# Electronics Large LED parrelel array help.

Discussion in 'Modding' started by frojoe, 17 Mar 2009.

1. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1
I need a parallel array of 20 identical leds. The source voltage is 4.5, and I'm going for a forward voltage of about 3.5. The parallel array wizard at linear1 is telling me to put a 56 ohm resistor on each of the leds, but due to space and heat I would prefer to place one resistor near the source, before the wires split to the parallel leds. The only advice I found on this is one mention of using an 8 ohm resistor to go from 4.5-3.5V, but that was in a particular instance(with only 8 leds). If anyone could point me in the right direction for calculating the resistor I will need, I would really appreciate it. I can do the math manually, but I can't quite figure out what to plug into my equations.

Also, I will be hooking 3 identical arrays to my source, each with its own switch. I will be running with any 1,2, or 3 of the arrays switched on at once. Is there anything else I need to do to make this work? Thanks.

-Joe-

Joined:
23 Sep 2004
Posts:
616
2
Joe,

The math you are looking for is:
20 leds * 20ma = 400ma
4.5 vIn - 3.5v = 1v R drop
1/.4 = 2.5 ohms @ 1/2 watt
This is not a common value, but hooking 4x 10 ohm 1/4 watt resistors in parallel will give the same value

Traditionally, this is not accepted as a good idea because if any of the LEDs do not draw exactly 20ma, the other LEDs will have to draw more. If one LED goes out, for instance, the rest of the LEDs will now run at 21mA. This can cause another burn out, after which the whole string will run at 22mA, etc.. and can cause a cascade which burns out the whole string. The benefit of using one resistor per LED is that if one burns out or misbehaves (say draws only 10 mA), the other 19 still get the proper current. It would get annoying using 60 resistors though.

As far as heat, the same amount of heat will be produced by dropping 1 volt over 20x 56 ohm resistors as dropping 1 volt over 1x 2.5 ohm resistor, because power = voltage * current, and has nothing to do with resistance. 1v * (20*.02A) = 400 mW of power dissipated as heat, regardless of the configuration of the resistors.

The "correct" way to do such a circuit is to wire all twenty in series, with a fV of 70v, then use a DC-DC converter with a current feedback loop to give it exactly 20mA. But that may be a bit overkill for your application.

JSD

3. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1
Thank you, that explains it perfectly. As far as multiple resistors, I know that is the correct way, however I really don't have room for it. I'm glad heat is not different either way though, and with this setup the resistor will be removed from the confined space, reducing the chance of heat related issues. I was thinking multiple resistors would waste more heat, because that just seemed to make sense, but looking at the math, there is no difference.

Because The leds are rated for 30mA, I'm hoping that is enough head room so any one led burning out won't take out the whole string. Also, I forgot to mention my source is 3 D batteries, so a 70v source would be hard to obtain.

As far as using this one source, for 1,2, or 3 arrays on at once, is there anything I need to do to isolate each array? It seems to me that this setup could mess with the voltages or currents when running different numbers of the 3 arrays at the same time, but I don't really know what I'm talking about. Thanks again,

-Joe-

Joined:
23 Sep 2004
Posts:
616
2
If you are running it off a battery, you don't want to be dissipating 400mW * 3 as heat! Much better solution is to get two 1A rectifying diodes which drop .5v each in series before your parallel LED array. This would eliminate the need for resistors all together, and dissipate much less heat.

The entire circuit would look something like this (pardon the ASCII art):
Code:
BAT + = 4.5v
BAT - = GND
1N4001 or similar
+B  |-- SW1 --|>|----|>|----o ARRAY 1 o---|  B-
+A--|-- SW2 --|>|----|>|----o ARRAY 2 o---|--A-
+T  |-- SW3 --|>|----|>|----o ARRAY 3 o---|  T-
vdrop         .5     .5        3.5

5. ### cpemmaEcky thump

Joined:
27 Nov 2001
Posts:
12,328
55
No free lunch. Same heat produced (VI) whether you drop volts by a diode or by a resistor. And, if it's possible, using diodes is even worse than using a single resistor to work several parallel strings.

A resistor drops a voltage proportional to the current through it; a diode's voltage drop is much less variable.

6. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1
So is there anyway to isolate the 3 arrays so that each one will always see 4.5 v, or should I just bite the bullet and use 3 separate sets of AA's, instead of one set of D's. Also, if I did this, would it be better to use resistors or rectifying diodes? Thanks again for all of the help, simple circuits I get, but as the layers get added on I lose track of what to do.

