power loss in 7805 used as current regulator

[jtaylor]

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...

 

[Tam/WB2TT]

"jtaylor" <jtaylor@hfx.deletethis.andara.com> wrote in message
news:gxY6c.7799$kc2.188303@nnrp1.uunet.ca...

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...


All of the 20 ma (actually more like 30) flows through the 7805, plus about

another 5 ma. Using your 20 ma, total power drain on the battery would be 9
x (20 + 5) = 225 mw. I hope this is for something that gets turned on
infrequently, for a few minutes at a time.

Tam

 

[Rom]

On Sat, 20 Mar 2004 10:33:20 -0400
"jtaylor" <jtaylor@hfx.deletethis.andara.com> wrote:

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a
100ohm resistor to limit the current to 20ma, how much power is lost
in the regulator?

This needed for estimate of battery life...

Hi! I'm a sort of french folk

(sorry for my bad english writing)

The power lost in the regulator (heat) is the difference between the
power supplied by the battery and the power fed to the load.
The current flowing through the load is (as far as I know) the same as
the input's.
So the power lost should be 80mW.


Hope it helps
--
Romain

 

[Richard Rasker]

On Sat, 20 Mar 2004 10:33:20 -0400, jtaylor wrote:

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...

Just the current consumption is enough - battery capacity is expressed in
Ah (ampere-hours) or mAh (milli-ampere-hours), so the power loss in the
7805 is not relevant.

If a new 9V-battery has a capacity of 200mAh (which I think is
reasonable), drawing 20 mA from it will give you 10 hours of battery life.

However, an 7805 will also draw 5mA on its own, so that gives you
approximately 8 hours of battery life.

Mind you, this capacity can vary quite a bit (not just between brands and
types, but also between identical batteries), and it's not a "hard" value:
it decreases with the amount of current drawn. So a 200mAh battery will
definitely not give you 1 hour at 200mA, but more likely something like
15 - 30 minutes.

The other problem is that this capacity is usually not mentioned in any
way (which is kind of funny when you think about it - with anything else
you buy, the manufacturer or supplier *must* specify exactly how much you
get).

About the 7805 and LED combination: are we talking about a white LED here,
since you implicitly indicate a 3V voltage drop [(5V - 3V)/100R = 0.02A]
If you just use the 7805 to supply the LED circuit with 5V, you can of
course do away with it altogether, and simply hook up the LED directly to
the 9V battery, with a 330 ohms series resistor - this will also save you
the 5 mA for the 7805 right away. Furthermore, you could consider using a
low-current LED, which has maximum brightness at a much lower current,
although these are still exceedingly expensive in white and blue.


Richard Rasker

--
Linetec Translation and Technology Services

http://www.linetec.nl/

 

[Spehro Pefhany]

On Sat, 20 Mar 2004 10:33:20 -0400, the renowned "jtaylor"
<jtaylor@hfx.deletethis.andara.com> wrote:

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

An alkaline battery might be specified to have "end of life" at
0.8V/cell or 4.8V. The 7805 will drop out at maybe 1.5V typically, so
the LED will begin to get dimmer at around 6.5V battery voltage, well
above the "end of life" voltage. It will still be dimly lit at 4.8V,
however.

Also the 7805 draws about 5mA just for its own needs, so the actual
current will be LED+5mA (typical).

This needed for estimate of battery life...

All the information you need is here (or other battery manufacturer
data)

http://rocky.digikey.com/WebLib/Panasonic/Web%20data/Panasonic_Alkaline_Hdbk_03-04_v1.pdf

and here:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf


You can see from the graph on page 10 (constant current*) of the
battery data that you can expect a battery life of around 18 hours
from a Panasonic Industrial alkaline cell under those conditions and
at 20°C.

* the 7805 will draw constant current of about 25mA at the input pin
until it begins to drop out.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[Rom]

On Sat, 20 Mar 2004 16:05:53 +0100
Rom <romain-4news@ssji.net> wrote:

So the power lost should be _180mW_. (sorry, I forgot a digit)
But there is some leakage current, so you should consider adding about

10% to 20%...

--
Romain

 

[John Woodgate]

I read in sci.electronics.design that jtaylor <jtaylor@hfx.deletethis.an
dara.com> wrote (in <gxY6c.7799$kc2.188303@nnrp1.uunet.ca&gt about 'power
loss in 7805 used as current regulator', on Sat, 20 Mar 2004:

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...


Well, it isn't needed for that purpose, but still....

