LED boost from a uP pin

I remember a while back someone posted a lil circuit for getting a boost voltage for driving LEDs out of a low voltage uP pin. I need to drive two high-brightness yellow LEDs from a single 3-3.5 volt output with ability to PWM the intensity (doesn't need to be different PWM values for both)

So just maybe silicon diode anode to the supply, the two LEDs in series with a boost cap from one output pin to the junction, and a PWM capable output pin sinking current to ground from the lower diode cathode?

I feel like there's some detail I'm missing

Reply to
bitrex
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Why not put them in parallel, each with its own resistor?

Cheers

Phil Hobbs

Reply to
pcdhobbs

In this app I'd prefer not to burn power in current-limiting resistors, I don't have much to start with. But I'd also be burning power in the flying diode and charging/discharging a cap, too. Don't know off the top of my head which is worse. My intuition tells me the former, but my intuition is often wrong...

Reply to
bitrex

The Osram oranges that we use need about 1.9 volts to be bright, 10 mA. You should measure yours... they could be very different.

From a 3.3v supply, a series resistor per LED will work, assuming your uP port can supply enough current; if not, add a transistor or mosfet. Efficiency will be Vled/Vcc, ballpark 60%.

A simple charge pump won't help. The only simple thing that will improve efficiency is a switcher, with an inductor. A boost switcher, with the LEDs in series, would work.

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This is ideally 100% efficient.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

Cool, I can do that. The yellow LEDs I have quantity of are not much to write home about; they're bright enough for my application at 1.95v, 2 mA but surely not as bright as the Osram oranges. Current consumption goes up rapidly after that, at 2v it's up to 8 mA with very little increase in brightness.

The emitted light is probably closer to green than orange, maybe around

580 nM.

Any suggestions on inductor values/switching freq to hit a ~2 mA target? I have some 1mH ~30 ohm DCR inductors on hand, don't know if that'd be too large

Reply to
bitrex

Not large physically, that is, they're through hole about the size of a TO-92 package

Reply to
bitrex

The Osrams are OK at 1 mA, 1.77V, and blinding at 10 mA, 1.88V. They probably level off brightness at some higher current, but I don't know where. 1 mA is too much for some apps.

If you want to save power, maybe get better LEDs.

You could Spice it. 1 mH will charge up at 3.3 amps per millisecond, so the ON time will need to be short, a couple of microseconds maybe. If you go constant ON time, you could vary the frequency to control brightness. Or vice versa.

More L would be better at this low current, but 1 mH might be OK.

Just using resistors is simpler! The swwitcher might be 75% efficient, and the resistors might be 60%.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

High brightness seems incompatible with drive from uP pin. do you mean high efficiency?

+V -->|--+ | pin --||--+-->|-- ground. both diodes could be LEDs if they are dark at 1.8V PWM isn't going to effect brightness much, changing frequency will be more effective. or you could add a series resistor and get more control via PWM
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Reply to
Jasen Betts

PWM control of LEDs burns the same amount of power, but this time burns more of it the LED Coulombs is Coulombs. You need a series inductor the get increased efficiency.

IME some LEDs do not like rectifying square waves.

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Reply to
Jasen Betts

Good idea. Could you link to some favorites? I need four that are more yellow than orange, and one "warm white"

Probably. Well I'll try it both ways.

Reply to
bitrex

Eh, it's just what it says on the package. Marketing term. They're not the old-timey yellow type with the diffused enclosure.

Ok, I can modulate frequency easy enough.

Reply to
bitrex

These appear to be just barely visible in a "normally indoor lit" room at about 1.72

Reply to
bitrex

I tested around 50 of these cheapo Shenzen-specials with just a current-limited bench supply and meter and tried to gedankenexperiment the probability distribution in my head of current draw at 2 volts; at that voltage it seems like a skewed Gaussian where there were many test units that were hitting 8 mA fairly well-matched to within a few dozen uA of each other, about a half a dozen which were drawing 9 or 10 mA, and a couple outliers far to the right drawing 11 or 12 mA. Only one drew less than 8 mA, at about 7.7

Reply to
bitrex

My impression was that the mfgr already pre-sorted the "good ones" ;-)

Reply to
bitrex

The led's are diodes, current exponential on voltage, negative temp coefficient. Current won't be very repeatable with constant voltage drive. So drive them constant-current, or with enough series r to make the current pretty much constant.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

The difference between Vcc and the LED voltage is bridged by ohmic losses in the pin drive circuit. May as well use a series resistor.

That circuit can drive an LED when the pin swing is too little voltage for the LED, if the upper diode is a schottky. I don't think there's any benefit in the 2-led version.

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Reply to
John Larkin

Chinese xmas light manufacturers just run a whole string off two AA batteries with the LEDs in parallel with a single low-value resistor in series with the anodes of like 20 of 'em, probably by "hand matching" and binning similar brightnesses at lower currents.

They picked this batch over, lol

Reply to
bitrex

I don't think this circuit is doing what you think it is doing. Try tracing out the action as the I/O pin cycles. The voltage on the capacitor is the wrong polarity when switching between the two LEDs. Pull "pin" down and the end of the cap at the junction gets a plus charge. Pull "pin" up and the cap is now *increasing* the voltage on the LED connected to ground. In steady state the voltage on the cap is always the wrong polarity immediately after switching until the current reverses the charge on it. This is just using the MCU I/O as your current limiting resistor. How is that going to save power? In fact, it may be stressing the I/O pin with excess current surges.

Or am I missing something?

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Reply to
rickman

If the processor gets locked up or stuck in an interrupt or during some protracted boot process or during debugging, it could become 100% efficient at smoking the fet or the power supply...

I wouldn't depend on a processor to keep the pulse with short enough under ALL normal and fault conditions.

Reply to
mike

If the uP pin latches in the low state then nothing happens. If it latches in the high state then the DCR of a small inductor will limit current thru the FET to a safe value.

Reply to
bitrex

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