LEDs Series-Parallel OK?

Hi Watson- Not sure if I missed somewhere in your posts but did you ever run any of your LED projects in the pulsed mode similar to strobbing as used in 7-segment displays as used in calculators? Efficiency of LED's goes up significantly in the pulsed mode. Early HP calculators such as the HP35 Series went another step farther to improve battery life by using inductors to discharge into the LED's & avoid losses in resistors. Still have my HP67 running. Got it in 1975 while at HP. Part of my time there selling the Opto line.

Cheers, John Stewart

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One string _can't_ draw more current than the other, since they're in
series.  What will happen is that the resistance of the two will be
different, and the one with the lower resistance will drop less voltage.

Before you make that claim, I suggest you read the truth about pulsing LEDs on Don K's LED pages.

"Furthermore, the usual ultrabright white LEDs, as well as gallium nitride or indium gallium nitride greens, blue-greens and blue LEDs of wavelength 460-475 nM have maximum efficiency at lower currents. It works against you to pulse these. Same for "low current red"/GaP red/"697 nM" red LEDs."

BTW, my question still remains unanswered.

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Not any more!

Yes, it's fine. If you could lay them out without the middle bus it would be a bit better:

The reason being that there's more voltage across the resistor(s) and therefore the current will tend to be more even, even at the end of battery life. If you want to see how much margin you have, just put it on your bench supply and run the voltage down to 7V or 8V. As well as the lamps getting dimmer, you'll probably see some differences in brighness between the LEDs.

Best regards, Spehro Pefhany

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I believe it should work fine. I have done similar.

The first white LED illuminator I built had some LED failures after it ran a while. I operated the LEDs near their max current spec. I attributed the failures to that, which may have been wrong - but white LED illuminators I've made since then running at lower currents have had no problems. Maybe I had a bad batch of LEDs, or maybe the whites are more sensitive than others.

Steve J. Noll | Ventura California | | The Used High-Tech Equipment Dealer Directory |

| The Peltier Device Information Site: |

That is a very good link. Guess I been away too long. I left HP & LED's in 1984 to join the sales group at R&S. Only LED's I saw after that were on a front panel!!

"Furthermore, the usual ultrabright white LEDs, as well as gallium

The answer is pretty much straight forward. All you need to know is the Temperature Coefficient of Forward Voltage. If it is negative as in a bipolar transistor you will have trouble due to current hogging. If it is positive as in a power FET then you will be OK.

My recollection is that all the LED's I was selling had a -ve coefficient. Some of the devices you are working with may be different. Does your supplier publish any kind of data sheet you could refer too?

Cheers, John Stewart

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How do you suppose that his circuitry will allow current hogging to
occur?

John Stewart wrote: [snip]

Supplier won't even tell the buyer who the manufacturer is, let alone the specs. I guess I shouldn't expect too much for $.25 apiece.

A common occurence, since a -ve temperature coefficient of voltage will reduce the Vf as the temp rises. Whichever LED in a paralleled circuit has the lowest Vf to begin will take the most current & temp will rise as a result. If you are running near the limit of If as Watson is, temp will rise at that

junction & more If will flow in the hottest LED. It's a good example of +ve FB. Destruction follows in a little while.

You might be successful if the LED's Vf's were carefully matched.

-ve temperature coefficient of voltage was a real problem for designers in the Ge transistor era. You needed all manner of "save your ass" circuits. Si is somewhat more forgiving, but lookout!!

Cheers, John Stewart

In article , snipped-for-privacy@sympatico.ca mentioned...

[snip]

It's negative, and can be seen as the LED warms up and the V drop drops. But the 51 ohm resistors give a volt drop at 20 mA, and 1.5V at 30 mA. So those balance the currents in each parallel array.

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Looks good Watson. In my mind I thought you were going to parallel LED to LED without benefit of limiting R's. As you are doing should & does work AOK. Cheers, John

In article , snipped-for-privacy@austininstruments.com mentioned...

