BME211HW5 - Homework #5 Solutions P6.3-1 P6.3-3 The...

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Unformatted text preview: Homework #5 Solutions P6.3-1 P6.3-3 The voltages at the input nodes of an ideal op amp are equal so va = -2 V . Apply KCL at node a: vo - ( -2 ) 12 - ( -2 ) + = 0 vo = -30 V 8000 4000 Apply Ohm's law to the 8 k resistor io = -2 - vo = 3.5 mA 8000 P6.3-5 The voltages at the input nodes of an ideal op amp are equal, so va = 0 V . Apply KCL at node a: v 12 -0 -0 - o - - 10-3 2 = 0 3000 4000 vo = - 15 V Apply KCL at the output node of the op amp: v v io + o + o = 0 io = 7.5 mA 6000 3000 P6.3-6 The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal so va = 2.5 V . Apply Ohm's law to the 4 k resistor: v 2.5 ia = a = = 0.625 mA 4000 4000 Apply KCL at node a: ib = ia = 0.625 mA Apply KVL: vo = 8000 ib + 4000 ia = ( 12 103 ) ( 0.625 10-3 ) = 7.5 V P6.3-8 The node voltages have been labeled using: 1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal. 2. KCL 3. Ohm's law Then v0 = 11.8 - 1.8 = 10 V and 10 io = = 2.5 mA 4000 P6.3-9 KCL at node a: va - ( -18 ) v + a + 0 = 0 va = -12 V 4000 8000 The node voltages at the input nodes of ideal op amps are equal, so vb = va . Voltage division: vo = 8000 vb = -8 V 4000 + 8000 P6.4-1 vb - 2 vb v +5 1 + + b = 0 vb = - V 20000 40000 40000 4 The node voltages at the input nodes of an ideal op amp are equal so 1 ve = vb = - V . 4 ve v -v 10 + e d = 0 vd = 10 ve = - V KCL at node e: 1000 9000 4 KCL at node b: P6.4-2 Apply KCL at node a: 0= va - 12 v v -0 + a + a va = 4 V 6000 6000 6000 Apply KCL at the inverting input of the op amp: v 0 -0 - vo - a + 0+ = 0 6000 6000 vo = -va = -4 V Apply KCL at the output of the op amp: 0 -v v io - o o = 0 + 6000 6000 v io = - o = 1.33 mA 3000 P6.4-3 Apply KCL at the inverting input of the op amp: - 0 s - 0 v v - a - = 0 R2 R1 R va = - 2 vs R1 Apply KCL at node a: 1 1 1 va -v0 va va - 0 R R +R R +R R + + = 0 v0 = R4 + + a = 2 3 2 4 3 4 va v R4 R3 R2 R R2 R3 4 R3 R2 =- Plug in values yields vo 30+ 900+ 30 =- = -200 V/V vs 4.8 R2 R3 + R2 R4 + R3 R4 vs R1 R3 P6.4-9 vb + 12 vb + = 0 vb = -4 V 40000 20000 The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = -4 V . KCL at node b: The node voltages at the input nodes of an ideal op amp are equal, so vd = vc + 0 104 = -4 V . KCL at node g: v - vg vg 2 - f + = 0 vg = v f 3 3 20 3 10 40 10 2 vf . 3 The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = 2 vf v -v KCL at node d: 3 0= + d e3 = + 20 103 20 10 20 103 20 103 vd - v f vd - v f vd - Finally, ve = vg = 2 16 vf = - V. 3 5 6 24 v f = vd = - V 5 5 ...
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