Lecture 241 Circuits with Dependent Sources Strategy: Apply KVL and KCL, treating dependent source(s) as independent sources. Determine the relationship.

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Presentation transcript:

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.

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.

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

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

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

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

Lecture 247 Answer V s = V

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

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

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

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 

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

Lecture 2413 Apply KCL at the Top Node 5mA = V/6k  + 5  V + V/3k  5mA = 1.67  V + 5  V  V V=5mA(1.67    ) V=5V