Zener Diode Voltage Regulators

Six worked problems on the shunt Zener regulator: testing whether the diode is in breakdown, finding the output voltage and branch currents, establishing the load and line ranges over which regulation survives, designing the series resistor and the diode power rating, and quantifying line and load regulation from the Zener dynamic impedance.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

A silicon Zener diode with \( V_Z = 10~\text{V} \) and a power rating of \( P_{Z,\max} = 400~\text{mW} \) is used in the simplest shunt regulator: a series resistor \( R_s = 1~\text{k}\Omega \) runs from a \( +16~\text{V} \) supply to the top of the Zener, the Zener cathode sits at that node and its anode goes to ground, and the load \( R_L \) is connected in parallel with the Zener.

1. With \( R_L = 2.2~\text{k}\Omega \), decide whether the Zener conducts, then find \( V_o \), \( I_R \), \( I_L \), \( I_Z \) and \( P_Z \).

2. Repeat for \( R_L = 1.2~\text{k}\Omega \).

• \( V_i = 16~\text{V} \), \( R_s = 1~\text{k}\Omega \), \( V_Z = 10~\text{V} \), \( P_{Z,\max} = 400~\text{mW} \).

• Ideal Zener model: an open circuit below \( V_Z \), a \( 10~\text{V} \) battery once it breaks down.

Step 1 — The ON/OFF test

Remove the Zener and find the open-circuit voltage that would appear across it. The remaining circuit is just \( R_s \) and \( R_L \) in series:

The diode conducts only if this \( V \ge V_Z \).

Part (a) — \( R_L = 2.2~\text{k}\Omega \)

so the Zener is ON and clamps the output at \( V_Z \).

(a) Zener ON: \( V_o = 10~\text{V} \), \( I_R = 6.00~\text{mA} \), \( I_L = 4.545~\text{mA} \), \( I_Z = 1.455~\text{mA} \), \( P_Z = 14.55~\text{mW} \) — only 3.6% of the 400 mW rating.

Part (b) — \( R_L = 1.2~\text{k}\Omega \)

The Zener never reaches breakdown, so it is OFF — an open circuit. The output is simply the divider voltage and the whole current flows through the load:

(b) Zener OFF: \( V_o = 8.727~\text{V} \), \( I_Z = 0 \), \( I_R = I_L = 7.273~\text{mA} \), \( P_Z = 0 \) — the circuit no longer regulates.

Lowering \( R_L \) pulls the unloaded node voltage down; once it falls below \( V_Z \) the Zener drops out and the "regulator" degenerates into a potential divider. Everything that follows in this set is about keeping the operating point on the right side of that boundary.

A \( 12~\text{V} \), \( 400~\text{mW} \) Zener diode is supplied from a fixed \( 20~\text{V} \) source through a series resistor \( R_s = 220~\Omega \); the load is in parallel with the Zener. Determine the range of load resistance \( R_L \) and of load current \( I_L \) over which the output stays regulated at 12 V. Also find the power dissipated in \( R_s \).

• \( V_i = 20~\text{V} \) (fixed), \( R_s = 220~\Omega \), \( V_Z = 12~\text{V} \), \( P_{Z,\max} = 400~\text{mW} \).

• Ideal Zener; \( I_{Z,\min} \approx 0 \) is not assumed to be the binding limit — the diode's own rating is.

Step 1 — The resistor current is fixed

While the Zener regulates, the node voltage is pinned at 12 V, so

This current is shared between the Zener and the load: \( I_R = I_Z + I_L \). The whole problem is deciding how far that split may be pushed either way.

Step 2 — Smallest \( R_L \) (largest \( I_L \))

The Zener stops conducting when the load takes the entire \( I_R \), i.e. \( I_Z = 0 \). Equivalently, the unloaded divider voltage must still reach \( V_Z \):

Step 3 — Largest \( R_L \) (smallest \( I_L \))

As the load is lightened the surplus current is dumped into the Zener, and the power rating sets the ceiling:

Step 4 — Power in the series resistor

\( 330~\Omega \le R_L \le 3.96~\text{k}\Omega \), i.e. \( 3.03~\text{mA} \le I_L \le 36.36~\text{mA} \); \( R_s \) must be rated for \( 290.9~\text{mW} \) (use a \( \tfrac12~\text{W} \) part).

