Before anything else: this page is for low-voltage work
Everything here assumes a low-voltage supply on a bench — a DC brick, a battery, a lab supply feeding regulators on a board. Anything connected to household mains is a different problem with a fatal failure mode. Mains-side work is licensed work in most places for good reason, it is governed by codes this page does not cover, and none of the arithmetic below applies to it. If your question involves what happens on the primary side of a transformer, in a wall box, or anywhere upstream of the low-voltage output, this is not the page and no page is a substitute for a qualified electrician. Buy a finished, certified supply and start your design at its output terminals.
Adding the rails up
Each rail's output power is its voltage times its current. Sum those and you have what the circuit actually consumes, which is usually a smaller number than people expect: a microcontroller board, a display and a handful of sensors rarely reach 5 W. What makes a supply bigger than that total is everything between the input and the rails.
A linear regulator passes the load current straight through and drops the excess voltage as heat. Feeding a 5 V rail at 800 mA from a 12 V input, the regulator dissipates (12 − 5) × 0.8, which is 5.6 W, while delivering 4 W. It draws 800 mA from the input regardless of the output voltage — that is the defining property, and the reason input current for a linear rail equals output current. A switching regulator instead trades current for voltage at some efficiency, so it draws roughly output power divided by efficiency divided by input voltage, which for the same rail at 88 percent is about 380 mA.
The consequence is that the same load can need a 1 A supply or a 3 A supply depending only on the regulator topology behind it, and the wasted power lands inside your enclosure either way.
Headroom, and what it is protecting against
| Headroom | Suits | Reasoning |
|---|---|---|
| 10-20% | Steady, fully characterised loads | Component tolerance and measurement error only |
| 20-30% | Most general designs | Room to add a rail later, and supplies run cooler below full rating |
| 50% or more | Motors, relays, radios, anything with surges | Average current understates the peaks badly |
Two arguments carry the headroom. A supply run near its limit runs hot, and heat is what kills the electrolytic capacitors inside it, so a supply at 60 percent load typically lasts much longer than one at 95 percent. And loads grow — the rail you add in six months has to come from somewhere.
Headroom is not a substitute for handling peaks. A motor stalling or a radio transmitting draws multiples of its average for a short time. If those bursts are brief, bulk capacitance near the load supplies them and the supply never sees the peak. If they last, the supply has to carry them, and a supply that current-limits or hiccups mid-burst produces a rail collapse that resets the microcontroller watching it.
Batteries as the input, and what they add
Running the same budget from a battery changes two things. The input voltage is not constant — it falls through the discharge, and a linear rail that has ample headroom at full charge can drop out near the end. And the wasted power is now runtime you paid for, so a linear regulator burning more than it delivers roughly halves the useful capacity.
Lithium cells deserve specific caution, stated plainly rather than buried. A cell that is shorted, over-discharged, over-charged or physically damaged — pierced, crushed, dropped hard — can vent and burn, and once it does it supplies its own oxidiser, so it is not straightforward to extinguish. Charging most lithium chemistries below freezing plates lithium metal on the anode and damages the cell permanently, sometimes in ways that are not visible. A pack without protection electronics is a fire risk, not merely an unreliable one. This page gives no pack-assembly procedure and none should be inferred from it; if you are building packs rather than buying them with protection fitted, that is a subject to study properly first.
The number the bench gives you
Every figure on this page is an estimate resting on your estimates. Datasheet quiescent currents are typical values, efficiency varies with load and is usually worst at light load, and the current a subsystem actually draws is often two or three times what its documentation implies once it is doing real work. Put a meter in series with the input and measure it, and put a thermometer or a finger near the regulators after twenty minutes at full load. A calculator gives you a starting value; the bench gives you the truth.
From here: the heat sink calculator turns the regulator losses above into a sink specification, the PCB trace width calculator sizes the copper carrying each rail current, and the voltage divider calculator covers the monitoring taps you will want on the rails themselves. If the input is a battery bank rather than a brick, the battery runtime calculator takes the input power figure from here.
Questions people ask
How much bigger than my calculated load should the supply be?
Twenty to thirty percent above the calculated input power covers most designs, and fifty percent or more if anything on the board surges — motors, relays, radios, anything that switches a load. The reasoning is not mainly about safety margin in the electrical sense; supplies are usually protected. It is that a supply near its rating runs hot, heat ages the capacitors inside it, and a lightly loaded supply lasts far longer. The second reason is practical: designs grow, and the rail you add later has to come from the supply you already bought.
Why does my linear regulator get so hot?
Because it dissipates the voltage it drops times the current it passes, and that is often more than the rail delivers. A 12 V input to a 5 V output at 500 mA means 7 V across the regulator at half an amp: 3.5 W of heat to deliver 2.5 W of useful power. Efficiency is just Vout divided by Vin, so a 12 V to 3.3 V linear rail is under 28 percent efficient no matter what you do. The fixes are to lower the input voltage before the regulator, to use a switching converter, or to accept the heat and give the part a thermal path that can carry it.
Do I add the rail currents together to size the supply?
You add the input power of each rail, not the output currents, and then divide by the input voltage. Adding output currents across rails at different voltages is meaningless — 500 mA at 3.3 V and 500 mA at 5 V are not one amp of anything. Work each rail out as watts at its input, sum those watts, add headroom, and convert to input current at the input voltage. For linear rails the input current happens to equal the output current, which is why the shortcut sometimes appears to work and then fails as soon as a switching rail is in the mix.
What efficiency should I assume for a switching regulator?
Eighty-five to ninety percent is a reasonable planning figure for a buck converter at a sensible load, and the calculator defaults to 88. The important caveat is that efficiency is not one number: it falls off at light load, where the converter's own quiescent draw dominates, and it falls again at very high load. It also drops as the ratio between input and output voltage grows. If the design spends most of its life idle, the light-load figure matters more than the peak one, and some converters have a low-power mode specifically for that.
How do I handle a load that draws current in short bursts?
Decide first whether the burst is a capacitor problem or a supply problem, because the answer differs. Bursts lasting milliseconds — a radio transmit slot, a relay pulling in — are best handled with bulk capacitance close to the load, so the charge comes out of the capacitor and the supply only sees the average. Bursts lasting seconds, like a motor accelerating under load, have to come from the supply itself, and the supply must be rated for them. Measure with a current probe or a shunt and a scope if you can, because the average current a meter shows can be a small fraction of the peak and tells you nothing about whether the rail is collapsing during the burst.