Linear Regulator Formulas: Power and Efficiency
Linear regulator formulas: power dissipated, efficiency, minimum input from the dropout, and the resistors of an adjustable LM317-style regulator.
The formulas
| To find | Formula |
|---|---|
| Input current | Iin = Iout + Iq |
| Power dissipated | Ploss = (Vin − Vout) × Iout + Vin × Iq |
| Efficiency | η = Vout × Iout / (Vin × Iin) |
| Minimum input voltage | Vin min = Vout + Vdropout |
| Adjustable regulator output | Vout = 1.25 × (1 + R2 / R1) (ignoring the adjust-pin current) |
Iq is the quiescent (ground) current of the regulator, and Vdropout the voltage it needs between input and output, from the data sheet. A linear regulator passes the load current through and burns the voltage difference as heat, so its efficiency is close to Vout / Vin. Use the heat sink article to find the junction temperature from the power. The 1.25 V reference is the usual value for an LM317-type part, and R1 is often 240 Ω.
Worked example: 12 V to 5 V at 1 A
The quiescent current is 5 mA and the dropout is 1.2 V.
- Iin = 1 + 0.005 = 1.005 A, so Pin = 12 × 1.005 = 12.06 W and Pout = 5 W
- Ploss = 12.06 − 5 = 7.06 W
- Efficiency = 5 / 12.06 = 41.46%
- The smallest usable input is 5 + 1.2 = 6.2 V
Worked example: adjustable regulator for 5 V
R1 = 240 Ω.
- R2 = R1 × (Vout / 1.25 − 1) = 240 × (5 / 1.25 − 1) = 720 Ω
Loss for a 5 V, 1 A output
| Input voltage | Power lost | Efficiency |
|---|---|---|
| 7 V | 2.035 W | 71.07% |
| 9 V | 4.045 W | 55.28% |
| 12 V | 7.06 W | 41.46% |
| 15 V | 10.07 W | 33.17% |
Frequently asked questions
Why is a linear regulator inefficient with a high input?
The extra voltage drops across the pass element at the full load current, and all of that power is heat. A 12 V input for a 5 V output wastes more than half of the power.
When is a linear regulator the right choice?
For small currents, low noise, a small voltage difference or a simple design. For larger differences and currents a switching converter wastes less.
What is a low-dropout (LDO) regulator?
A regulator that works with only a few hundred millivolts between input and output, which raises the efficiency when the input is close to the output.
Does the quiescent current matter?
In battery equipment, yes, because it flows all the time. For a 1 A load it is small, as in the example.
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