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Step of 5 11.012E Refer to the example 10.5 in the text book. Determine the value of feedback resistance. Substitute 5 mA/V for = 1 = 200 Ω Therefore, the feedback resistance of the feedback transcondutance amplifier is 200 Ω Ω Step of 5 Determine the value of closed loop gain of the feedback transcondutance Substitute 200 V/V for 2 mA/V for and 200 Ω for A, = = 0.4 1+80 = 4.938 mA/V Therefore, the closed loop gain of the feedback transcondutance amplifier is 4.938 mA/V Step of 5 Determine the input resistance of the feedback transcondutance amplifier. Substitute for 200 V/V for 2 mA/V for and Ω for = = 8.1 MΩ Therefore, the input resistance of the feedback transcondutance amplifier is 8.1 Step of 5 Determine the output resistance of the feedback transcondutance amplifier. Substitute for 200 V/V for 2 mA/V for and 200 Ω for = = 1.62 Therefore, the output resistance of the feedback transcondutance amplifier is .62 MΩ Step of 5 The transconductance of the feedback transconductance amplifier is given by If the value of drops by 50% then, 8m2 = mA/V Determine the closed loop gain of the feedback transconductance amplifier. = = 0.2 = 1+40 = 4.878 mA/V Determine the percentage change in the closed loop gain. %A, = -1.23% Therefore, the percentage change in the closed loop gain is -1.23%