Electronics Chaser LED's, resisters and all things in between

Hello all.... This is StealthRT and for some reson my other account wont let me post (have no idea) so here goes my question for u all to help me out with...


ok, this is the chaser board (1) that i will be hooking the LED's up to on the light bar. the chaser i am going to use is the Verad Scanner (pic below) that has 7 LED's that go back and forth. Now the LED chaser board is pretty much layed out like i drew it above. On the light bar i will need 60 LED's to make it up. This light bar is for my cars break light in the wing. It has 60 LEDS in it already and i need to just make a new one from scratch (2). Space is very limited in this area so keep that in mind. I will need to attach 8 LEDS to EACH LED on the casher board. (7 x 8 = 56) My question for this mod is : What kind of resisters will i need to get to house enough power to light up 8 LEDS on a single LED on the board? I drew a pic of how i think i would do it (3) but i dont know. The drawing (7 & 8) show how i think it could be hooked up.. using 1 resister to spead out to all 8 LEDs OR useing seprate resisters for each of the 8 LEDS.

The LEDS i will be buying for this mod are as follows:

Red (3mm) | 660 Wavelength (nm) | Water Clear | 900 Iv (mcd)
Green (3mm) | 565 Wavelength (nm) | Water Clear | 150 Iv (mcd)
Yellow (3mm) | 590 Wavelength (nm) | Water Clear | 700 Iv (mcd)
(buying them @ eled.com)




The Green & yellow LEDS are going to be used with the other mod i need help on as well. (doing both at the same time )
The drawing (4) shows the Stereo VU Meter that i will use as a mod for my sound system in my car.

The drawing (5) shows what i will have it showing as... The LEDS will display "R/T Bumpin'". The board is called the Velleman K4305 VU Meter Kit. The R and galf of the "/" is split for left and right chanels. i have figured it up and this is what i come up with:

The "R/T" has a total of 62 LEDs
(devide that by 2 (left & right) that would give u 32)
the "Bumpin'" has a total of 62 LEDs.
(devide that by 2 (left & right) that would give u 32)

green LED 1L/1R (MAX 8 on each side)
green LED 2L/2R (MAX 8 on each side)
green LED 3L/3R (MAX 8 on each side)
green LED 4L/4R (MAX 8 on each side)
green LED 5L/5R (MAX 8 on each side)
green LED 6L/6R (MAX 8 on each side)
green LED 7L/7R (MAX 8 on each side)
yellow LED 8L/8R (MAX 8 on each side)
yellow LED 9L/9R (MAX 8 on each side)
red LED 10L/10R (MAX 8 on each side)

So given that 8 LEDS on the left side will be controled by ONE green LED on the Left VU board. etc etc
and the other 8 LEDS on the right side will be controled by ONE green LED on the Right VU board. The board can be seen @ HiTechMods

My quesion for this one is:
Like with the light bar for my car, what kind of resisters will i need for this mod? given the examples and drawings?



Well thats all i think i have for you all. I hope u understand my questions. Please if you dont, dont hesitat to tell me or ask me to explain it another way.

Thanks for your time,
David

 

Just to check that I understand you correctly... You're going to remove the leds on these circuits and replace them with 8 leds for each one?
If so, you shouldn't need to add a resistor to the led arrays at all, since the circuit already drives leds it has the resistors built in (does that make sense?).
Your only problem is that 8 leds will draw 8 times the current, and the circuit may only supply enough for 1 led. You might find that the leds are too dim. Try wiring those leds up temporarily and see if it's bright enough. If it's not you'll need to make a driver circuit, possibly using an optoisolator, maybe an additional driver transistor, and current limiting resistor off that.

 

Mizon,
misunderstood. I will be taking off the LEDS on the boards.. but soldering wires where the LEDS were. each LED on the board will have to power 8 LEDS. (1 LED = 8 LEDs) As my drawing shows (7 & 8) the first LED will power the 8 LEDS and so on.

David

 

Welcome to the boards, Wildman.

The kit uses LM3914 ics, which will only provide enough current for a single LED per output, and even then, at a low current in bar mode.

To drive 8 you'll need a suitable PNP transistor switch on each output.

Alternatively, you could run 4 or 5 2v leds in series off a 12v supply instead of each single. In this case, the 3914 provides the needed current from a single resistor, so you don't need to worry about values. Check the datasheet.

 

hey cpemma, thanks for the welcome.