You have 5 V/100 ohms = 100 mA, not 20 - no you don't, because the LED
has a voltage drop, which I could guess if I knew the colour. I'll
assume it's a red one dropping 1.6 V. So your voltage across the 100 ohm
is actually 5 - 1.6 = 3.4 V, giving 34 mA. Hot diode! Too much! Maybe
it's a blue LED at 2.6 V. That gives 24 mA.

Anyway, let's say 24 mA. You have 9 - 5 = 4 V across the 7805 and 24 mA,
so the power is 4 x 24 mW = 96 mW - not much. So I assume you are using
a 78L05, not a 7805, which will steal another 5 mA from your supply.

To determine the battery life, you divide the milliamp-hour capacity by
the current in milliamps. 24 mA is really too much for a 110 mAh PP3
battery, but if that's what you have, you divide 110 by 24 to get 4.6
hours.

Amp-hour ratings are based usually on the current being low enough for
the battery to last 20 hours, though; that's why 24 mA is really too
much. Still with a good battery you would get around 4 hours.


--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Spehro Pefhany]

On Sat, 20 Mar 2004 16:25:51 +0000, the renowned John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

Amp-hour ratings are based usually on the current being low enough for
the battery to last 20 hours, though; that's why 24 mA is really too
much. Still with a good battery you would get around 4 hours.

What kind of crummy batteries are they foisting off on you blowkes in
Old Blighty anyhow? A nice Japanese 570mAh battery gives you 18 hours
and costs a paltry 79p at Digikey, a dozen at a time. They have almost
84,000 in stock, as I sputter.

<http://rocky.digikey.com/WebLib/Panasonic/Web%20Photos/6AM-6PIX^1S.jpg>

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[Walter Harley]

"jtaylor" <jtaylor@hfx.deletethis.andara.com> wrote in message
news:gxY6c.7799$kc2.188303@nnrp1.uunet.ca...

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...

I'm assuming your circuit looks like this:

____
| | ___
9V ----|7805|----|___|---o---->|----.
|____| | |
| 100R | LED |
'---------------' GND

In this case, the LED gets 5V/100R = 20mA, plus the quiescent current of the
regulator, for a total of 25mA. If you wanted only 20mA, just increase the
R to 133 ohms.

But a better solution would be to use a chip that had a lower quiescent
current (so that more of the output current is under your control) and that
wasted less power in the resistor (so that you have lower end-of-life input
voltage). One such chip would be the LM317L. (Note the "L" in the part
number - this is the low-power version of an LM317.) Its quiescent current
is less than 100uA; and the voltage between adj and out pins is 1.2V rather
than 5V, so to get 20mA out, you would use a 24 ohm resistor.

Dropout voltage of the LM317L is around 1.5V at these currents. At 4.8V
(extreme end of life on battery), that still leaves you (4.8V - 1.5V - 1.2V
=) 2.1V for the LED. So you'll be good to the last drop of battery life.

Doing it that way does mean that more of your excess power is being burnt in
the regulator rather than in the resistor: at 9V, assuming the LED takes
around 2V, then the resistor gets 1.2V at 20mA and the regulator gets the
remaining 5.8V at 20mA. So the regulator is dissipating about a tenth of a
watt. Will it get too hot? The datasheet at
http://www.national.com/ds/LM/LM317L.pdf says thermal resistance of the
little TO92 package is 180 C/W, so we expect temperature to rise by about 20
degrees C at worst. That'll be warm, but still well within the 125C thermal
limit of the device.

 

[jtaylor]

Walter Harley <walterh@cafewalterNOSPAM.com> wrote in message
news:c3hvdh$un0$0@216.39.172.65...

"jtaylor" <jtaylor@hfx.deletethis.andara.com> wrote in message
news:gxY6c.7799$kc2.188303@nnrp1.uunet.ca...
I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...

I'm assuming your circuit looks like this:

____
| | ___
9V ----|7805|----|___|---o---->|----.
|____| | |
| 100R | LED |
'---------------' GND

In this case, the LED gets 5V/100R = 20mA, plus the quiescent current of
the
regulator, for a total of 25mA. If you wanted only 20mA, just increase
the
R to 133 ohms.

But a better solution would be to use a chip that had a lower quiescent
current (so that more of the output current is under your control) and
that
wasted less power in the resistor (so that you have lower end-of-life
input
voltage). One such chip would be the LM317L. (Note the "L" in the part
number - this is the low-power version of an LM317.) Its quiescent
current
is less than 100uA; and the voltage between adj and out pins is 1.2V
rather
than 5V, so to get 20mA out, you would use a 24 ohm resistor.

