Bulbs leds in series on electric bike

[laurel]

Hello, on my electric bike, I have a 36V battery that I want to use for lighting in 12V. To preserve the autonomy, I thought to mount 1 bulb + 2 leds in series, but I am told that the intensity will be determined by that of the least powerful component.
So my light bulb will only have the intensity of a led? So I am forced to use 3 bulbs of the same intensity?
Is there not a way to get around the problem?
Thank you in advance.

[Edit] completed the title [/ modif christine]

[Other]

Hello,
It sells small regulators (like a transistor with 3 paws they are polivalent in voltage it just leaves the 12 volts
Under 1 ampere,
Normally there is no need for any component on the + 36 volts, a wire on the - and the middle leg it comes out 12 volt or 7,5 or 14 volts depending on the model it does not cost a few euros
The piece must be installed on an aluminum sheet to absorb heat. There is a slight loss of energy.
Inform you in electronic parts shops I do not know if the No correspond with our
UA 7812
79M12 (TO-220 box)

Andre

[elephant]

Oulà! Attention to these poor diodes!



In direct current:

A diode blocks everything when it is plugged in reverse
it drives when it is plugged in the right direction and that the potential difference at its terminals exceeds a certain threshold (0,775 volts for the red led's, and I think it's much more for the white ones)
It is therefore possible to imagine operating 15 led's red in series on 12 volts, but that is not all: in each diode can only circulate at maximum the current mentioned in its sheet of characteristics, it is therefore necessary to put a resistor in series, Otherwise the diode will break. We must apply our good old ohm law:

I = E / R and its variants: E = RI, R = E / I

Taking of course the residual voltage: if the diodes in series cause 6 volts to lose, you have to make your calculations taking E = 6

Do not forget to adapt the maximum power of the resistor: if a current of 50 mA passes through the diodes, it must be able to dissipate, under 6 residual volts 300 mW, under 10 V residual: 500 mW Remarks that, with a single diode you spend unnecessarily the energy in the resistance)

Tell me about the voltage at the terminals of the white led's, I do not know it (and of course on the maximum current, which varies from one type to another)

this is valid when you use "pure" diodes, by those designed to be supplied directly under 12 volts which already include the resistance

[delnoram]

Why not use only leds, in series with each other or with a resistance?

Regulators we find this under ref. 7812 or MC7812TC for MOTOROLA

[laurel]

You mean that by simply inserting an electronic component into 0,70 € after my switch, I will convert my 36V to 12V?
This would solve the problems of serialization and I could put what I want in derivation?
It sounds so interesting to me that I can not believe it.
On the other hand, I guess if I use 3 4W bulbs with this system, I will consume 1 amp, while in series it would be 0,33 amp, right?
Another thing: according to this diagram I just found, it seems that the wiring is different from that indicated by André: the 12V output is not in the middle.
Thanks for your advices ; Do not hesitate to continue.

[elephant]

Actually, if my memories are good, when you face the marking of the 7812

on the left: entrance of +
Medium: to the ground or the midpoint of a potentiometer to make the voltage variable
right: exit +

Tape 7812 data sheet on google

Output point mid c 'and for series LM

It would still have to check if the 7812 accepts 36 V at the input, I'm not sure at all!

[delnoram]

If it can help you.

[Other]

Hello,
There are many models

The 317T 1,5 amperes 37 maximum adjustable volt
The 723 150 milliamperes 40 volts adjustable from 2 to 37 volts
The 7818 (276-17771) 1,5 amperes 40 volts supply ..
Shop you will find there are plenty of models ..
The simplest is 7812 if your lamps 12 volts
But leeds normally it works lower
You can also look for a 7805 = 5volts ect .. 7 volts of 8 volts of 9 volts
There are losses with regulators, less than series resistors
Most small devices that run on a voltage range from 6 to 28 volts is what it uses.
Certain lamp has leed in the industry are made with a series of leed and it walks directly on the 110 volts AC there is included a resistance is a diode, (a leed which pete finished, more lamps)

Andre

[elephant]

Thank you, delnoram, for taking the time to look.

That's what I feared, the 7812 and 7805 do not support more than 30 volts at the input.
when the LM317, (it is him whose output is in the middle and the reference point on the right), 37 volts, it's a fair bit, but nothing prevents to drop the voltage by placing 3 or 4 diodes ordinary series (1N4007) before the LM 317.

Having said that, I remind you that the excess energy has to be dissipated, whether in a resistor or a regulator.

if the construction of the battery pack allows it, why not make a "plug" requiring only n elements (n calculated according to the number of led's), the efficiency would be greatly increased because the dissipation in the resistor or the regulator would be diminished

[laurel]

Well, here is a living subject!
As I am a novice in both electricity and electronics, I am afraid I will not be able to implement your proposals.
The proof of my incapacity: I wanted to test a led on which I soldered a resistance supposed to make it work under 12V: not working. I may have burned it with the heat of the iron, or shorted it.
And then finally, I think I'm going to be forced to put three energy-consuming light bulbs; In fact, I did a comparative test of lighting halogen bulbs (20W) and leds (36 leds supposedly equivalent to 15W) last night.
The difference is enormous, both in terms of (predictable) color and density; I felt that I would need 3 LED bulbs to match a halogen, and still ...
A combination 2x36 leds + 1 halogen in series would certainly be effective, but again, I suppose that the intensity of the leds (3,5W absorbed) will condition that of the halogen?