Note the load may not be removed completely: with \( I_L = 0 \) the Zener would take the full \( 36.36~\text{mA} \) and dissipate \( 436~\text{mW} \), well beyond its 400 mW rating.

A \( 10~\text{V} \) Zener with \( P_{Z,\max} = 400~\text{mW} \) and a knee current \( I_{Z,\min} = 5~\text{mA} \) feeds a fixed load \( R_L = 4~\text{k}\Omega \) through a series resistor \( R_s = 1~\text{k}\Omega \). The supply \( V_i \) is unregulated. Over what range of \( V_i \) does the circuit hold the output at 10 V?

• \( V_Z = 10~\text{V} \), \( R_s = 1~\text{k}\Omega \), \( R_L = 4~\text{k}\Omega \).

• \( I_{Z,\min} = 5~\text{mA} \) (knee), \( P_{Z,\max} = 400~\text{mW} \).

Step 1 — The load current is fixed

While regulation holds, \( V_o = V_Z \), so

Now everything that \( V_i \) does shows up as a change in \( I_Z \) alone, through

Step 2 — Lower limit: keep the diode above the knee

Step 3 — Upper limit: keep inside the power rating

\( 17.5~\text{V} \le V_i \le 52.5~\text{V} \) — a 3.00-to-1 input range for a constant 10 V output.

The window is wide because \( R_s \) is large compared with what the load needs; the price is a maximum Zener dissipation of 400 mW at the top of the range, all of it wasted heat.

Two Zener diodes are stacked in series to obtain a non-standard output: a \( 6.2~\text{V} \) diode and an \( 8.2~\text{V} \) diode, each rated \( 500~\text{mW} \), with the anode of the upper one joined to the cathode of the lower one. The stack is fed from \( V_i = 24~\text{V} \) through \( R_s = 330~\Omega \) and drives a load \( R_L = 1.5~\text{k}\Omega \) in parallel with it.

1. Find \( V_o \), \( I_R \), \( I_L \), \( I_Z \) and the power in each diode.

2. Find the smallest load resistance that still keeps the stack in breakdown.

3. With the load disconnected, what is the largest \( V_i \) the stack can survive?

• \( V_{Z1} = 6.2~\text{V} \), \( V_{Z2} = 8.2~\text{V} \), each \( 500~\text{mW} \).

• \( V_i = 24~\text{V} \), \( R_s = 330~\Omega \), \( R_L = 1.5~\text{k}\Omega \).

Part (a) — Operating point

Series diodes carry the same current and their breakdown voltages add:

Check that the stack really is in breakdown: unloaded, the divider would give \( 24 \times 1500/1830 = 19.67~\text{V} > 14.4~\text{V} \), so it is. Then

(a) \( V_o = 14.4~\text{V} \), \( I_R = 29.09~\text{mA} \), \( I_L = 9.60~\text{mA} \), \( I_Z = 19.49~\text{mA} \), \( P_{Z1} = 120.8~\text{mW} \), \( P_{Z2} = 159.8~\text{mW} \) — both comfortably inside 500 mW.

Part (b) — Minimum load resistance

Regulation is lost when \( I_Z \to 0 \), i.e. when the load takes all of \( I_R \):

(b) \( R_{L,\min} = 495~\Omega \), corresponding to \( I_{L,\max} = 29.09~\text{mA} \).

Part (c) — Maximum input with no load

Because the current is common, the diode that reaches its rating first is the one with the larger voltage:

With \( I_L = 0 \) all of it flows through \( R_s \):

(c) \( V_{i,\max} = 34.52~\text{V} \), set by the 8.2 V diode.

Stacking Zeners is how odd voltages such as 14.4 V are made from stock parts, but remember the rating check runs on the highest-voltage diode in the string, not on the total.

