LED flash lite: Max Current vs forward voltage (dropping)

Discussion in 'LEDs and Optoelectronics' started by wannabegeek, Dec 17, 2011.

1. wannabegeek

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Aug 17, 2011
Hi all,

I've chosen a project to keep my studies of electronics up, which is designing a
really good LED flash light.

Imagine I have 5V battery with a 55 Ohm resistor giving me 90mA, this is pulsed at ~25 Hz with a 50% duty cycle square wave.

Over time, the battery drops to 4.2 V. Now my current is ~76mA. The data sheet be linked below, has a plot of the forward voltage required for a given current. At 4.2 V, the forward current is only 60mA. Am I understanding this right ???

So what happens to the extra current my circuit is trying to pump through the LED ? Does it create extra heat and thus wear and tear ....?

I'm trying to find a way to practice using a FET as a current source. Does it help to keep the current stead, if the device requires a certain voltage for a given current anyways...if I was operating at 10V and wanted a constant current down to 4.2 V, then a current source clearly makes sense. But operating down in the elbow of the diode's IV curve...I feel kinda lost...

TIA,
wbg

http://www.alldatasheet.com/datasheet-pdf/pdf/103642/MARKTECH/LM1PWH101.html

Last edited: Dec 17, 2011
2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
wannabegeek, you need to investigate constant current switch mode regulators. That's the most efficient way of achieving a "really good" LED flashlight.

Often flashlights employ currents in excess of 1A (into 3 to 5 watt LEDs)

Droppint the extra voltage via a resistor (or any other linear device) is going to waste a lot of power and generate significant amounts of heat.

3. wannabegeek

133
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Aug 17, 2011
That's great...I did find my way to those circuits...power switching is a great topic and really different from what I studied in school which was basic signal amplification and error analysis.

My new idea is to incorporate a flashing LED into a boost circuit. The human eye can only see 24 frames a second, so it shouldn't hurt to flash the LED at some fraction of the boost frequency. This will also cut down on power consumption and make the theory even more interesting.

I was also thinking of keep a FET current regulator so there's no resistors at, except for the inductor, in my circuit.

Thanks Steve.

Last edited: Dec 20, 2011
4. BobK

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Jan 5, 2010
You seem to think that you will save energy by flashing the LED. This is not the case. The eye is an integrating device, so if you power an LED 1/2 the time it appears exactly the same as if it is producing half the light, In fact, this is how you dim an LED. There is no benefit to operating it less than full time.

Good LED flashlights use a current controlled buck/boost converter that allow the same light output as the battery runs down.

Bob

5. BobK

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Jan 5, 2010
Also, LEDs produce less light per Watt at their max current than they do at a somewhat lower current, so boosting the current above the max and running a lower duty cyble will allow you to run more current but will not boost the efficient to achieve the same brightness.

Bob

6. wannabegeek

133
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Aug 17, 2011
Hi Bob, thanks for your thoughts...yeah I figured the boost converter thing and plan to use a JFET to control current with I_DSS = 30mA in stead of a resistor.

I have seen an LED data sheet which shows increase in luminous output increasing with current.

The eye is an integrator....OK, but strobe lights work...By integrator I am assuming you mean a time average.
I = Sum { f(t_i)/T_i }

If I flash a light at 100 Hz, with a 50% duty cycle I'm pretty sure it will look just as bight....but I'll look into it, since no one has done it, it you are probably correct.

cheers,
wbg

7. Resqueline

2,848
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Jul 31, 2009
A LED running at 50% current and 100% duty will be substantially brighter than one running at 100% current and 50% duty.
A linear current limiter (like a JFET etc.) will be a waste of battery power compared to a switchmode constant-current source.

8. BobK

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1,688
Jan 5, 2010
It will look half as bright, try it.

Why do say no one has done it? I and many others have. This is how you control the brightness of LEDs, by keeping the current the same and varying the duty-cycle. You can also make a 3-color LED look like any color by varying the duty-cyles of the R G and B LEDs.

Bob

9. BobK

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Jan 5, 2010
Of course the light output increases, but the efficiency does not, it tends to decrease with increasing current.

Efficiency = light output / power input.

Power input = Current * Voltage

The Voltage goes up with higher current, so, even if the light output goes up linearly with the current, the power in goes up faster than linear since both the current and voltage are increasing.

And, as Resqueline said, a JFET is no different than a resistor when used to limit current. A buck converter uses in inductor to limit current. An ideal inductor, unlike a resistor or JFET or MOSFET or BJT does not dissapate any power.

Look up buck converter and understand it if you want to learn how to operate an LED at with minimum loss.

