How to make 5 V from 1.5 V

An inductor resists any change in its current. Close a switch across it and the current from the cell builds up, storing energy in the inductor's magnetic field. Open the switch and the current has to keep flowing somewhere. The inductor raises its own voltage until the current finds a path, here through a diode into the output capacitor. Its voltage adds to the cell's, so the output can sit far above 1.5 V.

The fraction of each cycle the switch spends closed is the duty cycle, D. For a perfect converter V_out = V_in ÷ (1 − D). Real ones lose some power, so TI's design note folds the efficiency in: D = 1 − V_in × η ÷ V_out (Texas Instruments). With an assumed 80 % efficiency and a fresh 1.5 V cell, that is 76.2 %. The chip adjusts D by itself, cycle by cycle, to hold the output steady.

The MCP1640 is a synchronous boost chip in a six-pin SOT-23 package. Instead of a diode it uses a second transistor inside the chip, which wastes less power than a diode's forward drop would at these low voltages. The data sheet's own 5 V example uses 976 kΩ and 309 kΩ for the feedback divider; the chip holds its VFB pin at 1.21 V, so the output is 1.21 × (1 + 976 ÷ 309) = 5.032 V (Microchip).

The inductor sees more current than anything else in the circuit. While the switch is closed its current climbs; while it is open the current falls. The swing is ΔI = V_in × D ÷ (f × L) (Texas Instruments), which at 1.5 V, 500 kHz and 4.7 µH comes to 486 mA peak to peak. The average of that triangle is the current drawn from the cell. At the most the chip can deliver from a fresh cell, 85 mA, the average is 357 mA and the peak sits on the 600 mA switch limit.

At lighter loads the average falls but the swing stays the same, so below about 58 mA out the bottom of the triangle reaches zero and the current stops for part of each cycle. The chip then switches to its light-load (PFM) mode and pulses only as often as the load needs (Microchip). Either way, the peak never goes above the switch limit.

So rate the inductor for the switch limit, not the load. The limit is 600 mA at minimum but about 0.8 to 0.85 A typical, so choose a saturation current above that; 1 A is a comfortable rating. A shielded 4 × 4 mm part with low resistance does the job. The output ripple from the 10 µF capacitor alone is about I × D ÷ (f × C) = 13 mV at full load, small enough for a microcontroller.

Two limits decide when the converter gives up. The first is starting: the MCP1640 needs 0.65 V typical, 0.8 V at most, to start up, and once running it keeps going down to 0.35 V (Microchip). The second limit comes first in practice: the switch cannot stay closed for more than 88 % of a cycle (the data sheet's minimum figure). As the cell sags, D climbs.

The duty limit is reached at 0.75 V. Energizer rates its AA capacity down to 0.8 V (Energizer), so the converter uses almost the whole cell. Below that point there is very little energy left anyway. If the converter is switched off once the cell is below 0.8 V, it may not start again.

Power in equals power out plus losses, so a converter that raises the voltage 3.4 times must draw more than 3.4 times the output current from the cell. At 50 mA out, that is 210 mA from a fresh cell, and more as the cell voltage falls.

The chip limits the current through its switch to protect itself, at 600 mA at least (Microchip). TI's note turns that limit into the most output current the design can promise: I_max = (I_LIM(min) − ΔI ÷ 2) × (1 − D) (Texas Instruments). For one cell that is 85 mA when fresh and 62 mA at 0.9 V. Typical chips have a higher limit and will give more, but a design should not count on it. With two AA cells in series (3 V), the same chip promises 127 mA.

The cell itself also loses some voltage. A fresh AA has 150 to 300 mΩ of internal resistance (Energizer). At the 349 mA a nearly flat cell must supply, that drops 52 to 105 mV inside the cell before the converter sees anything. Short, thick wires and a good holder matter for the same reason.

An alkaline cell gives less charge the harder you drain it. Energizer's chart for the E91 shows about 3000 mAh at 25 mA but only about 1500 mAh at 500 mA, both measured down to 0.8 V (Energizer). A boost converter multiplies the drain, so it pushes the cell toward the bad end of that chart.

Putting the two together gives a runtime: the cell's capacity at the current it supplies, times its average voltage (1.2 V, our assumption) and the efficiency, divided by the output power.

A sensor node that averages 5 mA runs for about 114 h, close to five days. An Arduino Nano class board at around 20 mA gets about a day. At 50 mA it is down to 7.5 h. These are estimates from a chart and an assumed efficiency; measure your own load current before you trust them.

