---
title: "Design regulator for 16 V from 25 V"
subject: "Electronics Devices and Circuit"
chapter: 6
key: "semester_3/electronics_devices_and_circuit:dc-regulator-design-16v-25v"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/dc-regulator-design-16v-25v
site: BCT Engineering
---

# Design regulator for 16 V from 25 V

Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators

- Asked in: 2071 Shrawan, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/dc-regulator-design-16v-25v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6

## Question

Design a regulator circuit to obtain 16 VDC. Choose approximate values of the parameters. Input voltage is 25 V.

## Short answer

For a Zener shunt regulator, use Rs=150Ω and a16V,1.3W Zener for up to50mA load. For higher current, use LM317 with R1=240Ω, R2=2.83kΩ.

## Answer

**Requirement:** Output 16 V dc from a 25 V input. Choose a simple Zener shunt regulator for small loads, and give the LM317 alternative for larger loads.

#### Design 1: Zener shunt regulator
```text
 25 V o----[ Rs ]-----+-------+----o Vo = 16 V
                      |       |
                    (Dz 16V) [RL]
                      |       |
                     GND     GND
```

Assumptions: load current IL(max) = 50 mA, minimum Zener current Iz(min) = 10 mA (above the knee), Zener Vz = 16 V (standard BZX/1N-series 16 V, 1.3 W).

1. Current in the series resistor: Is = IL(max) + Iz(min) = 50 + 10 = 60 mA
2. Rs = (Vin − Vz)/Is = (25 − 16)/60 mA = **150 Ω**
3. Power in Rs: Is² Rs = (0.06)² × 150 = 0.54 W → use a 1 W resistor.
4. Maximum Zener current (no load): Iz(max) = (25 − 16)/150 = 60 mA, and Pz = 16 × 60 mA = 0.96 W → use a **1.3 W** Zener (or 2 W for safety).
5. Add C = 100 µF across the output to reduce ripple.

#### Design 2: LM317 (for current up to 1.5 A)
Vout = 1.25 (1 + R2/R1): with R1 = 240 Ω, R2 = 240 (16/1.25 − 1) = **2.83 kΩ** (use 2.7 kΩ + 130 Ω, or a 2.2 kΩ fixed with 1 kΩ trimmer). Vin = 25 V is above Vout + 3 V = 19 V, so regulation is maintained. The LM317 dissipates (25 − 16) IL, so for 0.5 A it needs about 4.5 W and a heat sink.

**Answer: Rs = 150 Ω and a 16 V, 1.3 W Zener for up to 50 mA, or LM317 with R1 = 240 Ω, R2 = 2.83 kΩ.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the Zener design, what current is used to find Rs?
   A. Is = IL(max) - Iz(min) = 40 mA
   B. Is = IL(max) + Iz(min) = 50 + 10 = 60 mA
   C. Is = Iz(max) = 60 mA with no load only
   D. Is = IL(max) = 50 mA
   Answer: B. Rs must pass the full load current and keep the Zener above knee at the same time. (Key point: Series resistor current)

2. What is Rs?
   A. 16/60 mA = 267 Ω
   B. 25/60 mA = 417 Ω
   C. (25 - 16)/50 mA = 180 Ω
   D. (25 - 16)/60 mA = 150 Ω
   Answer: D. The drop across Rs is Vin - Vz = 9 V at 60 mA. (Key point: Rs value)

3. Why is a 1.3 W Zener chosen?
   A. With no load, all 60 mA flows through the Zener: Pz = 16 x 60 mA = 0.96 W
   B. At full load Pz = 16 x 50 mA = 0.8 W exactly
   C. Pz = Is^2 Rs = 0.54 W
   D. The Zener must dissipate 25 V x 60 mA
   Answer: A. Worst-case Zener power occurs at no load, so the rating must exceed 0.96 W. (Key point: Zener power rating)

4. What R2 gives 16 V with R1 = 240 Ω in the LM317 design?
   A. R2 = 240 x 16/1.25 = 3.07 kΩ
   B. R2 = 240(25/1.25 - 1) = 4.56 kΩ
   C. R2 = 240(16/1.25 - 1) = 2.83 kΩ
   D. R2 = 240 x 1.25/16 = 18.8 Ω
   Answer: C. Using R2 = R1(Vout/1.25 - 1) with Vout = 16 V. (Key point: LM317 alternative)

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