Is this the PNP transistor that your talking about that i need?


And could you please show me how i would hook that up to each LED on the board?

Thanks for your time,
David

 

2N3906 would be a typical cheap PNP transistor suitable for this application.

The outputs of the 3914 are active-low, so when they go low, the LED attached lights. Instead of an LED, you connect the base of the transistor. Then when the output goes LOW, the emitter-base jusnction is forward biased, and collector current flows. So your LED array would look like two parallel strings of four LEDs, each string with a current-limiting resistor, connected between the collector and ground. Emitter connects to +12V in this design, that's so you can run those LEDs in series.

I'm too lazy to sketch this for you, but someone else surely can if you need to see a diagram.

 

Thanks Linear for the reply.

Yes, i do need a drawing... i will draw what i think you are talking about and post in a few mins... but from what u told me, i dont know where i need to put the PNP.....


Thanks,
David

 

Something like this would replace each LED:

Code:

               E----------------------------------o +12V
from 3914 o----B   +--/\/\/--|>|--|>|--|>|--|>|---o gnd
               C---+
                   +--/\/\/--|>|--|>|--|>|--|>|---o gnd 

R is gonna be (12 - 0.3 - (4*Vf) ) / 0.02

 

ok Linear,
Do u mean i would connect the PNP to the 3914 chip itself? Would i not connect 8 PNP's to each LED output?:

Code:

  
               
LED 1 -----(+)-->--(12v)-----------------------------o
        ---(-)        +--/\/\/--|>|--|>|--|>|--|>|---o
                      +--/\/\/--|>|--|>|--|>|--|>|---o

LED 2 -----(+)-->--(12v)-----------------------------o
        ---(-)        +--/\/\/--|>|--|>|--|>|--|>|---o
                      +--/\/\/--|>|--|>|--|>|--|>|---o

etc etc...

- Is "/\/\/" a resisters?
- If it doesnt hook up like that, how would i have 56 LED's (7 x 8)total for the project?
- Would i hook up the ground from the LED to the PNP?

Thanks for your time,
David

 

I suspect we may be talking at cross-purposes here. The Velleman kit uses a LM3914, so the led cathodes connect to its outputs and led anodes to +12v, and linears instructions are accurate if you're using that board.

However, if you're using this Verad scanner you'll have to give us some details of the chips on the board. The outputs could well go high rather than low, so a NPN transistor (eg 2N2222A, 2N3904) would be needed, still linked as linear said, also shown on this 4017 chaser circuit

So, if the existing leds are connected between +12v and driver chip, use PNP, if between chip and ground, use NPN.

 

Each output pin (there are 10) on the LM3914 gets one transistor. This just allows it to switch a greater current, i.e. sufficient for your big batch of LEDs. (Note that if three LEDs were enough to get you happy, then you wouldn't need the transistor at all. It's the parallel combination that's making us need to sink 40 mA and needs the transistor to switch that.) So each transistor controls the 8 LEDs attached to it, in response to the signal from the pin on the 3914. I only drew one, but you'd need 10 of what I drew.

/\/\/ is a resistor.

What you are referring to as ground from the LED, that ain't exactly what you think. The 3914 output goes LOW (that means ground, or 0V) to turn on the LED. The other side of the LED (anode) is connected to +12V.

What my transistor does is switch on its load (the stacks of LEDs) when the output pin on the 3914 goes low. The real ground connection is what I have labled gnd. The LEDs are connected between the collector and ground, and that ground is the circuit ground that the 3914 is referenced to.

Now, about yer chaser, like cpemma said, we can't design to it without knowing what is behind there driving the LEDs. But it's one of two cases--either active-high or active-low, and that means you need NPN or PNP transistors, respectively, to switch your stacks of LEDs.

cpemma, I omitted the base resistor in my l33t ASCII diagram, do we need him to use one? The 3914 sinks 8 mA typical (off the top of my head) per output pin, is that kind of base current gonna roast a 2N390x?

 

I think you're ok with no base resistor, in one way it's self-governing, as base current ~ collector current/h_fe in that config. With one, it could be a safety device if the led/resistor is shorted. Bill Bowden doesn't bother on the link I posted.

The 3914 output current is limited to whatever you set with that single resistor, I_led=12.5/R, so the usual 1k2 gives ~10mA max. For a tranny booster you could use 2k - 3k3

(btw, do you have ~ (approx sign) on US k/bs?)

