The Zener diode in the voltage-regulator circuit of Fig. 2-45 has a constant reverse breakdown voltage VZ = 8.2 V, for 75 mA ≤ iZ ≤ 1 A. If RL = 9 Ω, size RS so that vL = VZ is regulated to (maintained at) 8.2 V while Vb varies by ±10 percent from its nominal value of 12 V.

Question

The Zener diode in the voltage-regulator circuit of Fig. 2-45 has a constant reverse breakdown voltage [latex]V_Z = 8.2 ext{V}[/latex], for [latex]75 ext{mA} ≤ i_Z ≤ 1 ext{A}[/latex]. If [latex]R_L = 9 Ω[/latex], size [latex]R_S[/latex] so that [latex]v_L = V_Z[/latex] is regulated to (maintained at) [latex]8.2 ext{V}[/latex] while [latex]V_b[/latex] varies by ±10 percent from its nominal value of [latex]12 ext{V}[/latex].

Accepted Answer

By Ohm’s law
[latex]i_L = \frac{v_L}{R_L} = \frac{V_Z}{R_L} = \frac{8.2}{9} = 0.911 ext{A}[/latex]

Now an application of KVL gives
[latex]R_S = \frac{V_b - V_Z}{i_Z + i_L}[/latex] (1)

and we use (1) to size [latex]R_S[/latex] for maximum Zener current [latex]I_Z[/latex] at the largest value of [latex]V_b[/latex]:
[latex]R_S = \frac{(1.1)(12) - 8.2}{1 + 0.911} = 2.62 \Omega[/latex]

Now we check to see if [latex]i_Z ≥ 75 ext{mA}[/latex] at the lowest value of [latex]V_b[/latex]:
[latex]i_Z = \frac{v_b - v_Z }{R_S} - i_L = \frac{(0.9)(12) - 8.2}{2.62} - 0.911 = 81.3 ext{mA}[/latex]

Since [latex]i_Z > 75 ext{mA}, v_Z = V_Z = 8.2 ext{V}[/latex] and regulation is preserved.

Chapter 2

Q. 2.SP.30

Q. 2.SP.30

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