Dropout voltage of the LM317L is around 1.5V at these currents. At 4.8V
(extreme end of life on battery), that still leaves you (4.8V - 1.5V -
1.2V
=) 2.1V for the LED. So you'll be good to the last drop of battery life.

Doing it that way does mean that more of your excess power is being burnt
in
the regulator rather than in the resistor: at 9V, assuming the LED takes
around 2V, then the resistor gets 1.2V at 20mA and the regulator gets the
remaining 5.8V at 20mA. So the regulator is dissipating about a tenth of
a
watt. Will it get too hot? The datasheet at
http://www.national.com/ds/LM/LM317L.pdf says thermal resistance of the
little TO92 package is 180 C/W, so we expect temperature to rise by about
20
degrees C at worst. That'll be warm, but still well within the 125C
thermal
limit of the device.

Thank you.

Some other responders seem to have missed the "current-regulator" in the
post subject.

Yes it is for white LED. 20mA is recommended for maximum brightness and no
life reduction. The battery I'm using is a NimH, says 160mAh. From what I
know the voltage curve is quite flat on these, so a low-dropout regulator is
less of a concern. The 7805 I'm using actually says "78L05, " I'll start
looking to see if this is a low-power part.

 

[Ian Bell]

jtaylor wrote:

I'm a sort-of newbie.

If I have a 9-volt battery running an LED through a 7805 using a 100ohm
resistor to limit the current to 20ma, how much power is lost in the
regulator?

This needed for estimate of battery life...

The volts drop across the regulator multiplied by the current thru' it so in
your case 4volts x 20mA = 800mW.

Ian

 

[John Woodgate]

I read in sci.electronics.design that Ian Bell <itb@yahoo.com> wrote (in
<c3i8or$27onk1$1@ID-225948.news.uni-berlin.de&gt about 'power loss in
7805 used as current regulator', on Sat, 20 Mar 2004:

The volts drop across the regulator multiplied by the current thru' it
so in your case 4volts x 20mA = 800mW.

Only for sufficiently large values of 4 and 20. (;-)
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Ian Bell]

John Woodgate wrote:

I read in sci.electronics.design that Ian Bell <itb@yahoo.com> wrote (in
c3i8or$27onk1$1@ID-225948.news.uni-berlin.de&gt about 'power loss in
7805 used as current regulator', on Sat, 20 Mar 2004:

The volts drop across the regulator multiplied by the current thru' it
so in your case 4volts x 20mA = 800mW.

Only for sufficiently large values of 4 and 20. (;-)

Absolutely. If 4 gets too small then the regulator may no longer. ;-)

Ian

 

[Active8]

On Sat, 20 Mar 2004 15:12:43 GMT, Spehro Pefhany wrote:

<snip>

http://rocky.digikey.com/WebLib/Panasonic/Web%20data/Panasonic_Alkaline_Hdbk_03-04_v1.pdf

and here:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

Good link. A friend once told me he got the most use out of
Panasonic batts in his pager.

Too bad Digi-Key doesn't have a complete reference on the NiCds,
just individual data sheets.

<snip>
--
Best Regards,
Mike

 

[Walter Harley]

"jtaylor" <jtaylor@hfx.deletethis.andara.com> wrote in message
news:Ri07c.7949$kc2.190326@nnrp1.uunet.ca...

Yes it is for white LED. 20mA is recommended for maximum brightness and no
life reduction. The battery I'm using is a NimH, says 160mAh. From what
I
know the voltage curve is quite flat on these, so a low-dropout regulator
is
less of a concern. The 7805 I'm using actually says "78L05, " I'll start
looking to see if this is a low-power part.

It's still 5mA quiescent current. (Datasheet at
http://www.national.com/ds/LM/LM78L05.pdf.) That's okay, it just means you
don't have as much regulation of the output current; the quiescent current
will vary according to the Vin/Vout differential, and it's outside your
control.

Two other things to think about here:

1. If you're working with NiMH, at relatively constant voltage, and with a
known number of LEDs (one), why are you using a regulator at all? Why not
just use a series resistor, of (9V - 3V)/20mA = 300 ohms? If neither the
voltage nor the load are changing you don't need a regulator.

