| [ Total Views: 1245 | Total Replies: 21 | Thread Id: 108228 ] |
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 Craig S
A.k.a.: Craig
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| posted on January 26th, 2015 at 02:31 PM |
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LED Lights
Hi All,
Decided to change to LED lights in the hope they'd be brighter and draw less power. The brake and reverse lights work brilliantly, but I'm having
dramas with the indicators.
Here's what I've tried so far.
1. Replace one globe and the indicators flash much faster (no surprise there).
2. Replace both globes and only one of the individual LEDs light up in each globe.
So I decided to replace the flasher relay with an LED specific one. Admittedly it's a relatively cheap one from Ebay. I've tried it all combinations
with both the LED globes and normal globes and absolutely nothing happens. I'm guessing that the relay is a dud.
But, before I go out and buy another relay, I thought I'd ask if there were any other ideas of what I should try.
Car is an LBug/1303 so the relay sits behind the fuses under the dashboard. The relay was in the first position, whilst the Bentleys says it should be
in position 3 (again not a big surprise).
Thanks
Craig
1969 Beetle - June
1974 L Bug - Gwendolyn
2002 Bora V6 4Motion - Anita
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1303Steve
A.k.a.: Steve Carter
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| posted on January 26th, 2015 at 03:16 PM |
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Maybe the new relay has a different wiring configuration?
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ian.mezz
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| posted on January 26th, 2015 at 05:34 PM |
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I think it cause they use less power allowing them to flash faster, just like in a newer car when one globes blows the blinkers flash faster on the
side that the globes blown .
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Joel
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| posted on January 26th, 2015 at 07:48 PM |
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As Steve said its probably the terminal layout is wrong, I know when I fitted my LED flasher it was the wrong way around for VWs, i had to switch the
wires around in fuse box.
Apparently most Jap cars are the other way around to Euros.
If your relay is this layout its the wrong one, VWs need positive on the right.
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vlad01
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| posted on January 27th, 2015 at 11:17 AM |
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| Quote: | Originally
posted by ian.mezz
I think it cause they use less power allowing them to flash faster, just like in a newer car when one globes blows the blinkers flash faster on the
side that the globes blown .
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yes flasher cans are load dependent for timing. LEDs have 100s x less load than conventional lamps.
You can get electronic flasher cans that don't rely on load and are designed for LEDs
71 notchback,
Past owner of, 70 NB, 73 SB and 72 FB TLE |
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vwo60
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| posted on January 27th, 2015 at 07:32 PM |
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you can also fit a dummy load and use the standard flasher can. http://www.ebay.com.au/itm/LED-LOAD-RESISTORS-FOR-VEHICLES-NOT-LED-COMPATIBLE...
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psimitar
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| posted on January 27th, 2015 at 10:14 PM |
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If you have the old bulb and it ain't blown then measure it's resistance. Then buy a 1/2W resistor slightly lower than the bulbs resistance and wire
it in parallel to the LED bulb. This should pull the overall load down to where the relay flashes much closer to the original time period. 
madness is in the eye of the beholder  |
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AA003
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| posted on January 30th, 2015 at 07:39 PM |
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The bulbs resistance needs to be measured when the globe is hot as globes have a positive temperature coefficient and the resistance varies greatly
with temperature. Better to use ohm's law. The resistor that you will need is 36 watts as it is replacing two 18 watt globes.
Much easier to buy a dummy load as vwo60 said.
I read it on samba, so it must be correct.
Sometimes Volkswagen dealers sell spare parts. Amazing isn't it!
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psimitar
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| posted on January 30th, 2015 at 09:48 PM |
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Better for heat dissipation to wire the resistor at each bulb and not just one resistor for each side.
1/2W should be fine, of I've remembered my degree tech labs properly, but 1W to be safer as due to the flashing nature of the indicator the resistor
isn't having to dissipate as much heat.
The resistor method is just a neater solution than trying to wire the additional ordinary style bulbs in parallel with the LED ones.
Least the OP has a few options to try now
madness is in the eye of the beholder |
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AA003
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| posted on January 31st, 2015 at 06:19 AM |
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| Quote: | Originally
posted by psimitar
1/2W should be fine,
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.......and carry a fire extinguisher.
I read it on samba, so it must be correct.
Sometimes Volkswagen dealers sell spare parts. Amazing isn't it!
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vwo60
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| posted on January 31st, 2015 at 06:25 AM |
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The inline resistor is a no brainer and available over the counter, use the kiss principal.
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modnrod
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| posted on January 31st, 2015 at 09:48 AM |
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The extra resistor will also mean that the total current draw on the electrics will be the same as if you were using a normal bulb as well.
I agree with the KISS principle.......
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psimitar
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| posted on January 31st, 2015 at 07:33 PM |
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| Quote: | Originally
posted by AA003
| Quote: | Originally
posted by psimitar
1/2W should be fine,
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.......and carry a fire extinguisher.
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Love how friendly this forum is lately. Have you ever stopped to think that someone has been doing this for a while and has a bloody electronics
degree and therefore has some idea of what they're doing.
Stop being such an arse 
madness is in the eye of the beholder |
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vlad01
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| posted on January 31st, 2015 at 09:45 PM |
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| Quote: | Originally
posted by AA003
The bulbs resistance needs to be measured when the globe is hot as globes have a positive temperature coefficient and the resistance varies greatly
with temperature. Better to use ohm's law. The resistor that you will need is 36 watts as it is replacing two 18 watt globes.
Much easier to buy a dummy load as vwo60 said.
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18w??
last I looked 21W is the standard globe for indicators and brakes lights and such.
