# LED driving questions.

Discussion in 'Electronic Basics' started by Boki, Apr 19, 2006.

1. ### BokiGuest

Dear All,

My customer said:

The LED circuit should draw <20uA under "battery good" condition.

Does that mean my driving transistor should provide less than 20uA for
LED when the LED in turn off state?

Question2,
I have to guarantee 10mA for turn on LED, but my supply voltage is only
2.8V

The cross voltage of LED is about 1.9V...

I have only 0.9V for emitter ( pnp driving ) or collector ( npn driving
)

Even I use 330ohm resistor at collector with npn, the current seems
still too small...

Best regards,
Boki.

2. ### robGuest

An led requires a least 5mA to make it lite. so """"The LED circuit should
draw <20uA under "battery good" condition"" is not possible. the following
formula applies. R = (Vs - Vf)/If where Vs is the supply voltage, Vf is the
forward current of the led. If is current required to lite the led. eg Vs =
12V, Vs = 1.4 V, If = 5mA therefore R = 2120 ohms

3. ### Rene TschaggelarGuest

Naa.
A Led is clearly visible with just 1mA.
If switched on at 1mA, but 10uA average,
this would mean a 100:1 deadtime.

Further. The VCE of an NPN is as low as
50mV saturated. Better than a resistor
in terms of power is the use of PWM, a
coil and a cap, in case you have a spare

Rene

4. ### DeefooGuest

To me that means that the LED driver should consume as little power as
possible, so the LED has to be off in this condition and leakage currents
should not exceed 20uA.

You should not be using a bipolar transistor to drive the LED.

--DF

5. ### BokiGuest

Got you.

but I saw a lot of design use bipolar transistor...

do you mean use MOSFET to drive is better?

Best regards,
Boki.

6. ### DeefooGuest

Well, "should not" is a bit strong, but you are already in a "battery bad"
condition when you have to activate the LED so you have to limit the power
consumption as much as possible. If you can avoid a base current by using a
FET, that's good. BTW there exist low current LEDs and personally I almost
never need more than 2mA through a LED to get good visibility.

--DF

7. ### Fred BloggsGuest

This can only be talking about the LED driver circuit overhead current
draw exclusive of the actual LED current. The overhead consists of the
various bias and drive currents required to cause LED current to flow.
I interpret it to mean that <20uA in either LED ON/OFF state.
You mean "forward" voltage of the LED.
For a 10mA LED current and a 20uA maximum driving circuit current, you
will have to use a MOSFET with a maximum threshold Vgs less than your
worst case (minimum) available gate drive voltage, or a very high beta
transistor with gain > 10/0.02=500. A small MOSFET will just keep worst
case leakage under 20uA. The simplest circuit is the LED in series with
a current limiting resistor and MOSFET. At 2.8V you then have
2.8V-1.9V=10mA*(R+RDS,ON)=10mA*R most of the time, making R=90 ohms or so.

8. ### ChrisGuest

Hi, Boki. There are dozens of battery monitor ICs that will do the
job. Why not just go with a one IC solution?

Many have MOSFET outputs that can sink 10mA with an Rds(on) of less
than 20 ohms(if that's what the customer expects -- I'd go with a
couple of mA strobed at 1Hz if all you want is a visible LED
indication, and you want to conserve a low battery rather than drive it
into the ground).

If you need help in specifying the IC, possibly you could describe what
kind of battery you're trying to monitor? Also you might want to tell
wish list of any other features you could use? This wheel has been
reinvented so many times, manufacturers have added all kinds of bells
and whistles to differentiate their products. You can easily take
advantage of that, or just go with a bare-bones minimum if cost is your
only consideration.

Good luck
Chris

9. ### Fred BloggsGuest

If by this cryptic language you mean the LED is a LO battery indicator,
then the requirements are more challenging than first impression. The
20uA refers to the battery current draw by the *all* of the LO battery
indicator circuit in the inactive state. These circuits will have some
kind of threshold comparison for the LO battery condition and the
challenge here is that most comparator circuits exhibit a large increase
of bias current draw from the battery as the battery voltage slowly
drifts close to that threshold. There are several varieties of
specialized LO battery indicator ICs that overcome this drawback. Go to
http://www.maxim-ic.com and check them out.

10. ### Jonathan KirwanGuest

It is hard for me to imagine that you are supposed to keep the
quiescent current down below 20uA, if the LED is going to draw at
least 10mA when the battery is good. For one thing, 10mA is a lot to
waste on an LED if a circuit is running off of some batteries (and I'm
imagining a Lithium button, right now, which pushes the total circuit
draw to a few mA at most, anyway.) For another, 20uA is such a small
price to pay if you are already paying the price of 10mA for the LED
that I can't easily see why that would be such a concern as to get
spec'd like this.
If this is a circuit which must somehow indicate battery good by
lighting the LED, but can only draw 20uA to do that, you will have to
accept pulsing the LED. A 10mA guaranteed pulse current with an
average draw of 20uA implies a 500:1 period, in ideal conditions. And
if this is the desire, then that still leaves out what should be done
when the battery is determined to not be good. Would the circuit
simply be OFF, in this case?

