"Perfect" LED current limiting challenge

[rshayes]

You can use switching techniques to control current as well as voltage. For example, connect an inductor in series with the LED. Switch the input inductor lead to the power source. The current will increase as a linear ramp (approximately). When the current reaches a certain level, say 35 milliamps, switch the input to ground. The current will now start to decrease. When it reaches a lower level, possibly 25 milliamps, switch the input back to the supply voltage.

The switch to the supply voltage can be a PNP transistor. The switch to ground can be a diode. The control circuit can be a schmidt trigger made with a comparator. The current through the LED can be sensed by placing a series resistor in series with the ground lead. A voltage drop of 100 to 200 millivolts at the peak current should give an adequate signal.

Efficiencies in the 80 to 90 percent range should be possible for red LEDs. Higher efficiencies should be possible for other colors and white LEDs.

The trade off is complexity for effieciency.

[Chris Smith]

Robert

All modern Laser diodes incorporate a PIN diode in the package which receives light from the laser element [side ways or reflective] and one pin on the package is used as a feed back loop to control the light level, via external current limiting circuitry.

The actual value out put from the diode is irrelevant per se, as it acts as a threshold value to control your circuitry via this diode.

Pin diodes are susceptible to light, as long as they are not epoxy coated in a Light resistant [black] package. Clear pin diodes are available, and respond to light exactly the same way as a photo transistor, only faster because of their low capacitance value due to the small DIE size.

[ian]

I yi yi!!
Everbody is missing the point. I appreciate the efforts to understand though..........
Read this very carefully...............
A series circuit.........
1) A 9V battery inputs to a DC to DC converter who's output voltage is determined by a micro.
2) The output of the converter goes to a LED.
3) The output of the LED goes to the collector of a 2n3904.
4) The emitter of the 2n3904 goes to a 5 ohm resistor to GND.
5) The current into the base of the 2n3904 will be limited so that the maximum current that will flow from collector to emitter will be 30mA based on a nominal Hfe.

Based on this circuit are the following truths.
1) The 5 ohm resistor will not limit the current, but it will provide a voltage reference to the micro.
2) No more than 30mA will flow through the circuit regardless of the DC to DC output voltage because the transistor will limit the current.
3) If the micro goes haywire and outputs a maximum voltage and turns on the LED, the LED won't burn out because of the current limit.
4) The maximum voltage drop of the transistor could be as little as .1V IF IT IS IN SATURATION.
5) The maximum voltage drop of the resistor is .15V.
6)IF the combined voltage drop of the resistor+LED+transistor is 4V then the DC to DC converter could be programmed to output 4V. The efficiency of the DC to DC converter will significantly reduce the actual current draw from the 9V battery by as much as half.

How the circuit works..............
1) Turn on the LED via the 2n3904.
2) Set the DC to DC converter to output 3V
3) Read the voltage at the 5 ohm resistor.
4) If the voltage is too low the output of the DC to DC converter has not achieved the breakover voltage of the LED.
5) Increment the DC to DC output voltage.
6) GO TO 3
7) Once the breakover voltage of the LED is reached the transistor will fully conduct because it will be in saturation. At this point even a small increment in the DC to DC converter's output voltage will increase the current through the LED a lot. When the voltage drop at the 5 ohm resistor is >.1V then MORE THAN 20mA is going through the LED.
PWM the LED to approximate 20mA.
9) By now, under ideal conditions, the draw on the battery will be less than 10mA, the LED will be lit at an equivalent of 20mA.
10) This circuit will compensate for varying LED voltage drops, transistor Hfe, and varying battery voltage.
11) If the micro goes haywire and outputs the maximum voltage+turns on the LED, the LED will not burn out unless.................
If the expected Hfe is 100, and I get a transistor with an Hfe of 300, guess what?????? 90 mA through the LED!!!!!

[JPKNHTP]

-JPKNHTP

[L. Daniel Rosa]

This is the first I noticed mention that the battery is the 9 volt variety. This does complicate things a bit.

There are 80-90% efficient current regulators that are designed for white LEDs. I can only guess that you are using a microcontroller because there is something else that you are not mentioning, which also needs between 3 and 4 volts.

Since we are all missing the point, will you please give more background on the project? I would like to understand why voltage regulation with PWM and overcurrent protection is so important when current regulation would be simpler. I am especially curious about the use of a 9V battery when a 6V source would be simpler.

[josmith]

You may run into problems with this circuit because the pwm circuit averages high current bursts in time. If you use a constant current circuit on the end of it as a safety valve it will probably limit the bursts and kill your efficiency.For what it's worth you need to limit the voltage between the base of the transistor and ground. As the voltage drop of the emitter resistor increases the base emitter voltage will decrease. Variations in current gain will effect the circuit but not very much.
I think your whole concept may need rethinking. Usually safeguards on a digital circuit come in the form of some kind of "watchdog". For instance you could monitor the on time to the led with a seperate timer and shut the current off if it is longer than a set time.

