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