Need to turn small current into burst power

[EricFowler]

I have a need to turn a trickle coming from a battery into a short burst of power. The parameters are:
Battery puts out 0.25mA max at 3V
Circuit needs 20 mA at 3V for 20 mSec once every second.

What kind of circuit/component do I need?

 

[(*steve*)]

You need... A capacitor!

You also probably need something to flash the LED. The voltage will fall somewhat during that 20ms depending on the size of the capacitor.

Note that 50 x 0.25 is 12.5, so your battery cannot supply as much power as you require. Either reduce the current, or the on time, or allow for a lower average current. And there will also be losses, so you need to factor that in.

What are you actually trying to achieve?

 

[jackorocko]

Did you write that correctly? .25mA == .00025A????

If you did, maybe you could charge a cap and use a uC to control the firing. Really depends on how fast you can charge a cap in a second to provide the 20mA needed.

 

[EricFowler]

More on power supply

I was thinking "capacitor" (I ain't totally stoopid) but when I described this to a genuine Hardware Guy [tm] he said I needed more than that. My Hardware Guy said that a cap would not provide stable power for long enough ... he mumbled something about inductance ... that was all the advice one free beer could get me.

The problem is that I have (or hope to soon have) an IC that needs to do low-power (yes, I did mean 20 milliamps) burst transmissions about once a second or so, at regular intervals, and sleep in the meantime. It will be controlled by an 8 bit uC, and powered by a coin battery. I haven't chosen the battery yet, but smallest and cheapest is bestest, and the small, cheap batteries put out small trickles, fractions of a milliamp, maybe a millamp tops. Nothing touches 20 mA (the radio chip uses 18 mA TX, 17 mA RX, the uC a trickle, and I want to light an LED also). The radio and uC can take anything from 1.8V up to ... I think it was 6V.

So the problem is still there: how to get 20 mA for 20 msecs out of a battery that only puts out << 20 mA?


Anyone want to take a swipe at it? If there is a component for this I can google it.

Thanks

 

[(*steve*)]

Just a capacitor should be fine.

The voltage will fall, but in 20ms it won't fall much. 100uF may be sufficient. The coin battery can supply more current than 0.25mA. If you have 1.2V of margin (as you suggest) it shouldn't be a problem.

 

[jackorocko]

http://data.energizer.com/PDFs/cr2320.pdf Yep, I agree with steve.

 

[BobK]

I have a need to turn a trickle coming from a battery into a short burst of power. The parameters are:
Battery puts out 0.25mA max at 3V
Circuit needs 20 mA at 3V for 20 mSec once every second.

What kind of circuit/component do I need?

Sorry, what yoiu ask is impossible.

.25ma * 1sec = 0.25 ma sec

20ma * 20mSec = 0.4 ma sec

The battery cannot provide the energy needed.

That said, what kind of a battery can only put out .25ma?


Bob

 

[(*steve*)]

Check the spec sheet. The battery has its rated life is a load of 0.25mA is applied.

The specs also indicate that the battery can supply higher current pulses.

The life will be a little shorter, but it should work fine.

But yes, my initial post pointed out that 0.25mA for a second is less than 20mA for 20ms.

 

[BobK]

Check the spec sheet. The battery has its rated life is a load of 0.25mA is applied.

The specs also indicate that the battery can supply higher current pulses.

The life will be a little shorter, but it should work fine.

But yes, my initial post pointed out that 0.25mA for a second is less than 20mA for 20ms.

Sorry, I made the mistake of not reading all the previous posts carefully before my answer.

I see now that the battery can provide 6.8ma at 2.7V in pulsed mode. This should be more than sufficient to charge a capacitor for your 20ma bursts.

You can size the capacitor based on the voltage drop you can tolerate during the burst. I.e.:

dV/dT = I / C

If you want to allow for a drop of 10% (0.3V) in 20ms you can calculate it as follows:

C = I / (dV/dT)

C = 0.020 / (0.3 / 0.020) = 0.00133 F = 1033 uF. A 1000 uF should do fine since the battery is also supplying current.

Bob