capcitors for energy reserve

What voltage?

3300uF is around 0.003C or 3mA equivalent. It won't last very long on 120mA draw. 1F supercap is not too expensive, if you can limit it to 2.7V each.

Yeah, that is an issue, it is 9V.

Rich

Hark back to your sophomore electronics engineering courses:

dV/dt = i / C

For a constant current draw, you can change that to

deltaV / deltaT = i / C

Do some algebra, and you get

C = i * deltaT / deltaV

You've got 40mA in, 120mA out, which means you need 80mA out of the cap. You know what voltage drop you can stand -- plug that figure into the above equation along with the rest of your knowns, and you should get a minimum capacitance.

I must say, though, that if 3300uF isn't enough, then at those current levels there's something else going on. If your calculations show that

3300uF is way more than plenty, then your problem may be that you're using aluminum electrolytics where you need to use something with significantly lower ESR.

I'm doing something similar to you right now, except that my current draw is for a shorter period of time. I have my filter cap in my "raw" power line, ahead of my local regulator. With 5V on the "raw" side and 3.3V on the regulated side, I can accept a 1V drop during the 'on' time, as long as the cap charges back for the next cycle.

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Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

So, if you want to supply a 120 mA load for 150 ms, that is .12 A * .15 sec = .018 C. If you want to limit the voltage drop to, say, .1 V, then the calculation is (.018 C / 0.1 V) = C in farads, or .18 F That is 180,000 uF, which is a pretty big cap. Possibly a stack of a few supercaps would do it.

Jon

You can use 4 in series and shunt regulate them. However, you power budget is very tight and can't afford too much power lost. Any other way to reduce power usage in your circuit?

Or learn how to accept a 6V drop.

Whatever you do, opening up the amount of voltage drop you can accept will help your cause.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

Note that I messed up my in-the-head math: 3300uF is, indeed, pretty small for this unless you charge the cap with a boost circuit, regulate after the cap, and accept large voltage swings.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

Look up CAPx expensive but can source loads for miliseconds... used in GSM stuff

Thanks guys for the ideas. I used 3300 because that is what I had. I am now using ~7000uF and I see better results, the voltage drops from 9V down to ~5V so I am too close to the 5V minimum. Charge time is ~1.7s. The input current is limited at 40mA.

I think this will work with a larger cap but sure seem inelegant.

Rich

You have 10mA surplus during normal op, right?

Use that to charge a lower value capacitor with a boost converter

When you need the 180mA, draw that from a buck converter, so you can draw all the energy out og the capacitor

Cheers

Klaus

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all the energy out og the capacitor

Interesting idea Klaus. I could see the enable pin of the reg being trigge red by the higher current draw... More parts though ($).

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I am now using ~7000uF and I see better results, the voltage drops from 9V down to ~5V so I am too close to the 5V minimum. Charge time is ~1.7s. Th e input current is limited at 40mA.

aw all the energy out og the capacitor

gered by the higher current draw... More parts though ($).

Well, you dont actually need to do anything

Run the boost continously, draw all the current it can

Have a zener clamp the cap

And let the buck convert down to the needed voltage. The circuit will adjus t to the current needed

Cheers

Klaus

More optimal:

Use a fet to feed the cap and boost. When you need to draw high current, turn off the fet and let the boost drain the cap

Cheers

Klaus

Why not look at a NiMH (nickel metal hydride) battery? They'll deliver 120mA for 150msec without effort, and the voltage-versus-charge curve is exponential rather than linear.

9V is a very popular voltage level for battery packs, but the cells tend to be expensive.

A single cell and a switching charge discharge system might work.

A single 1.1V AAA cell costing a couple of dollars looks as if it might work. 850mAh is way more capacity than you need, but 150mA at 9V is 1.23A at 1.1V, and you need a tolerably big cell to sustain that current without getting warm.

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Bill Sloman, Sydney

built a circuit that limits the input current to 40mA. This runs the circuit and charges the cap.

for calculating required capacitance? Any ideas on implementation beside my current limiter plus cap?

the voltage-versus-charge curve is exponential rather than linear.

Farnell also stock a 3V Li-Mn coil cell

but the internal resistance - 80R is much too high

For ten times the price you can buy something a little bigger with an acceptable internal resistance

The trade journals have been talking up lithium iron phosphate batteries but I've no idea where you could buy any.

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Bill Sloman, Sydney

Often, the best place to fix something is "elsewhere". You're going to a lot of trouble to limit the current then store it. Think about fixing the thing that's limiting your peak current and put the storage there.

You did not provide enough info on needed life , cost restaints, etc. But a 9 volt alkaline battery might be the answer. Alkaline batteries can be charged if the open circuit voltage is above about 1.2 volts per cell.

Dan

I'm not the original poster, who was a bit economical in describing what he was doing. Alkaline batteries aren't designed to be recharged, and don't seem to be guaranteed when used that way.

And you've snipped most of what I posted, without marking the snip.

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Bill Sloman, Sydney

Assuming 100% efficient supplies:

(120mA)(150ms)(5V) = 90mJ

(30mA)(150ms)(9V) = 40.5mJ

Leaving you with a 49.5mJ deficit.

Energy in a capacitor: W = C * V^2 / 2. Let V = 9V (it could be anything if we're switching in and switching out), then

C = 2 * (49.5mJ) / (9V)^2 = 1222uF

That's the ABSOLUTE MINIMUM capacitance you can use if you only go up to

9V.

If you start surveying capacitors of any given dielectric, you'll soon find out that over a fairly wide range of voltage ratings, a capacitor that stores a given energy is about the same size as all the other capacitors that store the same amount of energy. Tantalums are smallish, aluminums are bigger, ceramics are bigger yet, but the volume/energy ratio is about constant.

If you take into account the supply inefficiencies, and the fact that it'll be hard to suck the last drop of energy out of a cap, you probably can't go much smaller than a 2200uF cap, and that's taking heroic measures with the switching supplies and all.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com