What do I need to run 200 LEDs on 110V

[NoTime]

I thought Kevin had decided to go with a 12 V DC supply, in which case none of that matters. This stuff about peak currents and duty cycles is just a side-discussion between Phrak and NoTime, not relevant to what Kevin actually plans to build.

Please correct me if that is wrong.

You are correct, it is not particularly relevant to Kevin's current plans.

 

[Redbelly98]

Mind you, I have nothing against side-discussions and I think they're partly why Physics Forums is such a fantastic place. But I did want to clarify that for Kevin's sake.

At any rate, the main choices now seem to be either:
1) build a 12V DC LED string, or
2) find something pre-made if available, for less money and takes less time than making one's own.

 

[kevin071572]

Try looking at Page 8.
There is enough information there to work this out.

Page 8? Looks like a flower petal with numbers around it.

I thought Kevin had decided to go with a 12 V DC supply, in which case none of that matters.

Redbelly- Thanks for keeping that clear, I was starting to wonder if that was all just to run on DC or not.


Like I said earlier, I knew this was the place to figure out this little wiring question. I just didn't know it would spark such a "discussion"

I now know that I can string 200 LEDs together and I know that I can run them on a 12 VDC power supply, which will make building it soo much easier.

 

[Redbelly98]

Page 8? Looks like a flower petal with numbers around it.

It's the page that says "8" on it, which is p. 9 in the .pdf file numbering.

 

[Phrak]

Try looking at Page 8.
There is enough information there to work this out.

If you can't tell me where you got a count of 44 from, then I don't see any point to continuing this conversation.

I'm sorry to have frustrated you NoTime. I think it has become rather pointless, but from page 8, I assume I can't rely on the Duty Ratio plot, and use the max rated current of 30 mA instead.

$$\frac{(1.414)120V - 1.4V}{44} = 3.82V$$

At 3.8V per diode the forward current is about 30 mA.

 

[NoTime]

Mind you, I have nothing against side-discussions and I think they're partly why Physics Forums is such a fantastic place. But I did want to clarify that for Kevin's sake.

Side discussions are fine by me.
Originally, I just wanted to clear up some ambiguities with Phrak's schematic for the benefit of other people that might be reading this.
LED's are a fairly popular topic.

Good point about the clarification.
It's a good thing you mentioned this

 

[NoTime]

Page 8? Looks like a flower petal with numbers around it.

The flower petal is a color map.
Here is one painted in with the colors.
http://www.google.com/imgres?imgurl...n&sa=X&oi=image_result&resnum=3&ct=image&cd=1

It's the next page, I was referring to.

Redbelly- Thanks for keeping that clear, I was starting to wonder if that was all just to run on DC or not.

Only the part about dynamic resistance is important to you.
You can use 12v, 3.6v and 30ma in the calculator link.
Your project will work, but the final measured current will be around 10ma rather than the expected 30ma.

Have fun

 

[NoTime]

I'm sorry to have frustrated you NoTime. I think it has become rather pointless, but from page 8, I assume I can't rely on the Duty Ratio plot, and use the max rated current of 30 mA instead.

$$\frac{(1.414)120V - 1.4V}{44} = 3.82V$$

At 3.8V per diode the forward current is about 30 mA.

That explains a lot!
Nice touch with the bridge diode drop, but I will note that the nominal line voltage, in the US, is 118v rather than 120v. It also varies some.

I don't think this side discussion is pointless.
Since you showed your work it isn't frustrating either

Quiz questions:
Compute the RMS current for the 100ma 10% duty cycle square wave given in the data sheet.

Compute the duty cycle for your solution (120hz).
Assume a cutoff voltage of (where no current flows) of 3.819v.

Compute the duty cycle of the portion of a 60hz sine wave that exceeds the 120v RMS value.
Assume only the positive half of the full cycle conducts (no bridge or switch diode).

 

[NoTime]

You can use 12v, 3.6v and 30ma in the calculator link.
Your project will work, but the final measured current will be around 10ma rather than the expected 30ma.

I made a mistake here.
Use 20ma in the calculator link.

