  I + 100 V f = 0 Get V f in terms of I: V f + 4k  I + 100V f = 0 V f = 4k  I

5. Lecture 245 Inverting Amplifier Solve for I: I = 1.961mA Solve for V f : V f = -0.194V Solve for source voltage: V s = -19.4V

6. Lecture 246 Amplifier Gain Repeat the previous example for a gain of 1000

7. Lecture 247 Answer V s = -19.94V

8. Lecture 248 Another Amplifier 1k  + - 4k  10nF + - + - VfVf V s =100V f 10V  0  I Find the output voltage V s for this circuit, assuming a frequency of  =5000

9. Lecture 249 Find Impedances 1k  + - 4k  -j2k  + - + - VfVf V s =100V f 10V  0  I

10. Lecture 2410 Another Amplifier Apply KVL around loop: -10V  0  + 1k  I + 4k  I - j2k  I + 100 V f = 0 Get V f in terms of I: V f - j2k  I + 100 V f = 0 V f = j2k  I

11. Lecture 2411 Another Amplifier Solve for I: I = 2mA  -0.2  Solve for V f : V f = 39.6mV  89.8  Solve for source voltage: V s = 3.96V  89.8 

12. Lecture 2412 Transistor Amplifier A small-signal linear equivalent circuit for a transistor amplifier is the following: Find V 3k  6k  + - V5mA 5  10 -4 V

13. Lecture 2413 Apply KCL at the Top Node 5mA = V/6k  + 5  10 -4 V + V/3k  5mA = 1.67  10 -4 V + 5  10 -4 V + 3.33  10 -4 V V=5mA(1.67  10 -4 + 5  10 -4 + 3.33  10 -4 ) V=5V