I'm working on a RF telemetry side-project that requires an amplifier IC to be run for about 5 seconds bursts between charges. The system is portable and runs off of a 3.3V battery that can supply about 100 mA. The IC runs at 10V and draws 1.5 Amps, so 15 Watts. I can't simply use a boost converter on the device's battery because it can't supply nearly that much power. So I need to store the necessary charge and then dispense it at a regulated voltage.

The output voltage of my charge storage system should be well regulated at 10 volts, and I need it to be able to supply a constant 1.5 Amps for at least 5 seconds, after which I will turn it off and recharge it.

The goals here are to minimize size and amount of time it takes to charge the thing. I assume the solution is either going to be some sort of battery or capacitor with some circuitry to regulate the voltage output.

I don't know very much about practical charge storage and power regulation, but if someone knows a good general approach I can look into the details.

edit: Assuming money is no object, anyone interested in elaborating on using a super-cap?


2 Answers 2


15 W \$\times\$ 5 s = 75 J. If you want to store that in a capacitor you'll need a capacitance of

\$ C = \dfrac{2 \times 75 J}{(10 V)^2} = 1.5 F \$

There are supercaps with even higher capacitances, but types which can supply a 1.5 A current are between expensive and Damn Expensive™.

A battery will be a better choice. For instance a LiPo followed by a boost converter, to have a nicely regulated 10 V, even at varying battery voltage. Use another boost converter to charge the battery between load bursts; I would shut-down the charger while you're transmitting.

If the 3.3 V can supply 100 mA that's 330 mW. Suppose the two boost converters have an efficiency of 85 % then to get the required 75 J output you'll need 105 J input. At 330 mW that will take 315 seconds, that's 5 minutes and 15 seconds. So you can transmit a 5 second burst once every 5 minutes and 20 seconds. If you need a higher frequency you'll have to find another power source.

I almost forgot: the battery has a less than 100 % efficiency too. You can't get every joule you put into it out again. So in practice the charging time will have to be longer than the 5 minutes.

I would use two LiPo cells in series. Then you'll have 7.4 V nominal, 8.4 V fully charged. That's relevant, because the closer the voltage to the 10 V the less current will be required from the battery. If you draw 1.5 A at 10 V a 3.7 V cell would have to supply 4.8 A if we take again the 85 % for the booster. Two cells in series will only have to supply 2.4 A.

edit 2
Dave's ultracap seems to be a good alternative, but he should mention that the 3.3 V to 2.5 V regulator also has to be a switching regulator, unless you're satisfied with a 75 % efficiency there. If we use eBay as a price reference you can buy a set of two 240 mAh LiPo's for 4.5 dollar, free shipping. That means they only will be discharged for 1 % during the 5 s burst. So while 14 dollar for the cap isn't bad, and definitely something to remember, the LiPo's are only a third of that.

It's also worth noting that stepping down from 3.3 V to 2.5 V and then up to 10 V is less efficient than up from 3.3 V to 7.4 V, and then to 10 V. That's because a switcher's efficiency is related to both the in-out voltage difference and their ratio.

  • \$\begingroup\$ Fantastic. I figured a capacitor wouldn't work out, haha. I will look more into charging and utilizing LiPo batteries. I've used boost converters before. Thanks! \$\endgroup\$
    – Keegan Jay
    Oct 6, 2012 at 19:48
  • \$\begingroup\$ @Jay - Thanks for the accept, but don't be too hasty with that. See my edit :-). \$\endgroup\$
    – stevenvh
    Oct 6, 2012 at 19:57
  • \$\begingroup\$ Does drawing less current from the battery just increase the lifespan, or what? \$\endgroup\$
    – Keegan Jay
    Oct 6, 2012 at 20:07
  • \$\begingroup\$ Err, also I'm a bit confused. Is the battery supposed to entirely discharge in the 5 seconds? That doesn't sound possible/safe. \$\endgroup\$
    – Keegan Jay
    Oct 6, 2012 at 21:32
  • \$\begingroup\$ @JayKeegan It can be done with caps, but you have to have the willingness to foot the bill for the caps. And no, discharging an entire LiPo battery in 3 seconds is not safe, but if you only need 1.5A at a time then a normal cellphone style battery will last an hour. Making a battery last a long time is very very complex compared to the problem here. \$\endgroup\$
    – Kortuk
    Oct 6, 2012 at 22:16

Not a supercap, but an ultracap, which is designed for high currents, might work.

Rough calculation: Ultracaps have voltage ratings on the order of 2.5V, so your 1.5A @ 10V translates (ignoring efficiency issues for the moment) into 6A @ 2.5V. 6A for 5 seconds is a charge of 30C. Let's say you can tolerate a voltage drop of 0.1V over the 5s, so the capacitor you would need would be 300F.

To charge this capacitor, you just need a 2.5V regulator driven from your 3.3V supply that's current-limited to 100 mA. To drive your amplifier, you need a boost converter (regulator) that converts the 2.4-2.5V from the capacitor to 10V @ 1.5A. You would enable the boost converter only when you need it, along with the amplifier itself.

Restoring the 30C @ 100 mA (using a linear regulator) would require 300 seconds of dead time between transmissions. Longer, once you take real-world efficiency numbers into account. But if you use a 3.3V-to-2.5V switching regulator, which should give you almost 130 mA out, reduces the nominal recharge time to 227 seconds.

I just checked Ebay, and you can get a single Maxwell 350 Farad Ultracapacitor, which is about the size of a D cell, for about $14.


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