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Step of 5 2.032P (a) Refer to op-amp circuit in Figure P2.32 in the textbook. The voltage at non-inverting terminal of the op-amp is zero. Since it is directly connected to ground. The voltages at inverting and non-inverting terminals of the op-amp are zero for ideal op-amp. Calculate the current through source. 1V Ω = 0.1 mA Therefore, the current, is 0.1 mA Apply Kirchhoff current law at inverting terminal of the op-amp. Therefore, the current, I₂ is mA Step of 5 Calculate the voltage using ohm's law. 10 kΩ Therefore, the voltage, is - V Calculate the current = 100 Ω Ω 10 mA Therefore, the current, is 10 mA Apply Kirchhoff's current law at node 0.1 mA + 10 mA = 10.1 mA Therefore, the current, is 10.1 10.1 mA mA Step of 5 (b) Minimum output voltage is -13V Write the expression for current = Substitute -1 V for -13 V for and 10.1 mA for in the equation. = 10.1 mA = 10.1 mA = 1.188 Therefore, the maximum allowable load resistance, is Step of 5 (c) The values of current, and current, are independent of load resistance. But current through the load resistor is sum of current, and current, So, when the value of load resistance is varied, the current through the load resistor remains same. Therefore, the load current, is, 10.1 10.1 mA mA Write the expression for current Substitute -1 V for Vx 10.1 mA for and 100 for in the equation. V Step of 5 Recall equation (1) Substitute -1 V for Vx 10.1 mA for and for in the equation. =-11.1V Therefore, the range of output voltage is -11.1 V to -2.01 V >