This will work out quite closely to 20ma (about 17ma) using the 68 ohm resistor the calculator will show.
The reason it works is because the 3.6v is Vf at 20ma.
The calculator shows standard resistor values instead of exact values, or why the final value is not exactly 20ma.


My mistake was to use the 3.6v (Vf @ 20ma) instead of V_threshold which is about 2.9v.
If you use the 47 ohm resistor then the final current will be around 25ma, somewhat distinct from 10ma.

 

[Phrak]

Talk to me as if I have 23 odd years in electrical circuit development or design in power conversion, instrumentation, controls, fixturing, and fixing other peoples design errors, neither pushing papers nor peripheral support, interspersed with firmware coding and software engineering in test, engineering tools, and simulators, with exerience in primary and root cause failure analysis.

 

[NoTime]

Talk to me as if I have 23 odd years in electrical circuit development or design in power conversion, instrumentation, controls, fixturing, and fixing other peoples design errors, neither pushing papers nor peripheral support, interspersed with firmware coding and software engineering in test, engineering tools, and simulators, with exerience in primary and root cause failure analysis.

Fair enough.
Part of what I said to you made me realize what I had forgotten.
So I give you credit for fixing my mistake.

Your number of 44 also solves for a 20ma RMS current, except you got the count right.

One of the good things about PF is that if you do get things wrong there are people to set you on the right path,

Thanks.

 

[Phrak]

Fair enough.
Part of what I said to you made me realize what I had forgotten.
So I give you credit for fixing my mistake.

Your number of 44 also solves for a 20ma RMS current, except you got the count right.

One of the good things about PF is that if you do get things wrong there are people to set you on the right path,

Thanks.

You're a better man than me, NoTime.

I've been too under the weather to respond lately. I may still be too dopey.

We don't really see this exexcise as the same problem. In my view, the objective is to obtain the most luminosity with the least component count and cost, without thermally degrading the components. Thermal degregation is everything. This means keeping the peak current within bounds while delivering the highest possible average current. These things are diodes, and these sort of curcits the dynamic impedence is not a useful number given the wide swing in bias voltage. The current through one of these LEDS varies roughly as the exponent of the bias voltage, I'm sure you know. Roughly,

$$I = exp(kE)-1$$

$$P = c(E exp(kE) - 1)$$

So as the voltage increases from zero the current creeps slowly up with a ugly shape peak in as the AC peaks. The idea in any design is to keep the peak instantanious power down. But it's a little more than that. Thermal degragation is usually about wire bond integrity and more importantly mirgration of dopants at the junction. In this case you might add degregation of the fluorescent dyes. Beats me. The curves you see in the data sheet only hint at the thermal limits. You might ask yourself why these boundries in their Allowable Forward Current area plots are drawn where they are. There are six different boundries going on; I don't know where they all come from.

You know about thermal circuits right? In any case this is all for general consuption anyway. The heat driven into the die is the instantanious resistive power lost integrated over time. The heat conducted out depends on the bulk thermal resistance as the heat is conducted away via leads and plastic case. So in the simple situation of single pulse power, for instance, the instantanious temperature of the die ramps up like voltage across a capacitor in parallel with a resistor, being charged with a current source. To make a long story short, after approximating the repetative current duty cycle if you know the normalized thermal frequency response curve you obtain the heating obtained as if it were heated by a constant DC bias.

As there is no frequency response curve provided by the manufacturer, we can only guess based upon where the boundries show up in the Allowable Forward Current plots. The thermal response curves are based upon duty cycle and frequency. As Nichea Corp was not gracious enough to let us know the test frequency used in obtaining their Duty Ratio vs. Allowable Forward Current plot, we can just guess or defer to experimentation. [Edit: If it's 10KHz or better which is pretty typical for strobed character/segment multiplexing, then it's useless. If it's 1000 Hz, which should be more typical for scrolling displays that use descete diode assemblies it could be somewhat usefull.] The latter (experimentation) is what I recommend. The 44 diode count solution I had is pretty dismal: assuredly within max allowable current spec, but probably about as bright as running on 10mA DC bias.