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Buck Boost Converter Calculator

Work out what a DC-DC step-up or step-down converter actually gives you. Enter your input voltage, the output voltage you want and the current you know, and get the current on the other side, the power on each side, the watts lost as heat and the duty cycle, instantly.

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Output current you get
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Converter type
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Input power
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Output power
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Lost as heat
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Duty cycle

A DC-DC converter moves power, not current. Input volts times input amps gives input watts, efficiency decides how many of those watts survive, and dividing what is left by the output voltage gives the output current. Step the voltage down and the current goes up; step it up and the current goes down. Duty cycle assumes a standard continuous-conduction converter and is a design starting point, not a substitute for the controller datasheet.

How to use this buck boost calculator

  1. Input voltage, the voltage of your supply, battery or panel.
  2. Output voltage, the voltage you want the converter to produce.
  3. What do you know, pick whether you know how much current goes in, or how much your load needs out. Most people sizing a supply know the load and want the input, most people testing a module know the input.
  4. Current, enter the figure for whichever side you picked.
  5. Efficiency, leave it at 90% unless the module datasheet says otherwise.

Why the current changes when the voltage changes

The single idea behind every DC-DC converter is that it moves power, not current. Power is volts times amps, and the converter cannot create any, so if it lowers the voltage it must raise the current to keep the product roughly the same. The word "roughly" is the efficiency: a real converter keeps 85 to 95% of the input power and turns the rest into heat.

The whole calculation is three steps:

Worked example: stepping 15 V down to 6 V

Say you have 15 V coming in at 10 A and you want 6 V out. The input power is 15 x 10 =150 W. At 90% efficiency, 135 W survives. Divide by the 6 V output and you get 22.5 A.

So dropping the voltage to 40% of the input more than doubled the current. This catches people out constantly: the module might be fine, but the output wiring, the connector and the inductor now have to carry 22.5 A rather than the 10 A going in. The other 15 W has to leave as heat, which for a bare module without a heatsink is a lot.

Worked example: boosting 5 V up to 16 V

Now the other direction. 5 V at 4 A is 20 W in. At 90% efficiency you keep 18 W, and at16 V that gives 1.125 A out. Raising the voltage cut the current to roughly a quarter.

The trap here is the opposite one. The output looks modest at just over an amp, but the input side is still pulling 4 A, so the supply, the input fuse and the input wiring are the parts under strain. On a boost converter, always check that your source can actually deliver the input current, because a battery or USB port that sags under load will drag the whole thing down.

Buck, boost and buck-boost

One thing worth being clear about: a linear regulator is not a converter. It throws the voltage difference away as heat, so the output current can never exceed the input current, and its efficiency is only Vout / Vin. Dropping 15 V to 6 V with a linear regulator is 40% efficient and wastes 90 W as heat, versus 15 W for a switching converter. That difference is why switching converters exist.

Note: This calculator is a planning guide. Real converters have minimum and maximum input voltages, a maximum duty cycle, and a current limit that is often lower than the arithmetic suggests once heat is taken into account. Always check the module or controller datasheet, and have anything mains-connected or safety critical reviewed by a qualified electrician.

Frequently asked questions

Does a buck converter increase current?

Yes. A buck converter lowers voltage and raises current, because it transfers power rather than current. 15 V at 10 A is 150 W; at 90% efficiency that is 135 W out, and at 6 V that works out to 22.5 A. The output side carries more than twice the input current, so size output wiring for the output figure.

How do I calculate output current from input current?

Multiply input voltage by input current for input watts, multiply by efficiency, then divide by output voltage: Iout = (Vin x Iin x efficiency) / Vout. For 5 V at 4 A boosted to 16 V at 90%: 20 W in, 18 W out, 18 / 16 = 1.125 A.

What is the difference between a buck and a boost converter?

A buck steps voltage down and current up, a boost steps voltage up and current down. A buck-boost does either, which is useful when the input voltage crosses the output as a battery discharges. All three follow the same power rule.

What efficiency should I assume?

85 to 95% is typical for a modern switching converter, and 90% is a safe default. Cheap modules can be nearer 80%, and efficiency falls at very light loads and large voltage ratios. A linear regulator is different: its efficiency is roughly Vout / Vin.

Where do the lost watts go?

Into heat, in the switching transistor, inductor and diode or synchronous FET. 150 W in at 90% efficiency wastes 15 W, which needs a heatsink or airflow.

Can I draw more output current than my supply can give?

Output amps can exceed input amps on a step-down, but the supply still has to deliver the full input power. If a step-down needs 150 W in, a source limited to 75 W will sag or cut out whatever the converter is rated for. Check the input current against your source.

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