71 notchback,
Past owner of, 70 NB, 73 SB and 72 FB TLE |
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AA003
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| posted on January 31st, 2015 at 09:51 PM |
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| Quote: | Originally
posted by vlad01
18w??
last I looked 21W is the standard globe for indicators and brakes lights and such.
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Yes I think that you may be correct but it would be close enough.
It's a lot more than some jerk saying that 1 watt is enough that doesn't have a clue.
I read it on samba, so it must be correct.
Sometimes Volkswagen dealers sell spare parts. Amazing isn't it!
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Joel
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| posted on February 1st, 2015 at 10:10 AM |
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6 volt were 18w is probaby what you were thinking of.
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psimitar
Son of Jim - Creator of Good
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| posted on February 1st, 2015 at 06:07 PM |
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| Quote: | Originally
posted by AA003
It's a lot more than some jerk saying that 1 watt is enough that doesn't have a clue.
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You truly are an A-hole sir. I don't lend my advice unless it is something that I know about. Maybe if you go and integrate the power of the tapered
square wave of the indicator time period to calculate the heat build up in the resistor then I'll happily eat humble pie but in the meantime you
should learn to be a more open minded individual and if you think someone is in error when they write something then inform them of their mistake in a
more polite and informed manner with technical reasoning instead of being so damned rude.
madness is in the eye of the beholder |
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AA003
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| posted on February 1st, 2015 at 06:35 PM |
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Craig,
The problem with using LEDs is that the original flasher relays relied on the current of the globes to make them flash. As LEDs are very efficient and
use only a tiny amount of power the current is not enough to make them flash correctly.
Craig you have four options.
1/ Use an LED flasher unit and mount it in the fuse panel and change the connections. The connectors in the fuse panel can be released by pushing a
small jeweler's screwdriver down the back of the lug to release it.
2/ Use an LED flasher unit and mount it remotely using a short loom with male connectors for the fuse panel socket end and female connectors to the
relay.
3/ Use a load resistor that is commercially available for this purpose.
4/ Use a load resistor that you obtain yourself and wire it in parrallel with the LEDs.
To select a suitable load resistor you will need to calculate the resistance using Ohm's law.
You already know that the nominal voltage is 12 volts and the globes are 21 watts each.
To calculate the current you can divide the wattage by the voltage and the multiply by 2 as they are connected in parrallel.
21/12 x 2 = 3.5 amps.
Ohm's law is R=V/I R = 12/3.5 which equals 3.4 ohms.
To calculate the wattage of the resistor you will need to multiply the voltage by the current which is 12V x 3.5A = 42 Watts
So you will need a 3.4 ohm 42 watt resistor.
I read it on samba, so it must be correct.
Sometimes Volkswagen dealers sell spare parts. Amazing isn't it!
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psimitar
Son of Jim - Creator of Good
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| posted on February 1st, 2015 at 06:44 PM |
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Thank you
The maths is correct for a constant load but I had LED bulbs back at Uni some 12yrs ago and did the calculations then and was able to get away with 1W
resistors. Could of been 5W but pretty sure 1 was enough per bulb and worked happily until I sold the Bug a year or so later
madness is in the eye of the beholder |
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hulbyw
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| posted on February 1st, 2015 at 07:53 PM |
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Craig, you can buy resistors at Repco specifically made for fitting LED indicators. It is a common practice on utes and 4by's to change at least the
rear light clusters to LED and Narva make a specific resistor to suit. Probably an expensive way to do it, but easy as.
Cheers
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vlad01
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| posted on February 1st, 2015 at 08:23 PM |
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| Quote: | Originally
posted by psimitar
Thank you
The maths is correct for a constant load but I had LED bulbs back at Uni some 12yrs ago and did the calculations then and was able to get away with 1W
resistors. Could of been 5W but pretty sure 1 was enough per bulb and worked happily until I sold the Bug a year or so later
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there is theory and there is practice.
In theory 42w sounds to be correct amusing the flasher needs all the load the lamps produce which all calculations are based on.
However the flasher in theory (if taken apart and circuit reverse engineered to derive all values for calculation) may only in reality require a the
said small load.
In practice it obviously works. 
its like a theoretical assumption that a type 3 6v fuel gauge will fry in 12v system (calculated to be 4 x the original wattage) works perfect in
practice with years of no trouble for many people I know.
71 notchback,
Past owner of, 70 NB, 73 SB and 72 FB TLE |
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psimitar
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| posted on February 2nd, 2015 at 12:03 PM |
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| Quote: | Originally
posted by vlad01
| Quote: | Originally
posted by psimitar
Thank you
The maths is correct for a constant load but I had LED bulbs back at Uni some 12yrs ago and did the calculations then and was able to get away with 1W
resistors. Could of been 5W but pretty sure 1 was enough per bulb and worked happily until I sold the Bug a year or so later
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there is theory and there is practice.
In theory 42w sounds to be correct amusing the flasher needs all the load the lamps produce which all calculations are based on.
However the flasher in theory (if taken apart and circuit reverse engineered to derive all values for calculation) may only in reality require a the
said small load.
In practice it obviously works.
its like a theoretical assumption that a type 3 6v fuel gauge will fry in 12v system (calculated to be 4 x the original wattage) works perfect in
practice with years of no trouble for many people I know.
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Indeed
As for a light bulb, the wattage not only relates to the power drawn but also to the light emitted. Being as you're a electronics guy too then you
are quite right that the actual load the circuitry sees is far less than the stated load.
Your 6v fuel gauge should work fine at 12v so long as the wire for the electro magnet windings inside the gauge can handle the extra current. Above
that the device will electrically work in the exact same fashion
madness is in the eye of the beholder |
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