Zener Regulator Under Load and High-Line Conditions
A student designs a shunt Zener regulator to provide 6.2 V to a 1.0 kΩ load from a nominal 12 V supply that can rise to 15 V. The chosen Zener diode is rated 6.2 V, 500 mW. The series resistor is 220 Ω. At 12 V with the 1.0 kΩ load connected, the student measures 6.2 V across the load and 5.8 V across the resistor. When an extra 220 Ω load is connected in parallel, the output falls to about 5.4 V. The student asks whether the Zener is safe at 12 V, what minimum load resistance keeps regulation, and why the extra load causes the output to fall. Assume an ideal Zener once breakdown occurs and neglect Zener dynamic resistance. The supply ripple is small compared with the differences being tested.
1. At 12 V with the 1.0 kΩ load, calculate the currents in the series resistor, load and Zener. Is the Zener power rating exceeded?
Read the worked answer
The resistor drops 12 - 6.2 = 5.8 V, soI_R = 5.8 / 220 = 26.4 mA.
The load current is I_L = 6.2 / 1000 = 6.2 mA.
The Zener takes the difference: I_Z = 26.4 - 6.2 = 20.2 mA.
Zener power = 6.2 × 0.0202 = 0.125 W = 125 mW, below the 500 mW rating. At 12 V and this load the design is safe and regulating, with a power margin of about 4:1.
2. Find the minimum load resistance that maintains regulation when the supply is at its worst-case 15 V. Explain.
Read the worked answer
At 15 V,I_R = (15 - 6.2) / 220 = 8.8 / 220 = 40.0 mA.
For the Zener to stay in breakdown, I_Z must be at least zero, so the load may draw at most 40.0 mA:R_L,min = 6.2 / 0.040 = 155 Ω.
At 12 V the limit would be 6.2 / 0.0264 = 235 Ω, but the 15 V case is worse. Any load below about 155 Ω can pull the Zener out of regulation when the input is high.
3. Why does adding the 220 Ω load in parallel make the output fall to about 5.4 V? Is the Zener necessarily damaged?
Read the worked answer
At 6.2 V, the extra 220 Ω would take 6.2 / 220 = 28.2 mA; with the 1 kΩ load, the total demand would be about 34.4 mA. But at 12 V the 220 Ω series resistor can supply only 26.4 mA, so the Zener turns off. The output is then set by the divider formed by 220 Ω and the parallel load 1 kΩ || 220 Ω = 180 Ω:V_out = 12 × 180 / (220 + 180) = 5.4 V.
This is loss of regulation, not necessarily damage. At 5.4 V the Zener is below breakdown, so its dissipation is low.