The booster: up to 28 V · 10 of 12

More volts, less current

A booster cannot make power, only trade volts for amps. What comes in is capped by the half-amp fuse, about 2.3 W, so at 12 V the booster has roughly 160 mA to give, and both sides share it. The back of the board says 2 A at 5 V; the fuse says about 380 mA.

Power in, power out

A boost converter trades voltage for current. It cannot make power: what comes out is what goes in, less what it loses as heat in the chip, the inductor and the diode. So the question is not "how many amps can it give" but "how many watts can it take".

On this board the answer is set by the fuse on the input. It holds half an amp, and the input after the diode is about 4.5 V at that current. That is about 2.3 W in. Assume the converter keeps 85 % of it — a planning figure, not a measurement — and about 1.9 W comes out.

More volts, less current
12 V · 161 mA
Output voltage12 V
Rails in use
Power in
2.27 W
Power out
1.93 W
Available
161 mA
About 161 mA at 12 V. The fuse lets about 2.3 W in; at 85 % efficiency that is 1.9 W out, and at 12 V that buys 161 mA in total. Enough for a small fan, a relay coil or a short strip of LEDs — not for a metre of 12 V LED strip, which wants amps. Ask for more and the fuse heats, trips, and the rail drops out.

At the voltages people ask for

Set toRoughly available, both sides together
5 V380 mA
9 V215 mA
12 V160 mA
24 V80 mA

Both sides draw from the same fuse. Run the left side at 12 V and the right at 5 V and they split the 1.9 W between them.

What that is enough for

At 12 V, 160 mA is a relay coil or two, a small 12 V fan, a piezo buzzer that wants more than 5 V, or a few centimetres of LED strip. It is not a metre of LED strip, a pump, or a motor under load. At 24 V, 80 mA is a bias voltage or a sensor that needs 24 V, not a load.

Ask for more and the fuse heats, trips, and the rail falls away, then comes back a minute later. A circuit that runs for a while and then dies, over and over, is a booster asked for more watts than half an amp of USB can bring.

About the silkscreen

The back of the board says MAX 2A@5V. That is the MT3608's kind of number: its switch can carry 4 A, and fed from a big enough supply the chip is capable of 2 A at 5 V. On this board, 2 A at 5 V would need well over 2 A from USB, through a fuse that holds half an amp. The number to plan with is the table above.

When it does not work

Will it run a 12 V LED strip?

A short piece. A metre of common 12 V strip draws somewhere around half an amp to an amp or more, and the booster has roughly 160 mA at 12 V. A few LEDs' worth — a few centimetres, or one segment — is what fits. For a whole strip, use a 12 V supply.

Will it run a 9 V motor?

A small one, lightly loaded, perhaps. A motor draws several times its running current as it starts and far more when it stalls, and a booster that runs out of input current simply lets its output fall. Motors are the load these boards are worst at; a proper supply and a motor driver are the right tools.

The booster gets warm at light load.

Mostly the inductor and diode. Warm is normal under load. Hot with almost nothing connected is not: measure the output, which may have been turned past 28 V, and check nothing on the rail is shorted.

Why does the fuse trip at 12 V when my load only draws 300 mA?

Because the fuse is on the input, at about 4.5 V. 300 mA at 12 V is 3.6 W out, which needs about 4.2 W in at the efficiency these converters manage — nearly an amp from USB. The fuse holds half that.

Where this goes next

Two resistors and a sketch that prints the rail voltage once a second.

Watch the rail with an ESP32

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