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oharag (Mechanical) (OP)
20 Sep 10 9:19
I had a question on this forum concerning a SM resistor in a LED circuit. It has since disappeared? There was a poster that posted an admonishment (to whom I do not know), and I checked today and it has since disappeared. Was there a violation in posting rules?
MacGyverS2000 (Electrical)
20 Sep 10 9:31
Without seeing the post I cannot say for sure, but an SM resistor in an LED circuit sounds like a hobbyist-level or student-level question.  If that's the case, it was probably removed for being a violation of the site's TOS (posts by engineering professionals only).

Dan - Owner
http://www.Hi-TecDesigns.com

CorBlimeyLimey (Mechanical)
20 Sep 10 9:41
oharag (Mechanical) (OP)
20 Sep 10 9:42
It was a novice question since I'm a Mechanical Engineer by degree.

As a Mechanical Engineer in a small company without much of an Electrical Engineer in house (supposedly coming) I wanted to seek the help of knowledgeable resources on this board. I thought this is what these forums allow - dissemination of information (both technical and informational) to help those in need regardless of knowledge standing. With this knowledge I would have designed a reliable and successful product.

I suppose Google is my friend then.
CorBlimeyLimey (Mechanical)
20 Sep 10 9:49
I did not see the removed thread. Try reposting, but this time explain your situation (as above) and maybe the person(s) who Red Flagged the thread will be a bit more understanding.
oharag (Mechanical) (OP)
20 Sep 10 9:56
@CorBlimeyLimey (Mechanical)

I thought it was specific and detailed. I did all the ground work up front. I just needed reaffirmation on the specifics of a SM resistor - how to calculate wattage for proper sizing.

I would have liked at least an explanation for the deletion. It was a bit sudden and abrupt.
 
MacGyverS2000 (Electrical)
20 Sep 10 10:23
I don't recall the thread offhand, but Google will give you a million hits for determining resistance values (and proper wattage) for LED circuits.

Dan - Owner
http://www.Hi-TecDesigns.com

VE1BLL (Military)
20 Sep 10 10:47
It was a very simple question, but I noticed the "(Mechanical)" and generously assumed it was probably a legitimate question. These forums tend to be on hair trigger for student posts at this time of year.

Did you note down the answer already?
 
MacGyverS2000 (Electrical)
20 Sep 10 11:21
I'll flag such simple questions if it's the OP's first post... but oharag has quite a few in the appropriate forums (I always check before RFing), so I doubt I'm the guilty party.

That said, in a nutshell figure out the power dissipated in the resistor (voltage times current, or current squared times resistance), double it for safety, then find the closest (but higher) wattage resistor.  For example, a 100 ohm resistor passing 100mA... that's 1W, so find a 2W resistor.

Dan - Owner
http://www.Hi-TecDesigns.com

VE1BLL (Military)
20 Sep 10 11:25
Also, if I recall correctly, I'd pointed out that the voltage across the resistor was a fairly small percentage of the total power supply, thus making the circuit potentially sensitive to errors in assumptions or changes in conditions.

 
unclesyd (Materials)
20 Sep 10 11:43
If you have any question about the validity of the question on an electrical circuit one might want to direct the OP to this site.

http://www.allaboutcircuits.com/
oharag (Mechanical) (OP)
20 Sep 10 13:14
Ok now we are getting somewhere.

@VE1BLL (Military)
I now understand your concern. And I'm as concerned as well smile

Here is my circuit in a nutshell:

3.2 Vf typ (3.6 Vf MAX)
.150 mA
24 volt CV PS supplying 3.3 A
7 series in parallel with 22 chains

I calculated using typical Vf as the following:
24/3.2 = 7.5 devices so I rounded down to 7 devices in a series chain.

7*3.2 = 22.4 volts required

24 supplied - 22.4 volts = 1.6 volts to dissipate.

Resistance = 1.6/.150A = 10.67 ohms

Wattage across resistor = 1.6*.150 = .24 W

So I found a SM resitor 10.7 Ohms at .25 W.

I have a concern with Vf variance. Would this circuit not work if I get Vf's higher than 3.2 Vf? So if I change my circuit to 6 devices per series chain I need 19.2 volts and need to dissipate 4.8 volts. This just seems like a waste.

I still do not understand Vf variance. If I design for 6 and dissipate 4.8 volts with a resistor - won't the series chain be starved if Vf goes higher? So how do you accommodate this variance?

I just have worries about this design (regardless if I design for 6 or 7). I know some about LEDs but I do not know what happens when I get Vf variance.

@MacGyverS2000 (Electrical)

By doubling the resistor value don't I starve the series circuit? I.e. I dissipate needed voltage (or wattage) that is required by the LED series circuit in order to light. If I double to 3.2 volts to dissipate I'm stealing this from the series chain.

In my design I'm being held to optimizing the efficacy of the entire system -> total lumens/total wattage. So I'm trying to maximize total lumens and minimize total wattage.
 
oharag (Mechanical) (OP)
20 Sep 10 13:39
Maybe this will answer my question:

Does a SM resistor go at the end of the series chain? I always assumed at the start of the series chain. I thought a resistor dropped the voltage prior to it entering the series chain - thus limiting how much voltage is supplied.

If it goes at the end then the required voltage needed by the LED series will be used by the time it gets to the resisor - thus the resistor may or may not need to disippate voltage depending upon what is left over.

Sorry for the "noob" questions. I have searched the internets to try to piece all this together.

I promise once you answer these concerns/questions I'll be on my way smile
MacGyverS2000 (Electrical)
20 Sep 10 13:41
You double wattage handling capability, not resistance.

