## Regulator using zener diode only

[caption id="attachment_714" align="aligncenter" width="502"] Regulator using zener diode only[/caption]

- Across RL we have: VO = VZ = ILRL (Equation 1)
- From current law: I = IZ + IL (Equation 2)
- From KVL along indicated path: VS = I*R + VZ (Equation 3)

Equation 1 tells that output voltage VO will always be constant = VZ.

Assume two cases:

- Assume supply current I change by dI

From Equation 2: ?I = ?IZ + ?IL

From Equation 1: ?VZ = ?ILRL ; or, ?IL = 0 (since ?VZ = 0)

Thus ?I = ?IZ. This shows that excess current is bifurcated through the zener diode.

- Assume load RL changes by ?RL (with VS constant)

Output voltage VO will remain constant, but change in IL will be compensated by change in IZ

From Equation 3: ?VS = ?I*R+ ?VZ ; or, 0 = ?I*R + 0 ; or, ?I = 0

From Equation 2: ?I = ?IZ + ?IL ; or, 0 = ?IZ + ?IL ; or, ?IL = - ?IZ

Thus if IL increases, IZ decreases and vice versa.

## Regulator using transistor and zener diode

[caption id="attachment_715" align="aligncenter" width="493"] Regulator using transistor and zener diode[/caption]

**Few points:**

Correlating VO and indicated path from point X to GND: VO = VX = VZ + VBE (Equation 1)

I = IB + IC + IL ; or, I = IC + IL(since IB is very small) (Equation 2)

The increase in VBE causes more collector current IC to flow.

- Assume current I increase by dI keeping VS constant (opposite analysis will take place of I decreases)?I is positive. VS – I*R = VX ; or, 0 – ?I*R = ?VX ; (since VS is constant)

i.e. VX = VO decreases on increase in I. (Effect 1: VO tends to decrease)Next, from Equation 1: ?VO = ?VZ + ?VBE ; or, ?VO = 0 + ?VBE ;

i.e VBE also decreases on decrease in VO

As VBE decreases, IC decreases.From Equation 2: ?I = ?IC + ?IL

If ?I = positive (assumed); ?IC = negative (as VBE decrease); so IL must increase.

The voltage across load VL = IL*RL increases. (Effect 2: VO tends to increase)The Effect 1 and Effect 2 neutralize and VO is constant.

- Assume supply voltage VS is increased keeping current I constantThe analysis will take place just as done previously.

VS – I*R = VX ; or, ?VS – 0 = ?VX ; (since I is constant)

i.e. VX = VO increases on increase in VS. (Effect 1: VO tends to increase)Next, from Equation 1: ?VO = ?VZ + ?VBE ; or, ?VO = 0 + ?VBE ;

i.e VBE also increases on increase in VO As VBE decreases, IC increases.From Equation 2: ?I = ?IC + ?IL

If ?I = 0 (assumed); ?IC = positive (as VBE increase); so IL must decrease.

The voltage across load VL = IL*RL decreases. (Effect 2: VO tends to decrease)

The Effect 1 and Effect 2 neutralize and VO is constant.

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