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Step 1 of 3 4.016E Consider a zener diode with nominal voltage, 10 V at zener current, of 10 mA. The diode has an incremental resistance of Consider the following expression for the change in zener voltage for the corresponding change in current: Here, is the incremental resistance. Consider that the diode current is halved (That is, decreased by 5 mA). Substitute 5 mA for change in zener current, 50 Ω for = 0.25 V Refer to Figure 4.17 in the textbook for the diode i-v characteristics. Observe that the relation between diode current and voltage is linear. If the diode current decreases, the diode voltage also decreases. Substitute 10 V for nominal voltage, V₂ , and 0.25 V for change in voltage, to find the expected voltage. =10-0.25 V Therefore, the expected voltage, V is, 9.75 V - Step 2 of 3 Consider that the diode current is doubled (That is, increased by 10 mA). Substitute 10 mA for change in zener current, 50 Ω for = 0.5 V The diode voltage increases with increase in diode current due to linear behavior of diode. Substitute 10 V for nominal voltage, V₂ , and 0.5 V for change in voltage, to find the expected voltage. =10+0.5 V Therefore, the expected voltage, V is, 10.5 V Step 3 of 3 Refer to Figure 4.18 in the textbook for the model of zener diode. Consider the following expression for the zener diode: Modify the expression for the knee voltage, Substitute 50 Ω for r₂ , 10 mA for I₂ , and 10 V for V₂ =10-0.5 = 9.5 V Therefore, the value of knee voltage, V₂₀ is 9.5 V