# Shunt Voltage Regulator – Working Principle

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. ?VZ = 0. In all cases, we indicate load resistance by RL.

## Regulator using zener diode only

• 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

### 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.