- Asked 8/28
- 2073 Shrawan · 4 marks
Design LM317 regulator 3.7 V to 12 V
Short answer
R1=240 Ω. For 3.7 V, R2 min=470 Ω; for 12 V, R2 max≈2064 Ω. Use 470 Ω fixed + 2 kΩ pot (set for 2064 Ω max). Input voltage about 16 V (≥15 V).
Design a DC voltage regulator for 3.7 V to 12 V output using LM317.
LM317 is a three-terminal adjustable positive regulator (1.25 V to 37 V, up to 1.5 A). Between OUT and ADJ it holds a constant reference Vref = 1.25 V, so with R1 from OUT to ADJ and R2 from ADJ to ground:
Vout = 1.25 (1 + R2/R1) + Iadj R2 (Iadj ≈ 50 µA, neglected)
Circuit
+--------------+
Vin o--+---| IN OUT |---+------+--o Vout
| | LM317 | | |
[C1] | ADJ | [R1] [C2]
| +------+-------+ | |
GND +-----------+ GND
|
[R2 = fixed + pot]
|
GND
C1 = 0.1 µF (input bypass), C2 = 1 µF (output).
Design for 3.7 V to 12 V
- Choose R1 = 240 Ω. The current in R1 is 1.25/240 = 5.2 mA, which is above the minimum load current of the LM317 (about 3.5 to 5 mA).
- R2 = R1 (Vout/1.25 − 1)
- Minimum output 3.7 V: R2(min) = 240 (3.7/1.25 − 1) = 470 Ω
- Maximum output 12 V: R2(max) = 240 (12/1.25 − 1) = 2064 Ω
- A potentiometer alone cannot give a minimum above 1.25 V, so use a fixed resistor in series with the pot:
- Fixed resistor = 470 Ω (a standard value)
- Potentiometer = 2064 − 470 = 1594 Ω → use the next standard value, a 2 kΩ pot, and set it so that the maximum output is 12 V
- Input voltage: the LM317 needs a dropout of about 3 V, so Vin ≥ 12 + 3 = 15 V. Use an unregulated supply of about 16 V dc (the maximum Vin − Vout is 40 V).
The pass transistor inside the IC dissipates (Vin − Vout) IL, so a heat sink is needed at high current.
Answer: R1 = 240 Ω, R2 = 470 Ω fixed + 2 kΩ pot, Vin ≈ 16 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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