Most practical way to drive 4 strings of 20 LEDs (600mA a string at 3.5V)?

Atom you do know the AMC34063 is the same thing as MC34063A just AMC34063 they get for penny's and stick them in most all car chargers. There a bunch of part numbers for these chips.

Yeah i assumed that but those car chargers seem to break down quick so i wouldnt recommend it. All my friends have cars and they tell me about there chargers dont work and can you fix it and stuff heh.. i tell them go buy a new one.

I think you missed the point these are in them to MC34063A this is the same chip AMC34063. And I have never found a bad chip most are blown fuse or broke wire or a weak spring. I get all of them I can find just for that chip. and the inductor. Most free and some in a bag full of old chargers.

Be a good chip for Triode he can boost 3.5 volts up to 12 volts if he wants and run it till is dead. But I wouldn't no that to LiPo Cells

You can get batteries for free from old phones at the "drop off" ponts in most phone stores.
Use two 600mAHr cells in SERIES. This will give you at least 7v2
Or 2 x 2 = 4.8Whr

There's always plenty of ways to do something. For the batteries, I think AA would be a good size and shape, or AAA, the good news is I can probably fit a lot of them in the space provided, I got a look at the costume she's making and the strands are going to go in some butterfly wings, the support tubes that house the LEDs are 3/4" inside diameter, I can't put batteries at the ends, but if I load them up in the bases of the tubes I could probably put like 12 AA's in there.

I think I will try using some of the power conversion chips Be80be mentioned, I've wanted to learn more about those anyway even if it doesn't work out for this project.

I'm thinking I misread this datasheet the first time though
"
The ULN2801A-ULN2805Aeach contain eight darlington transistors with common emitters and integral suppression diodes for inductive loads. Each
darlington features a peak load current rating of
600mA (500mA continuous) and can withstand at
least50V in the offstate.Outputsmay be paralleled
for higher current capability"

Based on that doesn't it sound like I could just use a parallel set of 20 LED's on it? 30mA @ 3.4V * 20 in parallel = 600mA @ 3.4V. Am I wrong, or could I just parallel the outputs of one chip by twos to get 4 outputs that handle 1A each, and power a string off of each pair?

LED circuits are much more efficient when the LEDs are in series.

So then you also recommend a boost circuit? It does seem like for these applications these are the way to go, and Be80be and Atomsoft know their stuff, so I'll see what a MC34063 can do.

So using the site Atomsoft posted (**broken link removed**)

If I split the strings into two sets of 10 each, which it seems I could easily do, the output voltage would only be 35, I could use 6 of the 1.3 volt batteries that fit in the tube in series for 7.8V for that the calculator suggests these values:
Ct=321 pF
Ipk=405 mA
Rsc=0.741 Ohm
Lmin=135 uH
Co=29 uF
R=180 Ohm
R1=1k R2=27k (35V)

I'm not sure how using the closest real values will effect this though, converting to nearest real values it would be

Ct=330 pF
Rsc=1 Ohm
Lmin=150 uH (will one of these weird little ones work?: https://www.taydaelectronics.com/servlet/the-468/150uH-Inductor/Detail)
Co=47 uF (not very close to 29, should I combine to get closer?)
R=220 Ohm
R1=1k R2=33k

While were on components, am I right in thinking that Ct is ceramic and Co should be electrolytic? And will a 1N4001 work for the diode shown at right?

So, what if I use one converter circuit to get a 35 volt output, use a ULN2803 after that (rated at 50V 500mA per gate), with each gate controlling a string of 10 LEDs in series, put the 7.8 volts from 6 AA size batteries through a 5V regulator to run the microcontroller, and connect the 8 control pins of the ULN2803 to 8 outputs on the microcontroller. I'm going to keep rewriting my plan till it's good.

the INDUCTOR must be able to handle the output current.. i still dont understand 2 sets of 10... is it like

Set 1:
ANODE ---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o CATHODE

Set 2:
ANODE ---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o---|>|---o CATHODE

? if so dont you simply add the dropout voltage ? What leds are you using most have a 1.5v dropout i think. which is minimum to turn on . 1.5v * 10 = 15v @ 20mA each... so 15v @ 40mA max
As for other values remember most passive components come as +/- 5% or something so you should be safe with closest value. If in doubt try to simulate it in proteus... thats what i do

Yes, the way you drew it is what I have in mind. It's just because if I put all 20 in series it would be 70V which the transistors I have can't handle. If I just use the minimum voltage wont the LEDs be really dim?

It is the current that controls led brightness. The voltage is minimum. I can place a LED on a 125v line with a good enough resistor it will be ok

Have a look at this and see if it would work

Where is master Lol if there was one I have not seen it number one I went with there figures of 1.5 witch is not right I new that it would take a supply at 15 volts to light the leds right

I think some one need to give a Led 101 here if a led forward drop of a red led is about 1.85 volts now if you string them like I showed you will be dropping about 1.85 a led

Let's do the math here 8 leds x 1.85 = 14.85 volts will light all the leds

8 leds 198 Milliwatts x 10 = 80 leds 1980 Milliwatts = 134 Milliamps my way

You are doing 2 things here...
if its 1.85v each red led then there are 10 LEDs in a row. So its 18.5v needed. Since its series the current stays the same as 1 led. So 18.5v @ 20mA will light up 10 Red LEDs with a 1.85v drop.

If its in parallel then its 1.85v needed to power the 10 LEDs but with 10 x 20mA - 200mA

For 10 in Series its : 18.5v @ 20mA
For 10 in Parallel its: 1.85v @ 200mA

P = Watts

P = I * V

For Series : .02 x 18.5 = 0.37 Watts
For Parallel: .2 x 1.85v = 0.37 Watts

370 milliwatts ... the same

Here 8 in Series the big DVM is reading the mA the yellow the volts. There is basically no current limiting resistor "10 ohms "

Have a look I would of done it with 10 but my power supply is only 15.5 volts at 3 amps.

Make you 9 more and that's 80 leds 1980 Milliwatts = 134 Milliamps my way have at it.

Atom led's in series don't work as simple as adding 20x10 = 200 here is why the inner resistance will limits the current.

Thats funny tho... If your DVM says 14.85v and its 13.4mA then the resistance should be more like 1.1K Im sure resistance for a led is different for different suppliers

Read my article on connecting LEDs

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The previous poster no idea of how to connected LEDs and is giving the wrong idea to beginners, like the original poster.

Ok the way i see it if you supply exact voltage than no resistance is needed...

If you supply 15v for 8 leds which use 1.85v each... (14.8v) thats only 2mV left so to calc that 20mA / 20mV = .02 / .002 = 10

So 10 OHMs in deed

So if your using 10 LEDs you need minimum of 18.5v if you supply 20v then you have 1.5v left and need a 75 Ohm resistor on each strip.

Dang from the page collins posted

Solution 0: 8 x 10 array uses 80 LEDs exactly
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms
+----|>|----|>|----|>|----|>|----|>|----|>|----|>|----|>|---/\/\/----+ R = 10 ohms

The wizard says: In solution 0:
each 10 ohm resistor dissipates 4 mW
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 40 mW
together, the diodes dissipate 2960 mW
total power dissipated by the array is 3000 mW
the array draws current of 200 mA from the source.

Click to expand...

And Collins said have a look at his page better remove that calculator it has

absolutely no idea how to connect LEDs in series/parallel

Click to expand...

This is even funny a Man talks about me and has my led cube in his news letter http://www.talkingelectronics.com/FrameworkPages/TE-front.html