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Circuits with Dependent Sources (2.8)

Circuits with Dependent Sources (2.8). Dr. S. M. Goodnick September 24, 2003. 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.

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Circuits with Dependent Sources (2.8)

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  1. Circuits with Dependent Sources (2.8) Dr. S. M. Goodnick September 24, 2003 ECE201 Lect-13

  2. 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. ECE201 Lect-13

  3. 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. ECE201 Lect-13

  4. Inverting Amplifier I 1kW 4kW 10kW – + – + – Vf 10V Vs=100Vf + • Apply KVL around loop: -10V + 1kWI + 4kWI + 10kWI + 100 Vf = 0 ECE201 Lect-13

  5. Inverting Amplifier • Applying KVL yielded: -10V + 1kWI + 4kWI + 10kWI + 100 Vf = 0 • Get Vf in terms of I: Vf + 10kWI + 100Vf = 0 Vf = -(10kW/101) I ECE201 Lect-13

  6. Inverting Amplifier • Solve for I: I = 1.961 mA • Solve for Vf : Vf = -0.194 V • Solve for source voltage: Vs = -19.4 V ECE201 Lect-13

  7. Amplifier Gain • Repeat the previous example for a gain of 1000 • Answer: Vs = -19.94V ECE201 Lect-13

  8. Another Amplifier Find the output voltage Vs for this circuit, assuming a frequency of w=5000 100nF I 1kW 4kW – + – + – Vf 10V0 Vs=100Vf + ECE201 Lect-13

  9. Find Impedances • Apply KVL around loop: -10V0 + 1kWI + 4kW I - j2kWI + 100 Vf = 0 -j2kW I 1kW 4kW – + – + – Vf 10V0 Vs=100Vf + ECE201 Lect-13

  10. Another Amplifier • KVL provided: -10V0 + 1kWI + 4kW I - j2kWI + 100 Vf = 0 • Get Vf in terms of I: Vf - j2kWI + 100 Vf = 0 Vf = (j2kW/101)I ECE201 Lect-13

  11. Another Amplifier • Solve for I: I = 2mA 0.2 • Solve for Vf : Vf = 39.6mV90.2 • Solve for source voltage: Vs = 3.96V90.2 ECE201 Lect-13

  12. Transistor Amplifier A small-signal linear equivalent circuit for a transistor amplifier is the following: Find VX + 510-4VX 5mA 6kW 3kW VX – ECE201 Lect-13

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

  14. Class Examples • Learning Extension E2.17 • Learning Extension E2.18 ECE201 Lect-13

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