Power dissipation of voltage regulators?

  • Thread starter Daniel Kelly (AKA Jack)
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Daniel Kelly (AKA Jack)

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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
 
Daniel Kelly (AKA Jack) wrote:

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
Click to expand...
The power wasted (dissipated) as heat in the regulator, expressed
in Watts, is simply the Voltage drop across the regulator
multiplied by the load current. (Ignoring a tiny bias current)

In your example, the drop across the regulator is (12-5)=7V; the load
current is 1A, so the power lost is 7x1=7W, all the while the power
deliveried to the load is is only 5x1=5W. The efficiency is only
5/(5+7)= 0.416 (42%).

You can use a "switching" regulator instead of a "linear" regulator
which typicallys yields efficiencies approaching 97%.

MikeM
 
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:c3snc0$m9a$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.

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
Click to expand...
Just finished a reply to your other post regarding battery packs, so I
assume that this enquiry is also about powering the camera from a battery.
to answer the question first:

A linear regulator such as a 7805 or similar will take an input of 12V and
output a stable 5V. If you are drawing 1A at 5V, you are also drawing 1A at
the 12V input. You will therefore have 5W power in the load and 7W of heat
generation in the regulator. The efficiency is 5/12 = 40%,

A switch mode regulator can do rather better but is a much more complex
device. Efficiencies can be in the 70% - 80% region (don't quote me here, I
don't have lot of experience with them). In either case, you are losing a
significant amount of power to the regulation.

If, as I suspect, you are planning on using a small lead acid battery, you
will find they don't perform well as a deep discharge device. In other
words, although they are a robust battery, they get upset if they are
discharged completely and fully recharged regularly. They much prefer to be
discharged by about 30% - 40% before a recharge, so your capacity is further
reduced.

Much better if you can find a battery pack that will offer the right voltage
rather than trying to regulate the wrong output. See my other reply which
is a sales pitch for NiMh batteries.
 
Excellent! Many thanks for your reply!

Jack


"MikeM" <trashcan@yahoo.com> wrote in message
news:c3sonh$oka$1@coward.ks.cc.utah.edu...
Daniel Kelly (AKA Jack) wrote:

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

The power wasted (dissipated) as heat in the regulator, expressed
in Watts, is simply the Voltage drop across the regulator
multiplied by the load current. (Ignoring a tiny bias current)

In your example, the drop across the regulator is (12-5)=7V; the load
current is 1A, so the power lost is 7x1=7W, all the while the power
deliveried to the load is is only 5x1=5W. The efficiency is only
5/(5+7)= 0.416 (42%).

You can use a "switching" regulator instead of a "linear" regulator
which typicallys yields efficiencies approaching 97%.

MikeM
Click to expand...
 
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:c3snc0$m9a$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.

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
Click to expand...

Depends on the type of regulator. If you use a linear type (such as the
7805) you will lose more than half of the power! It will be as if you were
powering a 12V 1A load, even though the actual load only needs 5V - the
remaining 7V will sit across the regulator which will then dissipate 7W as
heat (that's a lot of heat!). To maximise efficiency you need to use a
switching regulator. Check the following link for info and IC
recommendations.

http://www.national.com/appinfo/power/

cheers,
Costas
 
Looks like the MAX831 would do the job. You can download the data sheet from
www.maximic.com . If you don't want surface mount, check out the LM2598, etc
at www.national.com . What you want is called a buck regulator.

Tam
"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:c3snc0$m9a$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.

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
Click to expand...
 
Pick up an LM3477 evaluation board from
National for ~$15 complete:
http://www.national.com/an/AN/AN-1193.pdf
Change one resistor(Rfb2) in the feedback circuit to set the
output to 5V and you have an inexpensive
and efficient, 12V to 5V switching power supply!
Tom Fisher
Dallas,TX


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in message
news:c3snc0$m9a$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.

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
Click to expand...
 
Thanks everyone for your great input!

Jack


"TBFisher" <tbfisher@hotmail.com> wrote in message
news:rot8c.1856$6p7.564@newssvr23.news.prodigy.com...
Pick up an LM3477 evaluation board from
National for ~$15 complete:
http://www.national.com/an/AN/AN-1193.pdf
Change one resistor(Rfb2) in the feedback circuit to set the
output to 5V and you have an inexpensive
and efficient, 12V to 5V switching power supply!
Tom Fisher
Dallas,TX


"Daniel Kelly (AKA Jack)" <d.kellyNOSPAM@NOSPAM.ucl.ac.uk> wrote in
message
news:c3snc0$m9a$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.

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
Click to expand...
 
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