Power dissipation of voltage regulators?

[Daniel Kelly (AKA Jack)]

Please could you lend me some of your expertise?

I have a 12V 6Ah battery. I'd like to power a 5V 1Amp device for as long as
possible.

If I use a voltage regulator then will I lose a significant amount of power?
Do voltage regulators dissipate much power? I've looked on the spec sheets
but they don't tell me about power dissipation.

Thanks,
Jack

 

[CWatters]

"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:c3snc7$m9c$1@uns-a.ucl.ac.uk...

Please could you lend me some of your expertise?

I have a 12V 6Ah battery. I'd like to power a 5V 1Amp device for as long
as
possible.

This is related to your question on battery packs later right?

If I use a voltage regulator then will I lose a significant amount of
power?
Do voltage regulators dissipate much power? I've looked on the spec
sheets
but they don't tell me about power dissipation.

There are two types of regulator Linear and Switching.

If you use a linear regulator the power dissipated in the regulator (Pd)
will be:

Pd=(Vout - Vin )* I
=(12V - 5V)* 1
= 7W

Ouch! Thats more than your application takes (=5W)

The battery life BL in hours) will be...

BL =C/I

where C=capacity in AH, I = current in Amps

Notice that the voltage of the battery does not feature in the equation for
battery life! Consider what happens if you were to add a cell to the battery
to make 13.2V/ The power dissipated in the regulator goes up BUT the battery
capacity does not increase In short a 13.5V battery will run your video cam
for the same time as a 12V battery! or for that matter a so would an 8V
battery.

You could reduce the power dissipated with no effect on battery life by
reducing the number of cells in the battery. Most linear regulators need 3V
"headroom" (eg 8V in to produce 5V out). You could use a 7 cell battery =
8.4V in which case the power dissipated at 1A out would reduce to 3.4 W or
about half what it was at 12V. Since 1A is still being drawn from the
battery the battery life is unchanged. "Low drop out" regulators have a
lower headroom requirement and save more power.

Better yet use a switching regulator....

These convert one voltage to another without burning huge amounts of power.
A usefull figure is their power conversion efficiency E in %.

E = (Pout/Pin) * 100

Where

Pin = Power In
Pout = Power Out

A perfect switching regulator would be 100% efficient in which case..

Pin=Pout.

Lets assume you have one of these!

If you remember that Power = Voltage * Current

you can write...

Pin = Vin * Iin
Pout = Vout * Iout

If your switching regulator is 100% efficient then

Vin * Iin = Vout * Iout

and rearranging you get

Iin = (Vout * Iout)/Vin

or in your case...

Iin = (5 * 1)/12
=0.41A

eg using a perfect switching regulator has DOUBLED the battery life because
the current has reduced from 1A to 0.41A.

However in practice most switching regulators aren't 100% efficient. If your
switching regulator is 80% efficient then the power burnt in the regulator
is calculated as follows...

Remember that...

E=(Pout/Pin) * 100

so

Pin = (100/E) * Pout

The amount burnt in the regulator is the difference between what goes in and
what comes out so..

Pd = Pin - Pout
= ((100/E)* Pout) - Pout
= ((100/80)* 5) - 5
= 6.25 - 5
= 1.25W

eg Much better than the 7W burnt in the linear regulator on 12V

Hope that helps

Colin

 

[CWatters]

You also need to think about how you plan to charge the battery.

For example a charger designed to charge a 7 cell pack from a 12V car
battery is much simpler than one designed to charge 10 or 12 cells from the
same 12V car battery. This is because the voltage of a 7 cell pack is less
than 12V whereas a 10 or 12 cell pack needs more than 12V on charge.

Colin

 

[Daniel Kelly (AKA Jack)]

Hi Colin,

Thanks loads for your great replies! Very useful!

Thanks,
Jack



"CWatters" <colin.watters@pandoraBOX.be> wrote in message
news:7YF8c.49440$e52.3281975@phobos.telenet-ops.be...

You also need to think about how you plan to charge the battery.

For example a charger designed to charge a 7 cell pack from a 12V car
battery is much simpler than one designed to charge 10 or 12 cells from
the
same 12V car battery. This is because the voltage of a 7 cell pack is less
than 12V whereas a 10 or 12 cell pack needs more than 12V on charge.

Colin