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

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

This post is written by Sayantan Roychowdhury. If you like this article, please share it with your friends and like or facebook page for future updates. Subscribe to our newsletter to get notifications about our updates via email. If you have any queries, feel free to ask in the comments section below. Have a nice day!

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