LED indicators - what are load boxes for?

[sheepshearer]

:scratch:if you're replacing bulbs with LEDs, is a "load box" essential? exactly what does it do? why can't the LED light manufacturers make the LED replacements look (electrically) like a bulb then?

[Kato]

LED uses much less power than a conventional bulb therefore if you fit them without resisters (load box) they will flash way to fast and fail at MOT time.

[sheepshearer]

OK - so it's just limiting the current supplied by the flasher relay.

can you get relays that are rated to drive LEDs only?

[LVC]

OK - so it's just limiting the current supplied by the flasher relay.

can you get relays that are rated to drive LEDs only?

Yes, be aware that sometimes a LED relay will work instead of adding a load resisitor, sometimes one load resistor per side (LHS pair and RHS pair) is needed and occasionally one resistor per indicator :eyepopping:

On my S4 I tried all options and the one that was the easiest and cheapest was one resistor per side pair, located under the seat wired into the rear indicators. With 4 LED indicators fitted. For some reason (I'm not an eleccy techy) the LED rated relay didin't help much :dizzy:

[sheepshearer]

be aware that sometimes a LED relay will work instead of adding a load resisitor

yeah - that was what i was hoping.

kinda tempted by these (http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&item=320351504930) but i'm too skint really (especially to get matching fronts)

cheers LVC and Kato-san

[LVC]

Nice - I recall seeing them fitted before to a Monster - works well with the tail chop.

[sheepshearer]

yeah - couple of MOTMs back - would be a couple of hundred Euros new for back and front

[Gilps]

I fitted Rizoma leds to the rear a couple of years back and needed to fit a resistor to each one. I've just fitted Rizoma leds to the front and ordered more resistors. I thought I would just give it a try without the front resistors and they work fine so I guess the resistors in the rear are doing enough to damp down the current. I've still got the spare new resistors if you are interested. How does £10 posted for the pair sound?

[sheepshearer]

thanks for the offer - i'll bear it in mind.

it's just a 25W 6.8 ohm resistor AFAICS - some people are asking £15 so a tenner a pair is good. but you can get an LED rated relay for £12 and that's a cleaner solution.

further up the learning curve than i was this morning :)

[uksurfer]

i tried everything on my M600, resistors and relays, nothing worked, tried and tried again.
i was very disapointed with the stuff i bought, i also have 2 resistors and a led flasher relay in the garage, i cheated in the end!
i wired two bulbs in series across the front indicators, that did the trick and cost me nothing extra,
after spending ?28 in bits which dont work
________
Weed Vaporizer (http://weedvaporizers.org/)

[sheepshearer]

here's the science bit...

the way the flasher relay works is that it has a bimetalic strip that's sprung into a normally on position. when you bung on your indy, the bulb lights and the strip heats up, curls and disconnects. the lamp goes out. the strip then cools and flicks back to on again and the lamp lights, the strip heats up again...ad infinitum.

so... when indies go too fast it's because there's too much current going through that strip and it heats up quicker and disconnects sooner.

LEDs impose a very light load on the circuit compared to a bulb, so the current is high (like a short circuit would be) so we have to compensate with a resistor. but what value?

remember P=VI (power = volts*amps) and V=IR (volts = amps*resistance) then P = V*V/R and R = V*V/P

we have a 12 volt system and an indy bulb is rated at 21W, so R = 12*12/21 = 7 ohms near as dammit

hence the nearest prefered resistor value of 6.8 ohms supplied for the job of reducing flash rate. and it needs to be rated at at least 21W to lose the heat (25W is standard).

now i've seen sites where they hang that resistor in parallel with the LEDs - that cannot be right. we want this resistor in series to increase the resistance and reduce the current. adding a shunt resistor would just make things worse.

by putting the resistor on the common earth to a pair (left and right, not front and back) of indicators, it should be possible to get away with a single resistor as both sides won't be flashing simultaneously (no hazard function on my Monnie anyhow) - but a resistor on each side will also work.

erm - that's it - i think.

[sheepshearer]

think i might be talking bo11ocks about the parallel/series bit. i've seen parallel all over the place now as the solution - which means i don't understand after all :(

can someone explain?

[Gilps]

think i might be talking bo11ocks about the parallel/series bit. i've seen parallel all over the place now as the solution - which means i don't understand after all :(

can someone explain?

Wire it in line on the positive wire, usually somewhere near the indicator.

[sheepshearer]

so it SHOULD be in series then? in which case i was right in the first place. i've got an alternator problem on the car - i'll ask the auto electrics bloke when he takes a look.

[pfcdaz]

just put LED indies on the front, and the resistor had to be wired in parallel, worked like a dream, not sure about the physics of it all though.....

[Capo]

There is some confusion here.

There are basically three types of flasher relays.

Electro Mechanical
Thermal
Electronic.

Electro mechincal relays are not load sensitive .i.e the flash rate remains the same regardless of load connected to it.

Thermal relays are load sensitive, the flash rate increasing with load. i.e the loss of a connected incandescent lamp would cause the flash rate to decrease.

Electronic relays can be constructed either way, however in order to comply with regulations that require an indication of lamp failure, they are constructed to increase the flash rate in event of a lamp failure, this is accomplished by monitoring the resistance of the circuit.

LED’s draw a fraction of the current of an incandescent lamp and the relay interprets this as a lamp failure and increases the flash rate.

The obvious solution is to restore the resistance of the monitored circuit.

The incandescent lamps are rated at 10W each.

To dissipate 10W in a 13.8V system...

V=IxR hence I=V/R
P=VxI,
substitute
P=V2/R, hence R=V2/P

R=13.8x13.8/10 = 19 ohms or for ease of purchase 22ohm should be fine (8.7watts).

Wired in parallel with the LED the circuit will be restored to 10A
As for power rating, the 10 watts of power consumed by the resistor is subject to a 50% duty cycle and should be adequate for the task.

If you're a bit more adventurous you can just put the one resistor across the output of the flasher unit. Around 19ohms for one pair of indicators replaced or halve the resistance (twice the current therefore twice the power) and double the power rating for both front and back