-Joe-

Joined:
23 Sep 2004
Posts:
616
2
Incorrect. Let's learn something, shall we?

Resistors are intentionally inefficient conductors, but conductors none the less: their Fermi level still lies in the conduction band. Resistance in every conductor, including the resistor, is caused by the scattering of electrons due to imperfections in the ionic lattice. Scattering not only changes the direction of the electrons, but also transfers some of their energy to the lattice. The result is twofold: First, the electrons loose velocity parallel to the field of conduction, and each scattering event reduces the average velocity (i.e. voltage) of the current in that field. Second, the additional kinetic energy imparted to ions within the lattice is observed as an increase in average kinetic energy of the conductor, otherwise known as temperature. To increase resistance, we simply introduce a great deal of havoc within the ionic lattice - say, by adding atoms who's Fermi levels are well below the valence band - to encourage the scattering effect. Thus the heat dissipated by a resistor is directly proportional to the voltage drop (the amount of additional potential within the field required to overcome the resistance) and the current flow (Overlooking Hirschman uncertainty, we can assume the probability of an electron with a fixed initial velocity being scattered while passing through a fixed length of a homogenous conductor to be constant, thus more electrons will directly relate to more scattering). Thus, a resistor is a linear device which abides by Mr. Ohm's law.

The semiconductor is not a conductor, nor an insulator, but rather a non-linear device - that is it's Fermi level lies halfway between the conduction band minimum and the valence band maximum. This gives them the potential to carry charge not only by electrons like conductors, but by also by "holes" - places for an electron to go. By doping, or adding, atoms with higher or lower energy levels to the semiconductor we can increase the charge carrier concentration (N-type) or the increase the number of valence band accepting holes (P-type). A diode is the combination of a section of P-type semiconductor (usually silicon) with a section of N-type semiconductor (also usually silicon). When the P-N junction is first created, electrons in the conduction band of the N-dopped section diffuse into the holes of the P-dopped section. When an electron and a hole combine, they both vanish, leaving an positively charged ion on the N side and a negatively charged ion on the P side. The extent of this process is known as depletion width, and is limited by the electronic fields surrounding the positive and negative dopant ions which were left behind during the diffusion: when the opposing fields reach sufficient strength to prevent any additional electrons or holes from passing the junction reaches equilibrium. In the ionized depletion region around the junction there are no charge carriers (holes or electrons) and thus it behaves like an insulator.

When an external voltage is applied with the same polarity as the built in potential (negative to negative, for instance) the diode is said to be in reverse bias, and until the potential is great enough to induce breakdown the junction will continue to insulate. However when a voltage is applied with an opposite polarity as the junction potential, i.e. forward bias, and the external potential is great enough to overcome the internal field (around .6 volts for silicon diodes, .5 for Shockley rectifiers) the combination process can continue and there is a very sharp increase in current flow. The diode has "turned on". In this state the junction acts as a conductor with negligible resistance - on the order of 10E-8 - and very little energy is lost to heat. The observed potential difference is simply the force required to overcome the junction's depletion field and keep it "turned on".

So no. There is not the same heat produced when dropping V*I via a resistor and a diode. A resistor dissipates the energy as heat, the diode by quantum mechanics. I do appreciate the chance to dig into the old ECE101 book though.

Unnecessary. Use the diodes. See Forrest Mimms III book "Getting Started in Electronics" for an application example if you don't believe me.

JSD

8. ### cpemmaEcky thump

Joined:
27 Nov 2001
Posts:
12,328
55
So explain why one needs a 20A rectifier bridge bolted to a heatsink in a big power amplifier when, if "very little energy is lost to heat", four ordinary little 1N400x diodes would do...

Or why diodes occasionally burn out...

Last edited: 19 Mar 2009
9. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1
So two, 1-amp rectifying diodes in series. Will all 1-amp rectifying diodes drop the voltage by .5v, or is that one of their specs? I saw diodes like this at radioshack, but they say they drop 1.6v(alternating component of forward voltage[RMS]) at (the alternating component of forward current[RMS]). Is this the correct 1-amp rectifying diode?

And just so I'm clear as too what your saying, I can use one source without any special measures, and it will still work no matter what number of the 3 arrays are on at once?

