1 / 18

- Y Transformation (2.7); Circuits with Dependent Sources (2.8)

- Y Transformation (2.7); Circuits with Dependent Sources (2.8). Dr. Holbert February 13, 2006. - Y Transformation. A particular configuration of resistors (or impedances) that does not lend itself to the using series and parallel combination techniques is that of a delta ( ) connection

tuari
Download Presentation

- Y Transformation (2.7); Circuits with Dependent Sources (2.8)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. -Y Transformation (2.7);Circuits with Dependent Sources (2.8) Dr. Holbert February 13, 2006 ECE201 Lect-8

  2. -Y Transformation • A particular configuration of resistors (or impedances) that does not lend itself to the using series and parallel combination techniques is that of a delta () connection • In such cases the delta () connection is converted to a wye (Y) configuration • The reverse transformation can also be performed ECE201 Lect-8

  3. -Y Transformation a a Ra R1 R2 Rc Rb c b R3 c b ECE201 Lect-8

  4. -Y Transformation • To compute the new Y resistance values • For the balanced case (RY= Ra= Rb= Rc) RΔ = 3 RY ECE201 Lect-8

  5. Class Example • Learning Extension 2.17 • Learning Extension 2.18 ECE201 Lect-8

  6. 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-8

  7. 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-8

  8. Inverting Amplifier I 1kW 4kW 10kW – + – + – Vf 10V Vs=100Vf + • Apply KVL around loop: -10V + 1kWI + 4kWI + 10kWI + 100 Vf = 0 ECE201 Lect-8

  9. 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-8

  10. 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-8

  11. Amplifier Gain • Repeat the previous example for a gain of 1000 • Answer: Vs = -19.94V ECE201 Lect-8

  12. 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-8

  13. 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-8

  14. 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-8

  15. 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-8

  16. 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-8

  17. 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-8

  18. Class Examples • Learning Extension E2.19 • Learning Extension E2.20 ECE201 Lect-8

More Related