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Understanding Circuits with Dependent Sources: Inverting Amplifiers and Transistor Models

This lecture delves into circuits that include dependent sources, emphasizing the application of Kirchhoff's Voltage and Current Laws (KVL and KCL). It explores the analysis of inverting amplifiers using operational amplifiers and examines negative feedback mechanisms. Detailed examples illustrate circuit calculations, such as solving for output voltage and determining the relationship between dependent sources and their controlling parameters. The session also provides insights into small-signal transistor amplifier models, enhancing comprehension of electronic circuit behavior.

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Understanding Circuits with Dependent Sources: Inverting Amplifiers and Transistor Models

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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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