Thank you so much for all the time you've spent helping me with this little project, I get the leds in a day or two and I'm ready to get going.

-Joe-

Joined:
23 Sep 2004
Posts:
616
2
This, however, is an accurate statement. A diode will either drop it's forward voltage, or not conduct. However if you reduce the current through a resistor, it's voltage drop will decrease.

So what? Consider the example of a string of 20 LEDs, each rated to 20mA. Above I gave a very cursory explanation, but since I'm at it, here's how it really works.

Q.E.D. a 2.5 ohm resistor is required to operate each at 20mA in a parallel configuration from a 4.5 volt supply. If one LED burns out, the parallel string is now drawing only 380mA instead of 400mA. Since resistors are liner, V=IR applies, thus the voltage drop across the resistor is now 0.95 volts, making 3.55V. LEDs, being diodes, are not linear, and their current draw (after saturation) is exponential. A 0.05 volt increase above their forward voltage will cause the LEDs to draw a huge amount of current (on the order of amps) for as long as it takes (nanoseconds) for the resistor to resume it's 1 volt drop, which we know happens at 400mA. The problem is, there are only 19 LEDs left, so each LED is drawing 21mA in-order to keep the system in equilibrium.

Consider next a series of two diodes hooked to the same 4.5 volt supply. Each diode drops 0.5 volts, regardless of current flow or input voltage, so long as the input voltage is greater than 0.5V. With all 20 LEDs operating at a total of 400mA, the two diodes reduce the voltage by 1.0 volt. When one LED burns out, the diodes still reduce the input voltage by 1.0 volt - since the diodes are non linear, they still supply 3.5 volts to the remaining LEDs in the chain. The forward voltage of the LEDs is not exceeded, and they continue to draw 20mA each. In fact, the same two diodes could be used for a single LED, or 500 LEDs in parallel, with no adjustment to the circuit whatsoever, as long as the diodes were rated for the 10 amps the array of 500 would draw.

Note that diode's power rating is not about heat dissipation like a resistor's power rating, it is a measure of the junction area available for current to flow over. If too much current is allowed through a diode with a small junction, permanent damage to the junction will result, most likely in the form of the magic smoke and/or exploding bits.

EDIT/3:
Didn't see your questions at the time, but I think I answered them anyway. Or maybe I didn't. See below.

EDIT2:
The diodes you're looking at must be a bridge rectifier, which is actually 4 diodes in one case. You're looking for these guys http://www.radioshack.com/product/index.jsp?productId=2036268

Yes, the arrays will draw more current from the battery, which may deplete it faster (see alkaline discharge curve), but will not affect one another.

EDIT3:
A 20A rectifier on a heat-sink is less expensive than a 1000A without a heat-sink. As for every conductor, resistance is given by R= (length * resistivity) / (cross-sectional area). A 20A diode has a junction of a certain size, which sets it's cross sectional area. A 1000A diode has a wider junction, which reduces it's resistance proportionally. It is the same concept as a 18 gauge wire getting hot with 20 amps running through it, where a 12 gauge wire will get less hot. The same goes for the 1N4 series diodes, versus the 20A rectifier.

Two reasons. First, diodes, like all semiconductors, are subject to thermal runaway. That is, while the resistance of a conductor will increase as temperature rises, at high temperatures a semiconductor's resistance will decrease exponentially. Since the primary change of temperature in both cases is heat generated by resistive losses (no matter how small), and a semiconductor's V/I curve is non linear, more current will create more heat, more heat will lower the resistance and allow more current to flow, etc.. until the melting point of the substrate. This is why connecting an LED directly to 5 volts will cause the magic smoke. The linear resistor on the other hand will just get hot.

Second, diodes, like all semiconductors, have a lifespan determined by the integrity of the junction. Damage to the ionic lattice around the junction is a function of time - randomly scattered electrons may knock out electrons from much lower energy levels out of orbit, permanently charging a region that was previously a semiconductor - but also a function of temperature, since the additional kinetic energy creates more opportunity for damage to occur. Thus the legend of "for every degree C you run your CPU cooler at, you double it's life" or whatever, CPUs are just billions of diodes and transistors. If you were to substitute the 1000A diode into your power amp, it would most likely last forever; but at what cost?

Last edited: 20 Mar 2009
frojoe likes this.

