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Zener regulation circuit

Before I work on the simulations, I am struggling to understand the voltage regulated circuit theoretically. I try to vary the V1 DC supply, V2 provides a constant DC voltage of 12 V. The zener (D1) breakdown voltage is 5.1 V, while the forward voltage drop of D2 is 2.2 V.

I am trying to analyze the effect of varying V1 on the circuit voltages and currents, and I would appreciate if you could let me know whether my understanding is correct. All transistor references are made to Q1.

If V1 = 0.5 V DC, Ir1 = 0 A, Vbase = 0 V, Vemitter = 0 V, Iemitter = 0 V. This is because 0.5 V is less than the turn-on voltage of transistor (Vbe = 0). SO the transistor is in cutoff.

If V1 = 1.3 V, Ir1 = (1.3 - 0.7)/220 A, Vbase = 1.3 - (220 x Ir1) V, Vemitter=(1.3 - 0.7) = 0.6 V, Iemitter = 0 V because 0.6 V is less than 2.2 V, so D2 is off.

If V1 =4 V, Ir1 = (4 - 0.7)/220 A, Vbase = 4 - (220 x Ir1) V, Vemitter = (4 - 0.7) = 3.3 V, Iemitter = (3.3 - 2.2)/220 A.

If V1 = 7 V, Ir1 = (7 - 5.1)/220 V (because the zener is not in breakdown, so the voltage at the node is 5.1 V), Vbase = 5.1 V, Vemitter = (5.1 - 0.7) = 4.4 V, Iemitter = (4.4-2.2)/220 A.

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I am trying to analyze the effect of varying V1 on the circuit voltages and currents,

Until the zener starts to conduct, it appears as a high resistance device and Vb = V1. So the output voltage Ve = Vb - Vbe = V1 - 0.7v.

Once the zener starts to conduct, it clamps Vb to its reverse voltage Vz, so Ve = Vb - Vbe = Vz - 0.7v.

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  • \$\begingroup\$ Thanks for the precise explanation. So my answers to are correct, right? \$\endgroup\$ – abruzzi26 Feb 12 '17 at 19:56
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You have understood this circuit ok for large signals and low frequency.

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