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ECE201 Lect-81  -Y Transformation (2.7); Circuits with Dependent Sources (2.8) Dr. Holbert February 13, 2006.

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ECE201 Lect-81  -Y Transformation (2.7); Circuits with Dependent Sources (2.8) Dr. Holbert February 13, 2006

ECE201 Lect-82  -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-83  -Y Transformation a cb a bc R1R1 R2R2 R3R3 RaRa RbRb RcRc

ECE201 Lect-84  -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-85 Class Example Learning Extension 2.17 Learning Extension 2.18

ECE201 Lect-86 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-87 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-88 Inverting Amplifier 1k  +–+– 4k  10k  +–+– + – VfVf V s =100V f 10V I Apply KVL around loop: -10V + 1k  I + 4k  I + 10k  I + 100 V f = 0

ECE201 Lect-89 Inverting Amplifier Applying KVL yielded: -10V + 1k  I + 4k  I + 10k  I + 100 V f = 0 Get V f in terms of I: V f + 10k  I + 100V f = 0 V f = -(10k  101) I

ECE201 Lect-810 Inverting Amplifier Solve for I: I = 1.961 mA Solve for V f : V f = -0.194 V Solve for source voltage: V s = -19.4 V

ECE201 Lect-811 Amplifier Gain Repeat the previous example for a gain of 1000 Answer: V s = -19.94V

ECE201 Lect-812 Another Amplifier 1k  4k  100nF + – VfVf V s =100V f 10V  0  I Find the output voltage V s for this circuit, assuming a frequency of  =5000 +–+– +–+–

ECE201 Lect-813 Find Impedances 1k  4k  -j2k  + – VfVf V s =100V f 10V  0  I +–+– +–+– Apply KVL around loop: -10V  0  + 1k  I + 4k  I - j2k  I + 100 V f = 0

ECE201 Lect-814 Another Amplifier KVL provided: -10V  0  + 1k  I + 4k  I - j2k  I + 100 V f = 0 Get V f in terms of I: V f - j2k  I + 100 V f = 0 V f = (j2k  I

ECE201 Lect-815 Another Amplifier Solve for I: I = 2mA  0.2  Solve for V f : V f = 39.6mV  90.2  Solve for source voltage: V s = 3.96V  90.2 

ECE201 Lect-816 Transistor Amplifier A small-signal linear equivalent circuit for a transistor amplifier is the following: Find V X 3k  6k  + – VXVX 5mA 5  10 -4 V X

ECE201 Lect-817 Apply KCL at the Top Node 5mA = V X /6k  + 5  10 -4 V X + V X /3k  5mA = 1.67  10 -4 V X + 5  10 -4 V X + 3.33  10 -4 V X V X =5mA/(1.67  10 -4 + 5  10 -4 + 3.33  10 -4 ) V X =5V

ECE201 Lect-818 Class Examples Learning Extension E2.19 Learning Extension E2.20

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