Joined:
23 Sep 2004
Posts:
616
2
To answer my own question, here is a 1800A surge, 200A continuous bridge rectifier 844-200MT40KPBF for \$100
Here we have a 250A surge, 20A continuous bridge rectifier for \$1.70 821-TS20P06G
with a 4.75" by 3" by 3" heat sink for another \$13 567-421-K

The 200A would not require the heat sink, but would still cost \$85 more. If you could find a 24 pin connector that could take it, it would be possible to run 00 gauge wire to your motherboard, and it would absolutely reduce the heat coming from the 14 gauge wire, but it would make wire management slightly more difficult and would be much more expensive. You could not, however, substitute 34 gauge magnet wire for the 14 gauge wire currently in place, as it would melt the first time the computer turned on.

12. ### Sushi WarriorWhat's a Dremel?

Joined:
20 Mar 2009
Posts:
137
2
Hey, I'm new to the forums (well, been lurking for a while ) and I have a very similar problem. I want to set up one array of 31 green, 3.4v/20mA LEDs on either a 12V or 5V source (whatever works best) and also an array of 19 LEDs (same forward v and current) for backlighting my keyboard. For this I want to use USB power (5v, does anyone know how many amps?). What sort of a resistor/diode would I need to use only 1 at the start of the array? Also, is it possible to do the array in series? That would be much easier. Thank you oh-so much for the help

So to simplify - 1 array of 31 (3.4V/20mA) on 12v or 5v, and one array of 19 on 5v from USB (anyone know how many amps USB has?) done with resistors/diodes at the start of the array.

Joined:
23 Sep 2004
Posts:
616
2
If it were the good old days/this wasn't your first post I'd say but since I seem to be in the mood today...

USB 1.1 has a published limit of 500mA. That being said, it is often possible to draw more without problems, but the 19 LEDs @ 20mA each will consume only 380mA which should leave plenty of room for your keyboard itself. Since it's running off USB you could get away with using 19 individual resistors (80 ohm) a single resistor (4.2 ohm 1 watt ~ 2 parallel 10 ohm and 1 47 ohm is close enough) , but since we're on the topic of diodes you could also use three of the diodes listed above.

For the 31 - why 31? 30 is a nice enough number. Anyway, the standard resistor configuration for 12v is something like three in parallel with a 120 or 100 ohm resistor (safe vs adventurous). Again, if you can get 4x 1 amp diodes (for the 600mA, 3A or 5A would really be best) you could run 30x or 33x LEDs in a 3 by 10/11 matrix, that is with 10 sets of { 3 series LEDs } connected in parallel, fed by 10 volts from the diodes (four in series gives 12V_in-2Vd = 10V_out).

One important thing to remember when designing for things that are not battery powered is that power supplies have the potential to exceed specified voltage (while batteries are chemically limited to a maximum). This is especially true of startup voltages in switching supplies like the ones in computers. For this reason, if possible it's a good idea to include a filtering capacitor of some sort (~200uF) in your circuit between the LED + and ground, to absorb some of the startup and shutdown transients.

14. ### Sushi WarriorWhat's a Dremel?

Joined:
20 Mar 2009
Posts:
137
2
Thanks for the response, I could have looked it up but I am stuck on a crappy comp right now - it really sucks at opening multiple tabs (P4 and 512mb ram ). Thanks for the detailed response. I need 31 because it is for the fins at the front of my case and there are exactly 31 fins. Thinking of it now, I could use 30. I did plan to run them off case power, but due to the fact I'd need to order a new capacitor and am on a rather tight budget, I guess I'll just go with batteries. So then I'd be using a 9V battery -.-

Now, sorry to sound sort of simple, but I have decided to use EL wire instead. I was going to buy a lot of 50 LEDs due to the fact that I can't find EL wire in green in Canada where I live, but looking for LEDs brought me to eBay and going to eBay led me to finding some EL wire. It seems like a much simpler solution for the 30 LED array but I may or may not still use the 20 LED keyboard array. I figure I'll buy a big huge lot of 1N4001's - 100 to be exact, for just 3 dollars and then 50 green LEDs, so I'll have enough left over to do plenty of other stuff. Sorry to get you to do all that thinking for nothing

Thanks a lot for all the help

15. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1

Good, I bought the right ones when I went to radioshack yesterday. Thanks for all your help, and a few good physics lessons along they way(never a bad thing). I'l get started on this as soon as my leds arrive. Thanks!

