Battery and dc-dc regulator pulse current

[graemegets]

Hi,

I'm trying to understand the relationship between the current draw capability on the dc-dc converter (for both Buck and boost) and the battery.

The battery i'm looking at is the 3.6 v 8.5 Ah Tadiran Battery. The circuit needs 5v with a peak current of 1.3A ( idle is 145mA). the datasheet for the battery (http://www.tadiran.com/pdf.php?id=TL-5920) says it can supply 230mA continuous and 400mA pulse.

There are a number of buck-boost converters that say they can deliver 2A continuous. (usually with a min Vin voltage, though this is not always in the datasheet).

My question is : does the dc-dc converter manage the 1.3A peak current draw or must the battery also be able to deliver at that rate?  My assumption is that the DC-DC converter managed this but want to make sure about this.

A secondary question is : are two batteries  in series (7.2V) with a Buck regulator "better/more efficient" than a single 3.6v with Boost regulator? The device being run is the Iridium 9603 which will only be switched on and used every 1 hour to 2 hours (depending on config) and for a duration of about 1min each time.

Many Thanks

Thanks

[TheElectricChicken]

Hi,

I'm trying to understand the relationship between the current draw capability on the dc-dc converter (for both Buck and boost) and the battery.

The battery i'm looking at is the 3.6 v 8.5 Ah Tadiran Battery. The circuit needs 5v with a peak current of 1.3A ( idle is 145mA). the datasheet for the battery (http://www.tadiran.com/pdf.php?id=TL-5920) says it can supply 230mA continuous and 400mA pulse.

There are a number of buck-boost converters that say they can deliver 2A continuous. (usually with a min Vin voltage, though this is not always in the datasheet).

My question is : does the dc-dc converter manage the 1.3A peak current draw or must the battery also be able to deliver at that rate?  My assumption is that the DC-DC converter managed this but want to make sure about this.

That battery is not going to handle it. Take the output that you would like, say 5v 2a calculate power, which will be 10 watts in that case, then add some for the efficiency of the converter. if it was say 90% then it's about 2 watts or less as I don't do the math. So you need about 12 watts. Divide that by the voltage of the battery. which will be about 3 amps. your battery is just under a tenth that, so you'd need about a dozen in parallel or a bigger battery.

[TheElectricChicken]

So the rule for power is if the voltage goes up the current goes down and if voltage goes down current goes up.

So you want 2 amp at 5v which is 10w

at 3.6 v, 10 watts is 2.8 amps

plus, think of it as a commision by a middleman, the converter itself uses some power, say 10% so you redo the calculation for 11w

at 3.6 v, 11 watts is 3 amps

so you need a battery that can output 3,000 ma at least, that is, 3amps.

[TheElectricChicken]

A secondary question is : are two batteries  in series (7.2V) with a Buck regulator "better/more efficient" than a single 3.6v with Boost regulator? The device being run is the Iridium 9603 which will only be switched on and used every 1 hour to 2 hours (depending on config) and for a duration of about 1min each time.

Usually it is easier to reduce the voltage, however, you would look at these things on ebay for example and look at the efficiencies. It depends on the item itself.

All other things being equal the 2 batteries are better than one because you'd halve the current and therefor the losses through the wiring. Starting with a higher voltage often helps overcome many kinds of losses. For a minute each time you'd still need batteries that can handle the total. If it were pulses of milliseconds you could add a capacitor to a smaller battery...

[MarkM]

Yeah, what he said.  If you're stepping down voltage the input current is less than the output current.  With a booster the input current is greater than the output current.  It all comes down to Ohms Law.

Is there a specific reason to use that battery/battery chemistry?  Looks like they beefed up the specs to make it say 8.5Ah. 

[graemegets]

Thanks for all the info.

I read the power requirement again and see that the 1.3A peak draw is for 8.2ms -rest of the time its an average of 158mA. (see attached jpg)

I'm sure that this situation must crop up in all sorts of portable apps where there is a peak current draw for a short time. toys with motors etc.

Is there a specific reason to use that battery/battery chemistry?  Looks like they beefed up the specs to make it say 8.5Ah.

My thinking was to reduce the space required by using 3.6 volt batteries with high amp-hours (expensive, but in this case worth it) . I'm sure they do beef up the specs! I would probably use the 19ah version anyway and take 10% off what they say.

I'm open to suggestions

Many Thanks all

[MarkM]

If you look at the graphs on your cell datasheet, the curves look very odd.  The capacity goes way down as the load goes up.  Try to compare that datasheet to this one, which is a good old 18650 lion.  You don't really need one this beefy, but it's just for comparison.

http://www.powerstream.com/p/us18650vtc5-vtc5.pdf