Bob

10. wannabegeek

133
0
Aug 17, 2011
Hi all,

Thanks for your replies...

@Bob I meant to write, that since nobody uses a flashed LED to save power, it stand to reason that, this idea not a good one. There fore, you are likely to be correct based on your superior experience alone...and I'll try it myself ..

I looked at Buck Converter's but didn't see the use for one in my LED circuit until now...
So when I read that a JEFT acts as an Ohmic device, that means it wastes power in the current control process as a resistor would...I think ?

So a Buck converter is a switched current limiter...very cool. I understand that an Ideal inductor is all reactance and thus will not dissipate any power only store energy not being used.

I just got my basic learning circuit working, using ancient scope that's hard to use at times...

This project has a been a great learning experience so far, I thank you all for your input.

wbg

11. BobK

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Jan 5, 2010
Okay, I misunderstood.
Yep, acts just like a resistor, wasting power. The efficiency comes down to simply Vout / Vin.
Yep, the inductor dissapates no power except for that of the series resistance. Use a really large (dimensions, not inductance) inductor and this can be reducted to near zero.

The losses in the buck converter are based on the electronics to control it. They are typically 80-90% effiicient.

The other advantage of a buck converter is that you can control it based on current instead of voltage, which is a better way to control LEDs since the voltage / current ratio varies from one device to another.

Also, you might want to look at buck/boost converters. These will keep your LED running at full steam even if the battery voltage falls below the forward voltage of the LED.
Great, keep at it!

Bob

12. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Read this page. It is an easy to read explanation of how all the bits of an SMPS work, and includes a current limited design.

13. wannabegeek

133
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Aug 17, 2011
So now I need to read about and build a Buck Converter to regulate my LED current...

In addition, I have not had success looking up a MOSFET that turns on at low voltage.
What transistors do people typically use for a low consumption circuit like this.

I'd like to run with 2 AA batteries ~ 3.0 V, all the way down to 0.9 V. I have seen a cool 555-like IC that will operate at 0.9V.

I just don't know what FET will turn on that low...even after some boost, what I have seen, V_GS to turn on is more like 3V.

Thanks !

14. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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The most efficient way to regulate the current is by using PWM (or similar) with an inductor and some way of measuring the current, using this for feedback to the PWM circuit.

That is effectively a switchmode power supply.

In your case, you want to run the device from a lower voltage than the Vf of the LED. You have to consider the requirement to boost voltage (sorry, I linked you to a page with regulators to reduce voltage). This implies a higher average current from the power source than delivered to it (because you can't generate power from nothing).

A problem arises if you use a simple boost regulator in that if your input voltage exceeds Vf, there can be no current limiting.

In addition, as the battery voltage falls, the current will increase. Because you are nearing the end of the life of the battery, the internal resistance will be quite high and the voltage will be plummeting.

Getting a boost regulator to operate at 0,9V is one thing, getting it to *start* at 0.9 V is quite another. The regulator may require a higher voltage power supply (which could come from the output if it was running, but not when it's not).

I would probably opt for a higher voltage input (i.e. more, but smaller cells) which would make the design process a lot easier.

15. wannabegeek

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Aug 17, 2011
Thanks Steve for your thoughts. I read about PWM but didn't consider it too much, since it requires a solid reference voltage on a diff amp to trigger the PW length.

Now that I read all your concerns about voltage being high enough to run a 0.9 timer, I'm starting to rethink my design.

In addition to researching the buck/boost setup to regulate current and boost voltage, I'm thinking about using a button cell battery only as a voltage reference. If the button cell were hooked up to really high impedance devices for voltage reference only, I might be able to make a pretty good circuit....

I still haven't found a good switching transistor.

Last edited by a moderator: Jan 8, 2012
16. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You do not need a voltage reference for PWM control of the brightness of a LED. Relative voltages are fine (e.g. resistive dividers)

With higher input voltages your choice of transistor becomes far easier.

17. wannabegeek

133
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Aug 17, 2011
I have been looking at the Black Regulator's in Steve's links....they are super cool and I'll need to study them for a while to make use of the mono-stable capacitor timed design.

In the mean time, I noticed that the 2N2222 is used to switch. Looking at a data sheet,
the collector-emitter saturation for 30mA occurs at low voltage, like ~0.1 V.

But with BJT's, the base-emitter current is always 'on' right ....?
So even with a low base current requirement, am I losing too much current for even a moderate beginner low-consumption design ?

I like the Black regulator's use of a cap to control 'on' time. It's nifty ...to really understand it I'll need to do a mesh analysis of it...or stick it in a simulator. My sim right now is Qucs running on Ubuntu. I need to learn gEDA or something....

cheers,
wbg