-Joe-

16. ### cpemmaEcky thump

Joined:
27 Nov 2001
Posts:
12,328
55
And if you read the datasheet, maximum forward voltage for the bridge is 1.4V, maximum heat dissipation at 200A is 480W (Fig 5).

Heat Loss in a diode = Forward Voltage drop x Current, as I said originally.

17. ### frojoeWhat's a Dremel?

Joined:
17 Dec 2008
Posts:
135
1
cpemma, I understand that you think the diodes will give off the same amount of heat as a resistor. That makes sense, but I'm not going to step into that discussion, because it's out of my league. In your opinion, is there anything wrong with using the 1 amp diodes that theshadow linked too? I can't tell if you are implying that that is a bad idea, or just that it won't make a difference either way. Also, do you agree with his statement that it is ok to run the 3 arrays off of one source, no matter which ones are turned on at the same time? Thanks,

-Joe-

Joined:
23 Sep 2004
Posts:
616
2
I did read the datasheet; You have confused maximum dissipation with nominal dissipation. Maximum dissipation is a measure of current capacity as a function of junction temperature. It is a maximum value, so no matter what the power dissipation will not exceed V*I, however in normal operation (below 200ºc jT) the array is much more efficient (see the 1K/W scale on the next graph). A resistor dissipates V*I at all times.

I'm sorry if I have not explained the mechanics clearly. I would suggest getting a book on semiconductor theory, but I don't expect you will, so this research may help:
http://www.delta.com.cn/jimi2/fiveyears/Papers/IEEE/124.pdf

Joined:
23 Sep 2004
Posts:
616
2
Let's put it another way. The Watt is the SI unit for power, i.e. 1 Joule / Second.

1 amp flowing through a diode with a 1 volt drop = 1 watt for 1 second = 1 Joule.
1 amp flowing through a resistor with a 1 volt drop = 1 watt for 1 second = 1 Joule.

The Joule is the SI unit for work, as in Work = Force x Distance. It does not specify how that work is done.

In the diode, the 1 joule of work is done to overcome the potential difference across the P-N junction, that is: the force required for each electron to jump from a lower energy level to a higher energy level multiplied by the distance from the lower energy level to the higher energy level, multiplied by the number of electrons in one amp (6.24E18 or thereabouts) equals one joule. In an ideal diode there is no "heat produced", in a real diode the heat produced follows the same rules that a conductor abides by.

In the resistor, the 1 joule of work is done by the scattered electrons (i.e. any electron that does not travel parallel to the input and output terminals) on the lattice, that is: the force the scattered electrons exert on the atoms in the substrate multiplied by the distance from their intended path multiplied by the number of electrons in one amp equals one joule. The energy imparted to the atoms increases their kinetic energy, and is manifested as heat and heat alone.

So in the PN junction the majority of work done is parallel to the flow of current (a.k.a. efficient) and the only heat produced is by accidentally scattered electrons(which is unavoidable because no material is perfect). In a resistor all work done is by scattered electrons intentionally - resistors are formulated with impurities to encourage it - thus all energy expended produces heat.

One watt does not necessarily mean one watt of heat. In the junction of an LED, the work done by the much higher energy jump (3.5v VS .5v) releases photons. In a one watt electric motor, the work is done spinning the rotor. In a one watt radio transmitter, the work is done creating an RF electromagnetic field. In all of these cases, heat is produced, but if the device is efficient the ratio of heat to useful work is low. A resistor operating at one watt looses 100% of energy to heat, and performs no useful work whatsoever (unless it is an electric heater, where we regard the heat produced as useful).

Resistors are indispensable in electronics - don't get me wrong - but they are not ideal for shunting large amounts of current. While the diode does not "save any power", it does not dissipate it as copious amounts of heat, which is what I said originally. A better solution would be a LDO regulator in a constant current configuration, which could see 60% efficiency (40% loss vs. 100%) or better yet a boost-buck switching regulator which could get 85-90% (10x more efficient), but unless that efficiency is needed there is no justification for the added complexity.

Both the resistor and diode have the same component count and cost, and in the typically small confines of a battery powered device dissipating over a watt of heat continuously is not an ideal solution.

BTW I understand where you're coming from, a lot of the numbers and symbols are the same. However said "technobabble" is what separates ignorant assumptions from accurate scientific assertions; I'm not showing off or starting an online pissing contest. Take it and learn something or leave it and don't.

Joined:
5 Jun 2005
Posts:
495