1. ECE201 Lect-131 Circuits with Dependent Sources (2.8) Dr. S. M. Goodnick September 24, 2003

2. ECE201 Lect-132 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.

3. ECE201 Lect-133 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.

4. ECE201 Lect-134 Inverting Amplifier 1k  +–+– 4k  10k  +–+– + – VfVf V s =100V f 10V I Apply KVL around loop: -10V + 1k  I + 4k  I + 10k  I + 100 V f = 0

5. ECE201 Lect-135 Inverting Amplifier Applying KVL yielded: -10V + 1k  I + 4k  I + 10k  I + 100 V f = 0 Get V f in terms of I: V f + 10k  I + 100V f = 0 V f = -(10k  101) I

6. ECE201 Lect-136 Inverting Amplifier Solve for I: I = 1.961 mA Solve for V f : V f = -0.194 V Solve for source voltage: V s = -19.4 V

7. ECE201 Lect-137 Amplifier Gain Repeat the previous example for a gain of 1000 Answer: V s = -19.94V

8. ECE201 Lect-138 Another Amplifier 1k  4k  100nF + – VfVf V s =100V f 10V  0  I Find the output voltage V s for this circuit, assuming a frequency of  =5000 +–+– +–+–

9. ECE201 Lect-139 Find Impedances 1k  4k  -j2k  + – VfVf V s =100V f 10V  0  I +–+– +–+– Apply KVL around loop: -10V  0  + 1k  I + 4k  I - j2k  I + 100 V f = 0

10. ECE201 Lect-1310 Another Amplifier KVL provided: -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. ECE201 Lect-1311 Another Amplifier Solve for I: I = 2mA  0.2  Solve for V f : V f = 39.6mV  90.2  Solve for source voltage: V s = 3.96V  90.2 

12. ECE201 Lect-1312 Transistor Amplifier A small-signal linear equivalent circuit for a transistor amplifier is the following: Find V X 3k  6k  + – VXVX 5mA 5  10 -4 V X

13. ECE201 Lect-1313 Apply KCL at the Top Node 5mA = V X /6k  + 5  10 -4 V X + V X /3k  5mA = 1.67  10 -4 V X + 5  10 -4 V X + 3.33  10 -4 V X V X =5mA/(1.67  10 -4 + 5  10 -4 + 3.33  10 -4 ) V X =5V

14. ECE201 Lect-1314 Class Examples Learning Extension E2.17 Learning Extension E2.18