If one of the 7 LEDs is substantially lower Vdrop, and the other 6 are substantially higher, then the lower one could have substantially more current, and of course that causes escessive heat, and makes the LED's voltage even lower.. But the 51 ohm resistors help balance out major differences in each parallel string. Another factor is that the array is coupled thermally, so LEDs surrounding the hot LED would get warmer and equalize the current somewhat. But the array really needs even closer thermal coupling, and they call it a heatsink.

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In article , snipped-for-privacy@interlogDOTyou.knowwhat mentioned...

I brought it down to under 3V per parallel string, total current was

14 mA, and I couldn't see any diff in brightness between the LEDs. So the resistors apparently do a good job of evening out the current.

I haven't butchered up the traces yet, so all fourteen are still in parallel. Thanks.

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In article , snipped-for-privacy@dontspambig-list.com mentioned...

Yeah, heat's a killer, and that concerns me, since this array gets hot. But it's inexpensive, so it won't cost but a quarter if I have to replace an LED.

Thanks.

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I don't think destruction is possible with the current balancing resistors. But yeah, the added heat will shorten the life as compared to the other LEDs. But then, what's a couple thousand hours when they usually last for a hundred thou?

Well, then I could get rid of the resistors?

Not silicon here, gallium Arsenide! Or some other more exotic combination of indium, phospor, aluminum, etc.

I heard that some blue LEDS (white is a blue LED die) have a substrate of sapphire or something like that. So maybe they do have some sillycon!

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This works better then expected, although not great, due to the LED internal resistance.

In article , levy snipped-for-privacy@hotmail.com mentioned...

I stuffed aniother PCB while watching Startrek, with 10 LEDs, 5 in parallel over 5 in parallel. Instead of every other hole, i used every third hole, which put more space between the LEDs so they are running a bit cooler. Resistors were again 51 ohms. I had to make only one trace cut and one jumper. Again, the PC board was part of a Radio Snack 276-168 predrilled board. The PS is cranked up to 9.46V at 150 mA, or about 30 mA per LED. This works out nicely witht he

9VDC 200 mA wall wart I have. 7 LEDs in parallel would've been 210 mA, and they would've been closer together and hotter.

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--- You must have missed the part where I asked about current hogging in Watsons _circuit_. You may notice that he has current limiting resistors in series with each LED, which should prevent current hogging quite nicely.

Taking a look at

and assuming that it's a typical white LED reveals a Vf tempco of

-4mV/K°, so a change in temp from 25°C to 100°C (Tj max) would result in a Vf change of -300mV. Subtracting this from the nominal 3.5Vf results in a new Vf of 3.2V, which leaves 1.8V to be dropped across the 51 ohm series resistor, so the new current into the LED will be I = E/R =

1.8V/51R = 35mA. Over the limit, but not due to current hogging.

Then there's another snag, the tempco of the series R.

Looking at

we find a typical tempco of resistance of -250ppm/C° from room temp to

100°C.

Unfortunately, we don't have any thermal resistance figures for the resistor, but let's assume that with 1.8V across it and 35mA flowing through it (63mW) we get a temperature rise of 25°C over ambient, to

50°C. That will cause the resistance to change by -6250ppm, or -0.625% from its nominal 51 ohms to 50.7 ohms. No big deal, considering that the tolerance of the resistor itself is +/-5%.

But that 35mA _is_ 5mA over the top, so what to do? Change the value of the resistor to limit the current through the LED considering its highest permissible junction temp.

We know that with a tempco of -4mV/K° we can expect the drop across the LED to be 3.2V, leaving 1.8V to be dropped across the R, so for 30mA through the LED and the R, R = E/I = 1.8V/30mA = 60 ohms. The closest higher value standard 5% resistor is 62 ohms, which ought to do it.

But, considering the 5% tolerance on the resistor, 62 ohms could wind up being 58.9 ohms on the low end, which would allow 31mA to flow. Still over the top, so sticking with standard 5% parts would result in the next higher value being 68 ohms, which would allow 28mA to flow with a low end resistance of 64.6 ohms and 25mA to flow with a high end resistance of 71.4 ohms.

We haven't considered the variation in Vf due to everything _but_ the LEDs tempco, but perhaps that's best saved for another day, since the point here is that current hogging is _not_ an issue with Watson's circuit.

-- John Fields