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Step of 6 4.046P The variations in the voltage is 5 mV The diode current is, = (1) Consider the diode current is for the voltage variation of 5 mV. Find the value of the ratio Substitute 5 mV for for n and 0.025 V for in equation (1). 0.005 = 1.22 Step of 6 The percentage change of the current is % change = Substitute 1.22 for in equation. % change = 100 = 22% Step of 6 Substitute -5 mV for for n and 0.025 V for in equation (1). = -0.005 = = 0.82 Step 4 of 6 The percentage change of the current is % change = Substitute 0.82 for in equation. % change = 100 = 18% Hence, the percentage change in the current for both the positive and negative signal is 22% and 18% Step 5 of 6 Given that the current change is limited to 10%. Then the change in the current would be 90% to 110%. Therefore, the current variation would be within 0.9 to 1.1. Substitute the values 0.9 for for n and 0.025 V for in equation. 0.9 = In (0.9) 0.025 = 2.6 mV Step of 6 Substitute the values 1.1 for for n and 0.025 V for in equation. 1.1 = In(1.1) 1x 0.025 =2.4 mV Therefore, the maximum allowable voltage signal if the current change is to be limited to 10% is to 2.4 mV