Our take: one AA and a boost chip make sense for loads up to about 20 mA, or for anything that sleeps most of the time. Above 50 mA, two cells or a lithium cell will do the job better and last longer.

The joule thief is the circuit most people find when they search for this. It is a blocking oscillator: one transistor, one resistor and two windings on a small ferrite ring. When the transistor turns on, current builds in the collector winding, and the base winding drives the base harder. When the core can take no more, the transistor turns off, and the collapsing field throws the collector voltage up until the LED conducts. Then the cycle starts again, many thousands of times a second.

It lights a white LED from a cell that could never do it directly, and it keeps doing so from a cell most devices would call dead. But it has no feedback. Its output is a train of pulses whose height is set by whatever the load clamps them to. With an LED that is the LED's own voltage. With no load, the pulses climb until something breaks down, which can be the transistor. Do not feed a microcontroller or a USB device from it. If you need 5 V, use a chip that measures its output.

A boost converter is not the only route, and not always the best one. The voltages come from the nominal cell voltages, the current limits from the MCP1640 working above.

The LDO row shows why linear regulators waste battery life here. With a 0.3 V dropout, a 5 V regulator needs 5.3 V in, so each of four cells must stay above 1.325 V. On Energizer's test curves an alkaline AA spends more than half its service time below that, so much of the charge is never used. Three cells straight are fine for many boards that accept 4.5 V, and four NiMH cells at 4.8 V run most 5 V parts without any regulator at all.

Work down the chart; the first yes is usually the answer.

• Keep the loop from the cell through the inductor, the chip and the input capacitor as short as you can. Long leads at 500 kHz add noise and losses.
• Use ceramic capacitors (X5R or X7R) for both 4.7 µF and 10 µF; the data sheet warns that high-ESR types cut the output power.
• Measure the load current with a multimeter before you choose: a board that idles at 15 mA may draw 80 mA when its radio transmits.
• Check the inductor's saturation current, not just its inductance. A cheap 4.7 µH part rated at 0.3 A will saturate and the chip will hit its current limit early.
• Ready-made boost modules sold online often carry a different chip from the listing. Measure the output with your load connected before you trust it.
• For a project that sleeps, compare sleep current with the chip's own quiescent current: the MCP1640 draws about 19 µA when it is not switching.

• Switching regulatorsThe buck converter, and how the same three parts make a boost.
• Duty cycle and PWMWhat the 76 % on this page means.
• InductorsInductance, resistance and current rating on a real part.
• Inductor saturationWhy the peak current decides the inductor.
• Battery typesAlkaline, NiMH and lithium cells compared.
• Cells in series and parallelWhere the 3 AA and 4 AA rows come from.
• Linear regulatorsDropout, and why an LDO wastes these cells.
• Schottky diodesThe rectifier a non-synchronous boost would need.

• Microchip, MCP1640/B/C/D 0.65 V Start-Up Synchronous Boost Regulator, data sheet DS20002234D (pinout (SOT-23: SW 1, GND 2, EN 3, VFB 4, VOUT 5, VIN 6); start-up 0.65 V typical, 0.8 V maximum; 0.35 V input after start-up; VFB 1.21 V; RTOP = RBOT × (VOUT / VFB − 1), Example 2: 309 kΩ and 976 kΩ for 5.0 V; automatic PFM/PWM (MCP1640 and MCP1640C), 500 kHz PWM; NMOS peak switch current limit 600 mA minimum; maximum duty cycle 88 % minimum; 4.7 µH recommended; CIN 4.7 µF, COUT 10 µF)
• Texas Instruments, SLVA372D, Basic Calculation of a Boost Converter’s Power Stage (D = 1 − VIN(min) × η / VOUT; ΔIL = VIN × D / (fS × L); IMAXOUT = (ILIM(min) − ΔIL / 2) × (1 − D); an 80 % efficiency estimate as a reasonable worst case)
• Energizer, E91 AA alkaline product data sheet, form E91NA1018 (nominal 1.5 V; internal resistance 150 to 300 mΩ fresh; capacity on continuous discharge to 0.8 V at 21 °C, read off the bar chart: about 3000 mAh at 25 mA, 2500 at 100 mA, 2000 at 250 mA, 1500 at 500 mA)

Checked on 20 October 2026. Our own choices: 80 % converter efficiency (the estimate TI's note suggests), a 50 mA design load, 1.2 V as the average cell voltage over a discharge, a 0.3 V LDO dropout, and the Energizer chart values read to the nearest 500 mAh, with straight-line interpolation between its bars. Runtimes assume continuous drain at room temperature. This page rebuilds an older ElectronicsInfoline article of the same title; none of its text is reused.