A Zener Regulator That Regulates Only Some of the Time
A student builds a simple regulated supply: a 10 V unregulated DC input feeds a 100 Ω series resistor and a 6.0 V zener diode connected in parallel with the load. With a 200 Ω load the output is a steady 6.0 V. With a 100 Ω load the output falls to 5.0 V, and the student replaces the zener, only to see the same 5.0 V. He then raises the input to 12 V with no load connected; the zener quickly becomes too hot to touch although it is rated 250 mW. He has a box of identical 6.0 V zener diodes and a selection of resistors, and wants to know whether the circuit can be made to regulate down to a 100 Ω load without overheating.
1. Explain the 5.0 V output with calculations, showing that the zener is not faulty. Then find the minimum load resistance for which this circuit still regulates, taking the minimum zener current as 5 mA.
Read the worked answer
With 10 V in and 6.0 V across the load, the drop across the 100 Ω series resistor is 4.0 V, so
$I_S=4.0/100=40\ \text{mA}$.
At 200 Ω the load takes $6.0/200=30\ \text{mA}$, leaving 10 mA for the zener, which is above the 5 mA knee, so regulation holds.
At 100 Ω the load would need $6.0/100=60\ \text{mA}$, more than the 40 mA the series resistor can supply at 6 V. The zener therefore turns off and the circuit becomes a simple divider:
$V_{\text{out}}=10\times\dfrac{100}{100+100}=5.0\ \text{V}$.
This matches the measurement exactly, so the zener is fine; the load is simply too heavy.
For regulation, $I_S\ge I_L+5\ \text{mA}$, so $I_{L,\max}=40-5=35\ \text{mA}$ and
$R_{L,\min}=\dfrac{6.0}{35\times10^{-3}}=171\ \Omega$.
A 100 Ω load is outside the design range.
2. Calculate the zener dissipation at 12 V with no load and at 12 V with the 200 Ω load, and decide whether the 250 mW rating is adequate.
Read the worked answer
At 12 V input with no load, the whole series current flows through the zener:
$I_S=\dfrac{12-6.0}{100}=60\ \text{mA}$,
$P_Z=6.0\times60\times10^{-3}=0.36\ \text{W}=360\ \text{mW}$.
That exceeds the 250 mW rating, which is why the diode becomes too hot to touch; prolonged operation would destroy it.
With the 200 Ω load, the load takes 30 mA, so the zener carries $60-30=30\ \text{mA}$ and
$P_Z=6.0\times30\times10^{-3}=0.18\ \text{W}=180\ \text{mW}$,
which is within rating. The worst case is always maximum input with minimum load, and that is the case the student tested.
3. Choose a series resistor that allows regulation down to a 100 Ω load at 10 V input with the 5 mA zener minimum, then check the worst-case zener dissipation at 12 V no load. Advise the student on the trade-off.
Read the worked answer
At 10 V input and 6.0 V output, the load current at 100 Ω is 60 mA. Including the 5 mA zener minimum, the series resistor must pass
$I_S=65\ \text{mA}$, so
$R_S=\dfrac{10-6.0}{65\times10^{-3}}=61.5\ \Omega$, i.e. 62 Ω from the box.
Now check the worst case, 12 V input with no load:
$I_S=\dfrac{12-6.0}{62}=97\ \text{mA}$,
$P_Z=6.0\times97\times10^{-3}=0.58\ \text{W}$.
The 250 mW diode is badly overloaded, so a zener rated at least 1 W would be needed, with a suitable heatsink.
The trade-off is unavoidable in this shunt topology: lowering $R_S$ to supply a heavier load forces the zener to absorb more current at high input and light load. The student should either accept the higher-power zener, restrict the input to 10 V, use a higher load resistance, or move to a series-pass regulator where the control element does not have to swallow the full surplus current.