Zener Diode as Voltage Regulator

PHYp–n Junction Diodes

A Zener diode operated in reverse breakdown holds an almost constant voltage across itself, which makes it an ideal voltage regulator across a load. ISC numericals apply Kirchhoff's laws to the series-resistor / Zener / load network: find the series current, the load current, the Zener current as the difference, and the power dissipated in the Zener; or size the series resistor for a required load and Zener current.

Series-resistor current

I_S = \dfrac{V_{in} - V_Z}{R_S}

I_S

= current through series resistor,

V_{in}

= unregulated input,

V_Z

= Zener (breakdown) voltage,

R_S

= series resistance.

Load current

I_L = \dfrac{V_Z}{R_L}

I_L

= load current; the regulated load sees the constant voltage

V_Z

Zener current (KCL)

I_Z = I_S - I_L

I_Z

= current through the Zener; the series current splits between Zener and load.

Power dissipated in Zener

P_Z = V_Z \, I_Z

P_Z

= Zener power dissipation; must stay below the diode's rated power.

Sizing the series resistor

R_S = \dfrac{V_{in} - V_Z}{I_Z + I_L}

Rearranged from

I_S = I_Z + I_L

; choose

R_S

so both required currents are supplied.

The Zener must be reverse-biased beyond its breakdown voltage; only then does it clamp the output at $V_Z$ .
The regulated output voltage equals $V_Z$ as long as the input exceeds $V_Z$ and the Zener stays in breakdown.
The series resistor drops the excess voltage $V_{in} - V_Z$ and limits the current.
By KCL the series current splits: $I_S = I_Z + I_L$ , so $I_Z = I_S - I_L$ .
If the load draws more current, $I_Z$ falls to keep $I_S$ and the output steady (and vice versa).
A minimum Zener current is needed to stay in regulation; size $R_S$ to guarantee it at the heaviest load.
Check $P_Z = V_Z I_Z$ against the diode rating, especially at no-load when $I_Z$ is maximum.

Where the marks go

Using $V_{in}$ instead of $V_Z$ to find the load current $I_L = V_Z/R_L$ .
Adding $I_Z$ and $I_L$ wrongly — series current is their sum, Zener current is series minus load.
Forgetting the Zener works in reverse breakdown, not forward bias.
Computing Zener power with $V_{in}$ rather than $V_Z$ , or mixing mA and A in $P_Z = V_Z I_Z$ .

How the board asks it

Numericalkcl split $I_S = I_Z + I_L$
In a Zener regulator an unregulated supply of $V_{in} = 20\,V$ is connected through a series resistor $R_S = 200\,\Omega$ to a $10\,V$ Zener diode placed across a load $R_L = 2\,k\Omega$ . Calculate the series current, the load current and the Zener current.
Numericalsizing the series resistor at minimum input
A $9\,V$ Zener diode is to regulate a supply that varies between $15\,V$ and $18\,V$ while delivering a load current of $30\,mA$ . If the Zener requires a minimum current of $5\,mA$ to stay in breakdown, calculate a suitable value of the series resistance $R_S$ .
Numericalpower dissipated $P_Z = V_Z I_Z$ at no load
A $6\,V$ Zener diode is fed from a $12\,V$ supply through a $150\,\Omega$ series resistor. Calculate the power dissipated in the Zener diode when the load is disconnected, and state why this is the worst case for the diode rating.
Diagram / graphseries-resistor / zener / load network
Draw a labelled circuit diagram to show how a Zener diode is used as a voltage regulator, marking the input voltage, the series resistor, the Zener diode and the load resistance.
Give reasons $I_Z$ adjusts to keep output steady
Explain how a Zener regulator keeps its output voltage almost constant when the load current changes. Why must the Zener diode be operated in reverse breakdown?
Define / stateregulated output equals $V_Z$
State the condition on the input voltage under which a Zener diode acts as a voltage regulator, and write down the value of the regulated output voltage.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.