A zener diode forms an integral part of any voltage regulator. Before we go ahead, we know, that in all cases, the voltage across a zener diode will remain constant. i.e. ∆V_{Z} = 0. In all cases, we indicate load resistance by R_{L.}

### Regulator using zener diode only

- Across R
_{L}we have: V_{O}= V_{Z}= I_{L}R_{L }(Equation 1) - From current law: I = I
_{Z}+ I_{L}(Equation 2) - From KVL along indicated path: V
_{S}= I*R + V_{Z}(Equation 3)

Equation 1 tells that output voltage V_{O} will always be constant = V_{Z}.

Assume two cases:

- Assume supply current I change by dI

From Equation 2: ∆I = ∆I_{Z}+ ∆I_{L }From Equation 1: ∆V_{Z}= ∆I_{L}R_{L}; or, ∆I_{L}= 0 (since ∆V_{Z}= 0)

Thus ∆I = ∆I_{Z}. This shows that excess current is bifurcated through the zener diode.

- Assume load R
_{L}changes by ∆R_{L}(with V_{S}constant)

Output voltage V_{O}will remain constant, but change in I_{L}will be compensated by change in I_{Z }From Equation 3: ∆V_{S}= ∆I*R+ ∆V_{Z}; or, 0 = ∆I*R + 0 ; or, ∆I = 0

From Equation 2: ∆I = ∆I_{Z}+ ∆I_{L}; or, 0 = ∆I_{Z}+ ∆I_{L}; or, ∆I_{L}= – ∆I_{Z}

Thus if I_{L} increases, I_{Z} decreases and vice versa.

### Regulator using transistor and zener diode

**Few points:**

Correlating V_{O} and indicated path from point X to GND: V_{O} = V_{X} = V_{Z} + V_{BE} (Equation 1)

I = I_{B} + I_{C} + I_{L} ; or, I = I_{C} + I_{L}(since I_{B} is very small) (Equation 2)

The increase in V_{BE} causes more collector current I_{C} to flow.

**Assume current I increase by dI keeping V**∆I is positive. V_{S}constant (opposite analysis will take place of I decreases)_{S}– I*R = V_{X}; or, 0 – ∆I*R = ∆V_{X}; (since V_{S}is constant)

i.e. V_{X}= V_{O }decreases on increase in I. ()__Effect 1: V___{O}tends to decreaseNext, from Equation 1: ∆V

_{O}= ∆V_{Z}+ ∆V_{BE}; or, ∆V_{O}= 0 + ∆V_{BE};

i.e V_{BE}also decreases on decrease in V_{O }As V_{BE}decreases, I_{C }decreases.From Equation 2: ∆I = ∆I

_{C}+ ∆I_{L }If ∆I = positive (assumed); ∆I_{C}= negative (as V_{BE}decrease); so I_{L}must**increase**.

The voltage across load V_{L}= I_{L}*R_{L }increases. ()__Effect 2: V___{O}tends to increaseThe Effect 1 and Effect 2 neutralize and V

_{O}is constant.

**Assume supply voltage V**The analysis will take place just as done previously._{S}is increased keeping current I constant

V_{S}– I*R = V_{X}; or, ∆V_{S}– 0 = ∆V_{X}; (since I is constant)

i.e. V_{X}= V_{O }increases on increase in V_{S}. ()__Effect 1: V___{O}tends to increaseNext, from Equation 1: ∆V

_{O}= ∆V_{Z}+ ∆V_{BE}; or, ∆V_{O}= 0 + ∆V_{BE};

i.e V_{BE}also increases on increase in V_{O }As V_{BE}decreases, I_{C }increases.From Equation 2: ∆I = ∆I

_{C}+ ∆I_{L }If ∆I = 0 (assumed); ∆I_{C}= positive (as V_{BE}increase); so I_{L}must**decrease**.

The voltage across load V_{L}= I_{L}*R_{L }decreases. (The Effect 1 and Effect 2 neutralize and V__Effect 2: V___{O}tends to decrease)

_{O}is constant.

This post is written by Sayantan Roychowdhury. If you liked this article please share it with your friends. Like our facebook page and subscribe to our newsletter for future updates.