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
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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