[Engineer1138]

Originally posted by L. Daniel Rosa:
I would like to understand why voltage regulation with PWM and overcurrent protection is so important when current regulation would be simpler.

I'd like to know this too. I mentioned it twice, but it seems to have been overlooked.

I agree with josmith that the whole concept may need rethinking. I believe ian either can't deviate from this design for reasons he hasn't mentioned, or he's become attached to it.

Ian: you haven't mentioned why my ideas won't work. Three people have indicated that current regulation is probably the way to go.
What is it about the micro that makes you so concerned about it "glitching" or going haywire that a watchdog won't fix?

[ian]

Well, I may indeed have to give up on this, but to answer your question.......
again..........
A simple current limit would draw 20mA from the battery rather than the ideal, lesser, 10mA or whatever minimum current the DC to DC converter could create based on the maximum voltage differential achieveable. I'm trying to make the BATTERY last longer.
I don't understand why this is so hard to realize. I STARTED THIS POST WITH "I need a battery operated very efficient(bordering on perfect) circuit for diving a LED".
What could be clearer than that? I mean.....the post TITLE reads "Perfect LED current limiting". Whats so "perfect" about a >50% loss of battery power through an energy wasting limiter?????
I'm trying to use 4 "D" cells, but it seems the voltage differential of 3V isn't enough to to maximize the use of the DC to DC converter. But if I don't use a DC to DC converter, measurements show I'm wasting almost 50% of the battery power through the current limiter. The forward drop of a white LED is about 3.1V, a red 1.5V, fresh "D"s output 6.4V.
OVER HALF OF THE BATTERY POWER IS WASTED. I'm sorry I'm not some newbie asking about ohm's law, I guess this one was too much of a challenge for you guyz.
Kudos to stephen though. Stephen, if I switch off the current through the inductor and LED won't that create a voltage spike?

<small>[ December 04, 2005, 09:39 AM: Message edited by: ian ]</small>

[JPKNHTP]

-JPKNHTP

[ian]

Just because I ask for help doesn't mean I won't get frustrated with people who don't read the question and repeatedly ask me about what I've already stated.

[philba]

People can (and do) have any kind of reaction they want. I think what several are saying here is that whacking the people trying to help will probably not net you better advice/suggestions. Sometimes in life 'tis better to bite one's tongue but it's your right to say anything you want.

Also, if you are going to criticize the advisors, I suggest you be very sure that you posed the question/problem in a 100% clean and unambiguous way and that your information was complete. If not, you will be doomed to no one "getting it".

People will remember this and perhaps not be so ready to try and help next time. Frankly, after the first "you don't get it" response, I wasn't willing to give it a try.

This thread reminds me of the old office joke: The floggings will continue until morale improves.

[ian]

Maybe true philba, but this IS an interesting exercise. There's a gazillion battery operated LED projects. If the collaboration of this forum could come up with a good, efficient, circuit that saves significant battery power it would be an accomplishment. Much more of an accomplishment than explaining ohm's law to beginners who know nothing of electronics. While I don't knock this, I think a real challenge, and a useful application might breathe a little more life. Feel free to be left out of the project.

<small>[ December 04, 2005, 12:56 PM: Message edited by: ian ]</small>

[k7elp60]

Well as I understand it the collector of a transistor is a constant current source.
Another way to achieve what you want is to use a three terminal regulator such as a LM317LZ as a constant current source. Connecting a resistor between Vout and Adj. terminals connecting the LED between the Adj. terminal a ground, with the Vin terminal connected to Vcc. One T092 device and one resistor. All the excess voltage is dropped accros the regulator.
Iconstant=1.25/R or R=1.25/LED current. For 20Ma the resistor would be 62 ohms

[Robert Reed]

With all this talk of not using a current limiter device due to low efficiency (voltage loss only-no current loss), what about the efficiency of your DC-DC converter? Certainly there is some power loss there. There is no way to control current without current control circuitry and that has to consume power, albeit small.

[Chris Smith]

Gee, rocket science comes hard to some?

A pulse controller [buck/boost] and a cap kept at the exact value using a pin diode feed back to keep the pulse controller at the exact voltage seem soo hard?

Instead of arguing with others, why don’t you just follow the lead as soo many here have suggested? Im under the impression that it is you, that doesn’t get it.

Their suggestions go from fair, to better, and all better than your suggestion?

Keeping the current draw down is old news, soo old that you don’t seem to know of its existence, even when this forum has presented it to you on a silver platter.

I think you have a mind block, and like a mind Fart when it passes you will be able to think again?