Serious Newby

[SGluedMyFingers]

I am trying to gain general knowledge in electronics for lighting models.  I have tried to research as much as possible, but I have some questions:

It is my understanding that I can use different types of LEDs if the different types are kept in series, with their own resistor and the forward voltage of each series doesn't exceed the source voltage [right?].  I'm assuming that resistors control both voltage and amperage?

O.K.  If I put a blinking LED in one of the series, when it's not lit, does the forward voltage go up/down in the other series?   In other words, will it blow out other LEDs if the resistor in the series does not account for this change in voltage?  Again, I'm assuming that I will have to account voltage increase when figuring the resistor for each series [right?]

Alright then, last question (for now):  Let's say that I want to include an inverter and CCFL in.  Can this just be figured into the parallel array just like an LED series?  Is it THAT simple?

God, I hope I'm making sense.  I may have this all wrong. 

THNX in advance for any/all help.

Cliff

[SurJector]

I don't know anything about blinking LEDs, so that what I say here do not apply to them.

When you run several strings of whatever in parallel, they all get the same voltage. If your source is beefy enough for your application, you'll get what is written on the source. Suppose you have a 12V (regulated) power supply. Each blue/green/white LED needs 3V, each yellow/amber/red 1.8V. If you put 2 blues, one yellow and one red in series, you get 9.6V across the LEDs and you complete your string with a 120ohm resistor; the resistor will get the remaining 2.4V and since current-through-resistor times value-of-resistor equals voltage-across-resistor, you get 20mA through the resistor. If you have a simple series chain everybody in the chain gets the same intensity, i.e. 20mA. The physics is slightly more complex, but that's about right.

You can put as many thing as you want in parallel upto the limit of your power supply. You can put 200 strings (4A) if your supply can handle 48W. You can add a CCFL and whatever as long as you don't go over your power supply rating (you should obviously remain 10 to 20% below for your power supply safety because manufacturers like to overspec their goods). You can have some problem if you turn off one string that was consuming a lot of energy, because the power supply will regulate the voltage with a slight delay during which the voltage will be above what is expected. Here "a lot of energy" depends entirely on your specific power supply (if you turn on a string that consumes a lot of energy, that's not as important because you'll lose for a fraction of second some energy in the others, but that's usually no big deal; think of the car lights dimming when you start the engine).

[Rob]

Welcome to the boards.

It is my understanding that I can use different types of LEDs if the different types are kept in series, with their own resistor and the forward voltage of each series doesn't exceed the source voltage [right?].  I'm assuming that resistors control both voltage and amperage?

Quick rundown on series and parallel:
series elements all carry the same current. The forward voltages add.
parallel elements all have the same voltage across them. The currents add.

Your source voltage effectively limits the size of how many LEDs you can place in series. The cumulative forward voltage can't exceed your source voltage.

Okay, quick rundown on mixing types:
if your different type LEDs have the same current rating, they can be freely mixed in series. You want just one current-limiting resistor for the series combination.

series example:
2 red (2.1V @ 20mA) and one white (3.3V @ 20 mA) LED in series with a 12V supply:
cumulative Vf = 2.1 + 2.1 + 3.3 = 7.5V
R = (12 - 7.5)V / 0.02A = 225 ohms

The resistor limits the current to 20 mA for the whole series string.

Like in SurJector's explanation, you can add other loads in parallel. The limiting factor there is the current rating of your power supply. If your power supply was rated for 12V @ 500 mA, you could do 25 of the above strings in parallel. Or you could run 10 plus a 300mA CCFL.

[Rob]

Now, on the blinkers--

if you want a series string of LEDs to blink, you only need one flasher in the circuit. Every LED in series with it will blink synchronously.

[SGluedMyFingers]

O.K. I think I understood most of that.  Combined forward voltage of any series cannot exceed source voltage and must run at same current.  Combined currrent of each series must not exceed source current rating.  A resistor only controls/absorbs current in the series - not voltage, right?

That brings me to another question (please bear with me, I'm stupid).  I have a CCFL power unit with a 375mA current draw and a power supply rated at 350 mA.  That's what they sent me and it works (I would direct this question to them, but they closed).  But isn't the power unit exceeding the power supply's current rating?  Do I have something mixed up? What happens if the current of the array exceeds the rating of the power supply?  Does it all fry or does it just not work?

The power unit says it has a 180 volt output - that's to the bulb, right?

