DeCarlo Ch4 Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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Thevenin Probs, 7/11/01 - P4.1 - @R.A. Decarlo & P. M. Lin PROBLEM SOLUTIONS CHAPTER 4 S OLUTION 4.1 . First, find V out / V s for each circuit. Then solve for R knowing V out = P 10 14.142 V . (a) Writing KCL at the inverting terminal, 1/1 k ( v - - v s ) = 1/ R ( V out - v - ) V out / V s = - R / 1 k , since the inverting terminal is a virtual short. Solving for R = - V out 1 k / V s = 2.828 k . (b) Writing KCL at the inverting terminal, V s /1.5 k = ( V out - V s )/ R V out / V s = R / 1.5 k + 1, solving for R = 1.5 k ( V out / V s - 1) = 2.743 k . (c) From (a) V out / V s = - 12 k / R , thus R = - 12 k V s / V out = 4.243 k . (d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus V out = P 10 10 + 6 ( 29 = 22.627 V . Using the general form from (b), R = 400 V out / V s - 1 ( 29 = 1.410 k S OLUTION 4.2 . (a) First, find the voltage at the non-inverting terminal as v + = 1/ 2 V s . Then write KCL at the inverting terminal, and make use of the virtual short property, ( V s / 2)/10 k = ( V out - V s / 2) / 30 k V out / V s = 30 k (1/ 20 k + 1/ 60 k ) = 2 . (b) Relating the output of the amplifier to the output of the circuit, V out = V amp (500 / 800). Then writing KCL at the inverting terminal, V s / 400 = ( V amp - V s ) /1.2 k V amp / V s = 1.2 k / 400 + 1 = 4. Therefore V out / V s = ( V amp / V s ) ( V out / V amp ) = 2.5. (c) Note that since no current goes into the non-inverting terminal of the op-amp, the voltage at that node is –V s . KCL at the inverting terminal, - V s / 4 k = ( V out + V s )/ 20 k V out / V s = - 6. S OLUTION 4.3 . Write KCL for both terminals, ( V - - V i )/1 k = ( V o - V - ) / 2 k V - /1 k = ( V o - V - )/ 3 k Solving and doing the appropriate substitutions, V o / V i = - 8 . S OLUTION 4.4 . This is essentially the basic inverting configuration, which is defined as V o / V i = - 2 k / 1 k = - 2 .

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Thevenin Probs, 7/11/01 - P4.2 - @R.A. Decarlo & P. M. Lin S OLUTION 4.5 . (a) By voltage division V L = 1 V 100 200 = 0.5 V . Using Ohm’s law I s = I L = 1 100 + 100 = 5 mA . (b) No current flows in the input terminal of an ideal op-amp, thus I s = 0 A and V L = 1 V . From Ohm’s law I a = I L = V L /100 = 10 mA . S OLUTION 4.6 . (a) Using voltage division, V 1 = V s 32||(8 + 24) [32||(8 + 24)] + 8 = 2 3 V s V out = V 1 24 24 + 8 = 0.5 V s (b) By voltage division, V 1 = V s 32 32 + 40 = 0.8 V s V out = V 1 24 24 + 8 = 0.6 V s (c) Using voltage division, V 1 = V s 32 32 + 8 = 0.8 V s , as no current enters the non-inverting terminal of the op-amp. Due to the virtual short property, V out = V 1 24 24 + 8 = 0.6 V 1 . This is indeed the same results as (b), which should be expected because of the isolation provided by the ideal buffers. S OLUTION 4.7 . Write KCL at the inverting terminal, - V s 1 /1 k - V s 2 / 2 k = V out / 4 k V out = - 4 V s 1 - 2 V s 2 = 40 mV . S OLUTION 4.8 . (a) The voltage at the non-inverting terminal is V + = 3 / 2 V , KCL at the inverting terminal gives (1.5 - 2.5) /10 k = ( V out - 1.5)/ 30 k V out = - 1.5 V . The power is P = V out 2 / 500 = 4.5 mW . (b) The voltage at the non-inverting terminal is 3V this time, thus KCL (3 - 2.5)/ 10 k = ( V out - 3) / 30 k V out = 4.5 V . The power is P = V out 2 / 500 = 40.5 mW .
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