2. If you really want to get the most out of the battery, what you need is a
switch-mode regulator rather than a linear regulator. No matter how you cut
it, the linear regulator (or series resistor, if you go that route) is only
going to be about 33% efficient when dropping ~9v to ~3v. The switch-mode
regulator can do 80 to 90% efficiency, assuming you find one that is
appropriate to the amount of current you're drawing. Another poster
mentioned the Roman Black regulator circuit; and there have been a number of
recent threads here on the topic of buck regulators for driving white LED's.

 

[John Woodgate]

I read in sci.electronics.design that Walter Harley
<walterh@cafewalterNOSPAM.com> wrote (in <c3ior0$skg$0@216.39.172.65&gt
about 'power loss in 7805 used as current regulator', on Sun, 21 Mar
2004:

If you really want to get the most out of the battery, what you need
is a switch-mode regulator rather than a linear regulator.

If the OP doesn't know how to run an LED from a linear regulator, how do
you think he is going to cope with a far more complex switcher?
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Fred Bloggs]

jtaylor wrote:

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote in message
news:En3kPqCWoVXAFwRx@jmwa.demon.co.uk...

I read in sci.electronics.design that Walter Harley
walterh@cafewalterNOSPAM.com> wrote (in <c3ior0$skg$0@216.39.172.65&gt
about 'power loss in 7805 used as current regulator', on Sun, 21 Mar
2004:

If you really want to get the most out of the battery, what you need
is a switch-mode regulator rather than a linear regulator.

If the OP doesn't know how to run an LED from a linear regulator, how do
you think he is going to cope with a far more complex switcher?


At this level, with standard componenets, things look to be just cook-book
stuff. Buy the bits, build it, turn it on.

What is surprising to me is the wide range in estimates of power loss.

It is not power loss but energy loss. The battery supplies
Ebatt=Vbatt*Iled*Tdis joules and the LED dissipates Eled=Vf*Iled*Tdis
joules, so that (Vbatt-Vf)*Iled*Tdis is wasted, or as a percentage, only
Eled/Ebatt=Vf/Vbatt of the expended energy has been used to light the
LED at any one time- this would be 3.3/7.2=46% for the white LED and
NiMH battery you're using.

 

[John Woodgate]

I read in sci.electronics.design that jtaylor <jtaylor@hfx.deletethis.an
dara.com> wrote (in <E%g7c.8291$kc2.199102@nnrp1.uunet.ca&gt about 'power
loss in 7805 used as current regulator', on Sun, 21 Mar 2004:

What is surprising to me is the wide range in estimates of power loss.

Well, the 20 mA looked to be wrong, unless some assumptions were made.
People made different assumptions. Then the type of LED wasn't
specified, so people assumed different LED voltages. Then the regulator
tail current was added in, or not. About the only thing we can be sure
of is that it's less than 420 mW (100 mA through a short-circuited LED +
5 mA tail current)!

Then again, the OP didn't need to know it!
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Fred Bloggs]

John Woodgate wrote:

I read in sci.electronics.design that jtaylor <jtaylor@hfx.deletethis.an
dara.com> wrote (in <E%g7c.8291$kc2.199102@nnrp1.uunet.ca&gt about 'power
loss in 7805 used as current regulator', on Sun, 21 Mar 2004:


What is surprising to me is the wide range in estimates of power loss.


Well, the 20 mA looked to be wrong, unless some assumptions were made.
People made different assumptions. Then the type of LED wasn't
specified, so people assumed different LED voltages. Then the regulator
tail current was added in, or not. About the only thing we can be sure
of is that it's less than 420 mW (100 mA through a short-circuited LED +
5 mA tail current)!

Then again, the OP didn't need to know it!

LOL!!!

 

[James Meyer]

On Sun, 21 Mar 2004 14:07:01 GMT, Fred Bloggs <nospam@nospam.com> posted this:

It is not power loss but energy loss. The battery supplies
Ebatt=Vbatt*Iled*Tdis joules and the LED dissipates Eled=Vf*Iled*Tdis
joules, so that (Vbatt-Vf)*Iled*Tdis is wasted, or as a percentage, only
Eled/Ebatt=Vf/Vbatt of the expended energy has been used to light the
LED at any one time- this would be 3.3/7.2=46% for the white LED and
NiMH battery you're using.

Since the regulator used as a current source in series with the LED as I
believe it is, ALL the LED current is delivered from the battery and there are
no other currents drawn from the battery.

It seems to me to be easiest to measure the LED current and use that to
determine the length of time the battery will supply that current by dividing
the battery amp-hours by the LED current without resorting to any other tedious
calculations involving voltage drops or joules of energy.

Jim