Design. A 12 V, 40 mA maximum load must be supplied from an unregulated rail that wanders between 18 V and 24 V. A 12 V Zener is to be used as a shunt regulator, and the diode needs at least \( I_{Z,\min} = 5~\text{mA} \) to stay on the flat part of its breakdown characteristic. The load current may fall to zero.

1. Find the largest permissible series resistance \( R_s \) and choose a standard value.

2. Determine the worst-case Zener dissipation and specify a power rating.

3. Specify the rating of \( R_s \) and verify the design at the worst corner.

• \( V_i: 18 \to 24~\text{V} \), \( V_Z = 12~\text{V} \), \( I_L: 0 \to 40~\text{mA} \), \( I_{Z,\min} = 5~\text{mA} \).

• Ideal Zener (\( Z_Z \approx 0 \)); \( R_s \) chosen from the E12 series.

Step 1 — Identify the two worst corners

• Lowest current available: \( V_i \) at its minimum and the load at its maximum. \( R_s \) must still pass \( I_L + I_{Z,\min} \).

• Highest diode current: \( V_i \) at its maximum and the load removed. Everything \( R_s \) delivers goes into the Zener.

Step 2 — Size \( R_s \) from the first corner

Take the nearest standard value below this — going above it would starve the diode at the low-line, full-load corner.

(a) \( R_{s,\max} = 133.3~\Omega \); choose \( R_s = 120~\Omega \).

Step 3 — Zener rating from the second corner

(b) \( P_{Z,\text{worst}} = 1.20~\text{W} \); specify a 2 W Zener (about 1.7× margin, which also covers ambient-temperature derating).

Step 4 — Resistor rating and design check

Check the low-line, full-load corner with the chosen 120 Ω:

(c) \( R_s = 120~\Omega \), 2 W; the design holds at both corners.

The cost of the shunt topology is stark here: to deliver 0.48 W to the load the circuit burns up to \( 1.20 + 1.2 = 2.4~\text{W} \) at no load. A shunt Zener is only sensible for small or fairly constant loads — anything bigger wants a series pass element.

A Zener regulator uses \( R_s = 470~\Omega \) from a nominal \( 20~\text{V} \) supply. The diode is specified as \( V_Z = 10~\text{V} \) at a test current \( I_{ZT} = 20~\text{mA} \), with a dynamic impedance \( Z_Z = 7~\Omega \).

1. Replace the diode by its piecewise-linear model and find the no-load and full-load outputs for a load current that varies from 0 to 20 mA. Hence find the load regulation in per cent.

2. With the load fixed at 20 mA, the supply swings from 18 V to 24 V. Find the change in output and the line regulation in %/V.

• \( R_s = 470~\Omega \), \( Z_Z = 7~\Omega \), \( V_Z = 10~\text{V} \) at \( I_{ZT} = 20~\text{mA} \).

• Piecewise-linear Zener: an ideal battery \( V_{Z0} \) in series with \( Z_Z \), valid for \( I_Z \) above the knee.

Step 1 — Extract \( V_{Z0} \) from the data sheet point

The quoted 10 V already contains the \( I_{ZT}Z_Z \) drop, so

Step 2 — Node equation of the regulator

Writing KCL at the output node, with the load drawing \( I_L \),

Part (a) — Load regulation

The drop is exactly what the small-signal picture predicts — the load current change is absorbed by \( Z_Z \) in parallel with \( R_s \):

(a) \( V_{o,NL} = 10.0088~\text{V} \), \( V_{o,FL} = 9.8709~\text{V} \), load regulation \( = 1.40~\% \).

Part (b) — Line regulation

With \( I_L \) held constant the circuit is a plain divider for changes:

(b) \( \Delta V_o = 88.1~\text{mV} \) for a 6 V input swing; line regulation \( = 0.1487~\%/\text{V} \), i.e. a ripple attenuation of \( 1/0.01468 = 68.1 \).

Both figures improve as \( Z_Z \) shrinks and as \( R_s \) grows — but a larger \( R_s \) also shrinks the load range of Problem 2, so the regulator's line performance and its current capability pull in opposite directions.