Question Solved1 Answer 3.5 3.a) Design a noninverting amplifier that has a voltage gain of xy using an ideal op-amp. The input signal lie in 3.5 the range from 1 to IV. Use 5% tolerance discrete resistors for the feedback network. Xy=55 b) A circuit known as a summing amplifier in Figure 3b. (a). Use the ideal op-amp assumption to solve for the output voltage in terms of the input voltage and resistor values. (b). What is the input resistance seen by 2m V? (C). What is the input resistance seen by SmV? (d). What is the output resistance? 10kΩ 1kΩ 2mV 2k2 5mV A Var Ov Figure-3b

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Transcribed Image Text: 3.5 3.a) Design a noninverting amplifier that has a voltage gain of xy using an ideal op-amp. The input signal lie in 3.5 the range from 1 to IV. Use 5% tolerance discrete resistors for the feedback network. Xy=55 b) A circuit known as a summing amplifier in Figure 3b. (a). Use the ideal op-amp assumption to solve for the output voltage in terms of the input voltage and resistor values. (b). What is the input resistance seen by 2m V? (C). What is the input resistance seen by SmV? (d). What is the output resistance? 10kΩ 1kΩ 2mV 2k2 5mV A Var Ov Figure-3b
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Transcribed Image Text: 3.5 3.a) Design a noninverting amplifier that has a voltage gain of xy using an ideal op-amp. The input signal lie in 3.5 the range from 1 to IV. Use 5% tolerance discrete resistors for the feedback network. Xy=55 b) A circuit known as a summing amplifier in Figure 3b. (a). Use the ideal op-amp assumption to solve for the output voltage in terms of the input voltage and resistor values. (b). What is the input resistance seen by 2m V? (C). What is the input resistance seen by SmV? (d). What is the output resistance? 10kΩ 1kΩ 2mV 2k2 5mV A Var Ov Figure-3b
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(3) (a) Non-invorting amplifier:" gain, "xy=55The gain of the non-inverting auplifier is{:[G=1+(Rf)/(R_(i))],[=>G=1+(Rf)/(R_(i))=55],[(Rf)/(R_(i))=54]:}the standard resistor values with %5 tolerance which can suit to get the gain 54 by their ratio are15 Omega and 820 Omegalets take R_(i)=15 Omega epsiR_(f)=820 OmegaThe circuit is(b)Due to virtual ground concept,V_(N)=0 (positive terminal is grounded)KCL at Node V_(N) :{:[ ... See the full answer