If this is really a circuit driven off of a lithium button and that is
the reason that the battery good LED circuit must draw <20uA during
normal operation, then you will need to pulse and this will imply a
large R and a modest C to start. Something basically like this:

+bat
|
|
\
/ 22k
\
| LED 10
+---|>|--/\/\---------+-------------,
| | |
--- | |
--- 47uF ,---------, ,--------,
| | | | |
| | voltage |->-| SCR |
gnd ,--->| trigger | | switch |->--,
| | | | | |
| '---------' '--------' | 4.7nF
| | | ---
| gnd gnd ---
| |
'--------------------------------'
positive feedback

You can get this to run at under the 20uA and off of 2.8V, generating
pulses, and use normal, cheap BJTs throughout (such as 2N3906 and
2N3904) -- including the "SCR" which can be fabricated from the same.
In fact, I set down a possible circuit on the above model that in
LTSpice draws about 20uA, pulsing the LED at 2Hz or so with RC decay
shaped pulses having half-power widths of 1ms or so, with peak pulses
of 12mA. It uses (2) 2N3906 and (1) 2N3904 and a couple of 1N4148s.

I'm baffled why you can't ask the customer for more details, though.
That seems to be the place to go for clarification.

Jon

11. ### Jonathan KirwanGuest

Sorry, that should be large C and modest R.

Jon

12. ### Rich GriseGuest

As has been mentioned, 20 uA will not light an LED. This will work if
the LED off means "battery good" and the LED on means "battery low".
You must have missed the part where he says Vf is 1.9V and Vs is 2.8V. He
also seemed to indicate that he needs 10 mA ILED, so R = (2.8 - 1.9) / .01.
And, Vsat (of the driver transistor) can go as high as .3V at 15 mA for
the PN2222, so at 10 mA it will be less and that's the max so it could be
way less. So it should be R = ((Vs - Vf) - Vcesat)/If, or 60 ohms if your
Vcesat is .3, or 80 if it's .1. So, use a 75 ohm resistor.

Cheers!
Rich

13. ### Sjouke BurryGuest

Google for a LED switching supply.Those can
convert up from the powersupply, and sense the
resulting current at ground connection.
If a Led fails "shorted",the others just keep
on working.
They work by switching a coil current into
the stack of LEDS,and adjust switching freq.
to get the wanted current.
Note that it works without series resistor,
so you use your energy very efficiently.

15. ### BokiGuest

Do we have any more state ? : )

Agree.

Best regards,
Boki.

16. ### Fritz SchlunderGuest

Typical indicator LEDs won't light in any way usefully if at all at 20uA,
but the technology has been improving.

I've got some white 5mm "26,000mcd" 20 degree viewing angle clear LEDs from
here:

http://www.lck-led.com/

There is some noticeable part to part variation at extremely low currents,
but for the most part they seem to light quite visibly in subdued lighting
conditions all the way down to less than 1uA. They glow very nicely at
20uA, and in fact I use them as indicators at that current level, even for
quite bright office fluorescent standard lighting conditions. Higher
current than a few tens of microamps isn't really desirable for indicator
use since they quickly start to become dazzling to look at straight on.

There is one drawback however... Typical transistors/MOSFETs and whatnot
can often have fully off state leakage currents exceeding 1uA. This means
you can't always guarantee the LEDs to be fully extinguished when switched
with typical transistors/MOSFETs, unless you place a parallel resistance
next to the LED to soak up the leakage without producing visible light.

17. ### Fritz SchlunderGuest

Hmm... Redoing the math, I take part of that back. I use them at 200uA as
indicators in office standard lighting conditions. At 20uA they still glow
very much visibly, but only if you look at them within the 20 degree viewing
angle. As indicators I prefer to easily be able to establish if they are
lit or not when viewed on or off angle.

18. ### Spehro PefhanyGuest

Or drive with a CMOS output. Not like you'll be needing a big
transistor for 20uA!

Best regards,
Spehro Pefhany

19. ### Don KlipsteinGuest

I know of LEDs that light damn well and bright at half a milliamp!
Please mention upper limits and lower limits of voltage available to the
LED or upper end of the LED's dropping resistor! Also color requirements
- deep pure red can require a good couple milliamps but with lower voltage
drop in the LED! Worst is where the color is required to be yellow or
yellowish.

- Don Klipstein ()