ECE201 Lect-51  -Y Transformation (2.7); Circuits with Dependent Sources (2.8) Prof. Phillips February 3, 2003

ECE201 Lect-52  -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-53  -Y Transformation a cb a bc R1R1 R2R2 R3R3 RaRa RbRb RcRc

ECE201 Lect-54  -Y Transformation To compute the new Y resistance values For the balanced case (R Y = R a = R b = R c ) R Δ = 3 R Y

ECE201 Lect-55 Class Example

ECE201 Lect-56 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-57 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-58 Inverting Amplifier 1k  +–+– 4k  10k  +–+– + – VfVf V s =100V f 10V I Apply KVL around loop: -10V + 1k  I + 4k  I + 10k  I V f = 0

ECE201 Lect-59 Inverting Amplifier Applying KVL yielded: -10V + 1k  I + 4k  I + 10k  I V f = 0 Get V f in terms of I: V f + 10k  I + 100V f = 0 V f = -(10k  101) I

ECE201 Lect-510 Inverting Amplifier Solve for I: I = mA Solve for V f : V f = V Solve for source voltage: V s = V

ECE201 Lect-511 Amplifier Gain Repeat the previous example for a gain of 1000 Answer: V s = V

ECE201 Lect-512 Transistor Amplifier A small-signal linear equivalent circuit for a transistor amplifier is the following: Find V X 3k  6k  + – VXVX 5mA 5  V X

ECE201 Lect-513 Apply KCL at the Top Node 5mA = V X /6k  + 5  V X + V X /3k  5mA = 1.67  V X + 5  V X  V X V X =5mA/(1.67    ) V X =5V

ECE201 Lect-514 Class Examples