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Is it possible to take a lower DC voltage, such as the 1.5v from a AA battery, and step it up to a higher DC voltage, such as 5v, which keeps that 5v potential even when there is no load, and which doesn't consume any power when there's no load?

In other words, is there a thing (a board, a component, etc) which can take a AA battery and output 5v without the battery running down when the 5v-needing board is off or in low power mode?

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  • \$\begingroup\$ No. No free energy. You need to consume something (even if small) for the step-up to keep happening. \$\endgroup\$ – Fizz Sep 30 '15 at 23:18
  • \$\begingroup\$ Having said that, a battery discharges by itself over time, so if you can find a step-up method that consumes negligible power in comparison, you could call that (almost) no power lost. But I don't think we're there yet with 1.5V batteries vs SMPS. \$\endgroup\$ – Fizz Sep 30 '15 at 23:26
  • \$\begingroup\$ You could of course try an cheat and have some kind of very low power load detection circuit running at 1.5V (think how standby works on a computer). But the 5V won't be always on, just like the CPU ain't really on when it's in standby. \$\endgroup\$ – Fizz Sep 30 '15 at 23:29
  • \$\begingroup\$ Someone has actually though of the latter already (well, not at those voltages you ask) ti.com/tool/loaddetectpwrsupply-ref \$\endgroup\$ – Fizz Sep 30 '15 at 23:30
  • \$\begingroup\$ Well, if you wanted the 5V nominally there, you could add a supercap. \$\endgroup\$ – Daniel Oct 1 '15 at 0:14
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Any circuit that you put across the battery will draw some power, but with no load it can be very low. The problem comes when a load is applied and it needs to hold the voltage up. The more power it has to provide under load, the higher the quiescent current may have to be for it to work properly.

The 'thing' you are talking about is called a Voltage Booster or DC/DC Step Up Converter. The two most common ways to step up a DC voltage are:-

  1. Switched Capacitor

A capacitor is connected in parallel with the battery to charge it, then reconnected in series to double the output voltage. With no load the capacitor will stay charged for a long time, but under load it will start to discharge and so must be topped up periodically.

The switching circuitry will draw a bit of current even when there is no load. This quiescent current can be reduced by lowering the switching frequency, but then the output voltage will drop more as loading increases.

  1. Inductive booster

An inductor (coil of wire) is switched across the battery, causing an increasing current flow and building up a magnetic field in it. Then the switch is opened, and as the inductor's magnetic field collapses it tries to maintain the current flow - increasing voltage until it does. A diode steers this voltage to the output, where it charges up a capacitor.

An inductive booster must draw some current from the battery even with no load. However it may be able work in 'discontinuous' mode where it draws a short burst of current to charge the output capacitor, then waits for the voltage to drop a little. If capacitor leakage is negligible then the duty cycle (ratio of switch on to off time) can be very low, resulting in very little average current draw from the battery. Quiescent current draw will then be determined by how much power the circuit needs to work in this mode.

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I think you want an ultra low quiecent current boost converter.

Here is one from TI...

http://www.ti.com/product/tps61200

It can take an input voltage as low as 0.7 volts and boost it to a selectable output voltage between 1.8 and 5.5 volts. It only draw 55 micro amps (millionths of an amp) when there is no load on the output but can supply up to 600mA @5 volts when there is a load.

There are other chips like this that have even lower quiecent current draw (some lower than the internal discharge rate of a battery) but the typically have lower maximum current limits, so picking the right on depends on your application.

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I guess depending on the kinds of load you want to put on it and how stable the voltage needs to be, you can use a charge pump to triple the voltage to 4.5 VDC, you may even try quadrupling the voltage to 6VDC, but you would be really pushing the limits. I think you're best bet if you go this route would be to use 2 batteries and use the charge pump to double that voltage. This all of course depends on what you are using the potential for.

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