 

I got one of these ~ but I thought it let me refer to $HOME.

 

Alright guys, i'm getting better at getting a picture of what your talking about now. I have some questions though...

1) With the resister, what kind do i need to house 8 LEDS?

2) a max of 10mA? One LED that i am buying is rated at 20mA... Is this where the resister comes to play?

i will draw a scetch of "hopfully" the final outline of these projects to get appoved by u both before i invest in getting the products.

Thanks,
David

 

>1) With the resister, what kind do i need to house 8 LEDS?

Like I said above, R is gonna be (12 - 0.3 - (4*Vf) ) / 0.02

That's the voltage across the resistor (12 - the drop across the trnasistor minus the cumulative drop across the LEDs) divided by the current in that branch. Ohm's Law.

>2) a max of 10mA? One LED that i am buying is rated at 20mA... Is this where the resister comes to play

Not in this case. You are using the voltage from the 3914 output to switch the collectir current in the transistor on/off. It's on when the emitter-base junction is forward biased, which happens when the output of the 3914 goes LOW. That's what you want. Where cpemma and I had our little sidebar is because there will be a current, I_B that flows (out of the base and into the 3914 output) as a result of this, I just was making sure the design I posted didn't make the 3914 sink more current than it's rated for.

At the end of the day, a transistor is what you use here to switch a large current from a smaller one. We're using it like a relay, except that it's smaller, silent, cheaper, and more reliable.

 

All right u 2, here is my scetch, tell me if it's right or wrong.


The mod for the light bar is going to be connect to the VARAD scanner. Here is the pic for that scanner.


If you can not read it, it says:
Varad
IC13
0037 GM5

Doing the math for this resisters i came up with this:
(12 - .3 -(8*vf?)) / .02 = 185?

- So i need a resister thats rated @ 185? (250 range)
- Also, i did not enter anything for the VF in the equation.
- I also noticed that the Varad circut only had 9 outputs (18 total) when you thought it was 20 total.
- The VU meter (20 total) is the LM3514 as u already stated, i also have one from "Kits R Us" that uses the LM3515 chip. Any differnce just in case?
- I will also need to know which "legs" on the chip i need to solder the transisters too:
-The varad has 9 legs on each side.. so which side do i use and which ones do i need. Or do i just have to follow the paths to find that out?
-The LM3514 has 10 legs (as u stated) [the LM3515 has 9 legs though] so the same question applys here as did with the varad.

Thanks again,
David

 

okay, we have to know whether the varad thinger is active-high or active-low. You'll need to look at the side of the LED that isn't connected to the chip, and if it's connected to +V, then that's active-low, otherwise if it's connected to ground, that's active-high.

I have no idea what IC13 is, so you'll have to follow traces to know which legs connect to LEDs. For the 3914, there's a lovely data sheet that will tell you.

As for your sketch, you hooked up all the LEDs in parallel--that's wrong. You want 2 parallel strings of four in series, each string has its own resistor. Your R calculation is hosed as well. 47 ohms would be appropriate for blue LEDs with a Vf = 3.7V @ 20 mA.

Check the data sheet for the transistor for the pinout, EBC is the order from left to right if you're looking at the flat side of the package.

As for counting legs, I sure hope your 3914 has 18 legs, not 10 or 9.

 

Lets get the easy bits out of the way first.

Actually you used V_f=1 in the equation. Red, yellow and plain green LEDS take about 2.1v each, so you can't run 8 in series off 12v. That's why 2 strings of 4 in series have been suggested.

So (12-0.3-(4*2.1))/.02 = 165, nearest resistor is 180R.

(Your drawings show the leds in parallel and no resistor, go back and look at linears picture. )

The 3914 is a linear output scale (for voltmeters), the 3915 a log scale (for vu-meters).

Check the datasheet.

Looks like Varad have either relabelled a chip, or quite possibly had one made to their needs. Not that expensive if you want a few thousand.

So you need to trace where all the leds are connected.

What does this "scanner" do?

(edit: Sorry, linear, we've covered most of the same ground but I'm the slower typist )

 

The scanner is a knight rider-esque sort of toy.

 

Ok, heres another re-drawing with the added fetures:


As for the Varad, i traced the wire to an R4 on the pic and then it goes under the chip itself... without a trace coming out the other side or anything...




The /\/\/ will be 180 as cpemma requested.

David