I know, I know... more questions.  I don't want the LEDs to blink simultaneously, so I'll run them (about three) on three seperate series.  If I understood the above, that shouldn't effect the other series' voltage but may add current draw to the array.  Right?

Now (last one), let's say I'm using batteries for the power source.  The more parallel series that I have, the faster the battery wears down, right? 

Thanks guys.  I know this is elementary stuff...  it's just complicated.

[SurJector]

O.K. I think I understood most of that.  Combined forward voltage of any series cannot exceed source voltage and must run at same current.  Combined currrent of each series must not exceed source current rating.  A resistor only controls/absorbs current in the series - not voltage, right?

Yes and no. The LEDs (and all diodes for that matter) have a roughly constant tension between their poles, but should not be run with more current than their specified rating (lest you want to fry them). The resistor (R) has a highly variable tension (U) between its poles and the current (I) that goes through it can vary but there is the relation U=R.I (and the power P=U.I=R.I.I=U.U/R is limited, lest you want to fry it). The resistor does not control or absorb anything, but if you have a string:
Power_supply_positive-Red_LED-Blue_LED-Red_LED-Blue_LED-resistor-Power_supply_negative
such that the Power supply is regulated to 12V and the LEDs absorb 9.6V (3V for the blue and 1.8V for the red), then the resistor will get 2.4V between its poles, so that the current through the string will be 2.4/R. If you take R=120ohm, you get I=20mA.

I have a CCFL power unit with a 375mA current draw and a power supply rated at 350 mA.  That's what they sent me and it works (I would direct this question to them, but they closed).  But isn't the power unit exceeding the power supply's current rating?  Do I have something mixed up? What happens if the current of the array exceeds the rating of the power supply?  Does it all fry or does it just not work?

If the current exceeds the rating of the PS, it can fry (literally, it becomes too hot). Usually that's no problem for the equipment because at a certain point the current stops flowing. It can be a problem for the house though: it can catch fire (remember: it gets too hot).

The power unit says it has a 180 volt output - that's to the bulb, right?

You can think about current like water in a tube. The tension/voltage is the difference of height between two points: a 24V power supply is a pump that takes water at a certain level and pumps it 24m higher. The current is the quantity of water that goes through a tube. The LED would be a half tube with fixed length and angle, so that the height between its top and bottom part is fixed (3m for a blue, 1.8 for a red). The resistor is a tube of fixed length, but you can angle it as you wish, if the difference in height between both ends is high, the water flows faster, if the height is lower the water flows slower (that's only approximative). Inside a string, the water flows at a fixed rate (i.e. the intensity is constant).
You can add several tubes in parallel, that will not change the height of the pump (the tension is the same), but it will add to the total flow (the currents add).

To come back to your question, this should mean that there is 180V between the two poles of your power supply.

I don't want the LEDs to blink simultaneously, so I'll run them (about three) on three seperate series.  If I understood the above, that shouldn't effect the other series' voltage but may add current draw to the array.  Right?

Right. But you probably need to put a resistor in each string to have a low enough current inside.

Now, let's say I'm using batteries for the power source.  The more parallel series that I have, the faster the battery wears down, right? 

Yes (unfortunately, that'd be nice if the batteries did not wear out).

[Rob]

Let me take a swing at this--Just FYI, SurJEctor is outside the US, so while we are saying the same thing, his terminology is a little different at times.

O.K. I think I understood most of that.  Combined forward voltage of any series cannot exceed source voltage and must run at same current.  Combined currrent of each series must not exceed source current rating.  A resistor only controls/absorbs current in the series - not voltage, right?

You were doing good till the last sentence. The resistor is both limiting current, and dropping voltage. What my comments above wer intended to convey is that it's more sueful to think about an LED as a current-driven device (apply 20 mA current and it lights) versus a voltage driven device (apply 3.3V and it lights up). So when you select your resistor, you select it to limit the current to the LED. The implicit assumption is that your supply looks like a voltage source. Most of the time, that's okay.

That brings me to another question (please bear with me, I'm stupid).  I have a CCFL power unit with a 375mA current draw and a power supply rated at 350 mA.  That's what they sent me and it works (I would direct this question to them, but they closed).  But isn't the power unit exceeding the power supply's current rating?  Do I have something mixed up? What happens if the current of the array exceeds the rating of the power supply?  Does it all fry or does it just not work?