You have encountered the biggest downsides to controlling LED current with a simple resistor... changes in power supply voltage or LED Vf can create wide variances in current through the string.  It's a games of numbers... how much variance in current are you willing to allow for other parameter changes.  If you put 7 in a row, even a small spike of 2V in your supply will nearly the current in your string.  Losing 1V will cut your design current in half.  This is why no quality/efficient display circuit will control current in this manner (they'll use a current-controlled device).

Plenty of chips out there to do the job for you, but here's an example of a constant current source using a cheap and readily available part, the LM317 voltage regulator:
http://users.telenet.be/davshomepage/current-source.htm

 

Dan - Owner
http://www.Hi-TecDesigns.com

MacGyverS2000 (Electrical)
20 Sep 10 13:44
Doesn't matter what end of the chain it goes on... you need voltage dropped, it does the job wherever it may be placed.

Dan - Owner
http://www.Hi-TecDesigns.com

VE1BLL (Military)
20 Sep 10 13:58
Vendors (such as Deal Extreme, a Hong Kong based Internet retailer, for example) sell "LED driver" modules for a few dollars each ($2-$20 depending, typically with free world wide shipping).

 
itsmoked (Electrical)
20 Sep 10 15:01
You should not do it the way you're planning.

1) Fixed resistors mean large variability in string currents.

2) The first failed-shorted LED results in most or all the LEDs in a string going down in flames.  With a controller one shorted LED means that the remaining LEDs in a string continue with the original current - unaffected.

Keith Cress
kcress - http://www.flaminsystems.com

IRstuff (Aerospace)
20 Sep 10 16:21
Not to mention the fact that unless you get a higher performance resistor, its resistance will change over temperature, and a range of environmental temperature will affect the current as well.   

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

PHovnanian (Electrical)
11 Nov 10 21:09
Back to the original topic: I replied to a post regarding a Google translation of a Russian vendor's web site. The poster (probably Russian) was searching for the 'English name' of the piece of equipment. I replied and included a link to an equivalent American piece of equipment, which may have been interpreted as shilling for that company. The whole thread promptly disappeared.

They take those rules seriously.
itsmoked (Electrical)
11 Nov 10 23:45
PHov;  If you think a post like that might be seen as a shill-ing, you can state, "I have no association with this company."

Also, posts are often removed for an entirely different reason than you think.  There are certain IP numbers that will get nixed if they ever show up. That means you could see a post and possibly respond to it before the powers-that-be see it and remove the entire thread.

Keith Cress
kcress - http://www.flaminsystems.com

PHovnanian (Electrical)
12 Nov 10 11:21
Keith,
  Its probably the shilling that got me tossed. Why they deleted the original post I can't figure. The guy seemed to have a legit question, even if it was one of terminology and not straight engineering.

  I find it hard to believe that they remove posts based on IP addresses manually. That's a lot of work for something so easily automated.  
IRstuff (Aerospace)
12 Nov 10 14:54
Everything is on a case-by-case basis.  

Most likely, the OP was found out to be a student.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

itsmoked (Electrical)
12 Nov 10 14:56
I believe it is automated but requires individual confirmation.  It's not just a specific one it's adjacent  ones and that's why it's not automatic.

For instance a school might have hundreds of IPs that are in a block. Sometimes a banned individual tries to come back on all the adjacent IPs.  Or you could have a shill working from the same corporate block as the original spammer.

Keith Cress
kcress - http://www.flaminsystems.com

oharag (Mechanical) (OP)
7 Dec 10 13:13
I'm back smile

I got my boards in and they work fine.

Though I have one last question:

I designed in 11 SM resistors for each series chain to drop 1.6 volts. Did I make a mistake?

Why wouldn't I have just one SM resistor at the start of the circuit to drop 1.6 volts and the deliver the remaining 22.4 volts down stream in a series parallel design? I designed the circuit as I see depicted on the web.

Wouldn't it be the same if I just used one resistor and disapated .25 W * 1 device instead of .25W * 22 devices?

Again I am a Mechanical designing LED circuits. I really do not have an Electrical of knowledge on board. Hopefully this will change.
ScottyUK (Electrical)
7 Dec 10 13:30
Your could use one resistor but it would need to dissipate the total wattage of the 22 resistors. You might not be able to use a SMD type. Or maybe you could - depends on the resistor and the wattage.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

oharag (Mechanical) (OP)
7 Dec 10 14:16
@ScottyUK

Because I have to disappate 3.3 A * 1.6 VDC instead of .150 A * 1.6 VDC, right? Now I get it smile

Thanks for the answer.
ScottyUK (Electrical)
7 Dec 10 15:47
One thing to bear in mind is that with a single drop resistor shared among many LEDs you may not get equal sharing of the current because of slight variations between the individual LEDS. Others here know a lot more about LEDs than I do and will probably comment on how big a problem this would be in a practical design. The one-resistor-per-LED solution avoids this possible complication.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

PHovnanian (Electrical)
9 Dec 10 12:18
What ScottyUK said.

The LEDs in each series string will pass equal current, by definition. But between strings, differences in the sums of the LEDs' Vf drops could result in each string drawing more or less current. In most cases, differences between individual LED Vfs will average out within each string.

Another problem to consider is that your current regulating resistor is dropping only 1.6V out of a total string voltage of 24 Volts. The 22.6 V LED drop assumes some nominal operating temperature. As the ambient temp and/or the LED temperatures change, so will their Vf. And since the regulation slope of the resistor (the inverse of its resistance ) is pretty steep, a small change in the strings Vf will result in a large change in current.

Not knowing the device parameters (temp coefficients) and the thermal operating characteristics, I can't say whether this will be important. But this is the sort of stuff that people with advanced degrees in LED strings have to deal with every day.  

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