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Step of 8 4.047P All the diodes are in parallel combination and characteristics are also same. Calculate the current through each diode. 0.1 10 = 0.01 A Therefore, the current through the each diode is 0.01 A Step of 8 Calculate the small signal rd Substitute 0.01 A for 1 for n and 0.025 V for in equation. rd 1x 0.025 = 0.01 Therefore, the small signal for single diode is 2.5 Ω Step of 8 Calculate the equivalent resistance, R = rd N Substitute 2.5 Ω for rd and 10 for N in equation. 2.5 10 = 0.25 Ω Therefore, the equivalent resistance is 0.25 Ω Step of 8 When one diode is conducting the current through the diode is 0.1 Calculate the small signal resistance. rd 0.025 = 0.1 Therefore, the small signal resistance is 0.25 Step of 8 The parallel combination of the diodes can be considered as equivalent to a single diode having the junction area ten times as that of a single diode. So in both the cases the small signal resistances are same. Step of 8 Consider the resistance 0.2 Ω is connected in series with the diode. Since the small signal resistance, rd is obtained as 2.5 Ω which is to be connected in series with the resistance 0.2 The equivalent resistance of the 10 parallel-connected diodes is, N = Substitute the values 2.5 Ω 10 for N and 0.2 for R in equation. 10 0.27 Ω Therefore the equivalent resistance is 0.27 Step of 8 The small signal resistance connected, rd in series with the resistance, rs is shown in Figure 1. rd Figure 1 Step of 8 From Figure 1, the expression for series resistance is, Substitute 0.27 Ω for and 0.25 Ω for rd in equation. 0.27 0.25 + 0.27 Ω Therefore, the equivalent resistance of the 10 parallel connected diodes is 0.27 Ω and the series resistance is