It's tough to exactly specify what goes on if you connect a load in excess of the nameplate rating on any power supply. Partly because there are a lot of different power supply designs. What SurJector described is definitely worst case (house goes up in flames).   

A best case would be that the manufacturer was conservative when they developed the rating, and your light will work okay. You are only at (375/350 = 1.07) 7% over the rated load current. Outcomes in between may look like the power supply losing its ability to keep the voltage steady, or getting hot and failing early.

I would feel better if your lighting rig used a power supply rated closer to 500 mA.

The power unit says it has a 180 volt output - that's to the bulb, right?

Right. Those CCFLs take a fairly large AC voltage to light.

I know, I know... more questions.  I don't want the LEDs to blink simultaneously, so I'll run them (about three) on three seperate series.  If I understood the above, that shouldn't effect the other series' voltage but may add current draw to the array.  Right?

Right.

Now (last one), let's say I'm using batteries for the power source.  The more parallel series that I have, the faster the battery wears down, right? 

Right. In general, the load current goes up, battery life goes down. Batteries have a capacity rating in mAh (milliamp-hours) that indicates to some extent how much load current they can provide for how long.

Thanks guys.  I know this is elementary stuff...  it's just complicated.

[SGluedMyFingers]

You guys have no idea how much this is helping.

O.K., now it's getting clear.  The problem with the CCFL unit is that the POWER SUPPLY will run too hot because more current is going through it than it can handle.  In this case,   the danger is not to the CCFL POWER UNIT .  If anything, the power unit is not getting enough current - resulting in dimmer light? or is light brightness more related to voltage?

I think that the power supply is likely just adequate to run the CCFL - it was recommended by the manufacturer of the CCFL power  unit.  But, I wouldn't want to add anything to this without getting another power supply.

So, look at this:

-------------- (+) (power supply 12v rated @ 500mA) (-) -----------------------------------------------------------
l                                                                                                                                                              l
l-------------  (+) (CCFL 12v @375 mA) (-) ----------------------------------------------------------------------------l
l                                                                                                                                                              l
l-------------(+) (blinking LED 3.2v @20mA) (-) ----------1/2W 470 Ohms----------------------------------------l
l                                                                                                                                                              l
l-------------- (+) (blinking LED 3.2v @30mA) (-)----------1/2W 330 Ohms---------------------------------------l
l                                                                                                                                                              l
l---(+) (LED 2.3v @20 mA)  (LED 1.9v @20 mA) (LED 3.3v @20 mA) (-)---1/4W 270 Ohms-----------------l
l                                                                                                                                                              l
l------------------------------------------------------------------------------------------------------------------------------l

If I'm not lost, that should work fine (although the resistors should probably precede the LEDs in each series).  55 mA under power supply rating.

A friend with limited electronics experience said that I would have problems with this array because the current would follow "the path of least resistance" (in this case, the series with the 375 mA CCFL and the 30mA blinking LED) and the later series (20 mA blinking LED and 3 LED series @ 20mA) would not get enough current.  Is there validity to this claim?  I mean it can't be this easy can it?
 
THANKS GUYS!  I know you've already put a lot into this thread.  Sorry I'm a little thick.  Wait 'till I start wanting to play with ICs

[SurJector]

If I'm not lost, that should work fine.  55 mA under power supply rating.

It should.

A friend with limited electronics experience said that I would have problems with this array because the current would follow "the path of least resistance" (in this case, the series with the 375 mA CCFL and the 30mA blinking LED) and the later series (20 mA blinking LED and 3 LED series @ 20mA) would not get enough current.  Is there validity to this claim?

Your friend is wrong. If you wish to express it that way, most of the current will follow the "path of least resistance" i.e. 375mA will flow through the CCFL. Some will flow through each LED string exactly as you wish.

I mean it can't be this easy can it?

Yes it can ! You got it right !

[Rob]

To keep with the waterpipe analogy that SurJector introduced above, you can look at those resistors as flow control valves. You adjust the valves (select different resistances) to adjust the amount of water (current) flowing in each pipe (parallel circuit branch). The total water flowing can't exceed the pump's output (power supply current rating) wihtout losing pressure (drop in voltage at the power supply).

As for the resistors placement, they limit current exactly the same whether they are before or after the LED. Just like the valve in the pipe.

[SGluedMyFingers]

SurJector, Rob, I do appreciate it.  That was a whole lot faster than Trade School.

Cheaper too.

I will be back