On my car I have an indicator module that has a U2043B IC inside (datasheet attached).

The system has a bulb outage detection by using a shunt resistor in series with the incandescent bulbs. When it detects a bulb outage the flash rate doubles.

As I would like to move over LEDs to replace the bulbs, I would like to modify my indicator module box to accomodate this.

It's fairly straight forward how it all works, but I would like a little guidance from you guys (the experts) if my logic adds up.

The standard shunt resistor is 30 milli-ohm, which incidentally is the same resistance described in the datasheet calculations.

I was thinking if I replace the shunt resistor for a much higher resistance for the LEDs, say 800 ohms (100mA draw from LEDs) then this would be the simplest way to solve the problem....

I have also attached a photo of the insides of the unit for you to see how simple it looks and how things all relate.

Thanks!
Dominic

 

Where did get 800 ohms from? That's an 80 volt drop at 100ma. You sure you don't mean 80 ohms? If you're using 20 ma LED's 5 will get you to 100 ma. I don't think 5 LED's will be as bright as a 21 watt incandescent lamp.
With LED's, I'd skip the whole bulb failure monitoring anyways. Your car will be a compacted pile of scrap metal before you lose an LED.

 

yeah the problem i face is the rest of the indicator module handles other functionalities such as keeping the indicators flashing if the vehicle is involved in a crash or if the alarm goes off etc.

I would rather just do a simple modification to the circuit than spend time developing my own unit with a 555 or something.

As you say, it would be really rare for an LED to pop. Although i'm not bothered whether the system could detect a popped LED, but rather just fool the system into detecting a much smaller current draw or not at all.

 

Make a voltage divider and feed something greater than 81 mv to pin 7. That should fool the chip into thinking the LED's are drawing current.

 

As I would like to move over LEDs to replace the bulbs, I would like to modify my indicator module box to accomodate this.

Are you trying to replace, say, the turn signal bulbs with the newer LED "plug in" replacements?

 

That's right. I'm trying to avoid having to put large resistors in parallel with the led (to draw the 21w required to stop it going into bulb blown detect).

It's fatal discussing this with fellow owners about the idea, as they want me to do it to theres too!

Make a voltage divider and feed something greater than 81 mv to pin 7. That should fool the chip into thinking the LED's are drawing current.

that sounds like a plan.. why didn't i think of that.... thanks!

 

You're better off just buying the pre-made LED replacement bulbs however even those can present problems at times. Some flasher circuits depend on a certain amount of current draw to operate and will act as if a bulb is out (or not even work) when you replace the bulbs with the LED type.

Most auto stores have these bulbs, eBay, Amazon and a multitude of other websites and parts suppliers also offer them.

I know our motorcycles don't like them but we've figured a way around that by using a different flasher module that isn't current dependant.

 

motorcycle flasher units do have a problem with this and usually require a whole new unit. I think some of them even still have bi-metallic strips in them. old skool.

As for my project, I've bought replacement LED bulbs, so the are already moulded into a bayonet fitting. As mentioned before, on my flasher unit the double flash rate is a feature of the IC to detect a bulb out. I believe it already has it's own independant flasher oscillator.

 

Make a voltage divider and feed something greater than 81 mv to pin 7. That should fool the chip into thinking the LED's are drawing current.

Ok I finally got around to getting back to this project and this method doesn't seem to work.

I put various size resistors, even variable resistors to find a sweet spot, but nothing...

strangely, i disconnected pin 7 and it half works, in that the pulse speed is slightly faster than normal.

Another problem is the load. I tried putting some led clusters on the circuit, which draw around 120mA each, and basically the current wasn't enough to start the circuit.

I'm beginning to think, to remove the surface mount IC and make some small 555 timer based circuit to go in place of the old IC.

Unless anyone has any ideas?

 

Did you apply the 81mv between +V and pin 7? +81 to +12Vs, -81 to pin 7, just as it would be if it were generated from the 30 milli ohm sense resistor? A voltage divider could do that.

 

yup, that's the first thing i did was make a voltage divider.

I even went to the extent of making a variable voltage divider.
The chip does use pin 6 to measure the supply voltage. Although it shouldn't make a difference if the voltage divider is taken off the same power supply.

Your idea should work with no problems, so i'm mythed as to why it's not working.

 

Trying to make your own flasher could be a real can of worms. Many 555 timers are limited to a Vcc of 16v, some go to 18v, but I don't know of any that are rated for up to 60v transients that you may very well experience.

The transients happen during "load dumps", like when you turn your headlamp off. It takes time for the generator output to stabilize; and during that time the voltage spikes can get quite high.

 

*Update* I've got it working... there was a surface mount diode on the pcb right next to the relay, which i didn't see, which had been causing it not to work for some strange reason... anyway it works!

Now the next problem, without sufficient load, the circuit won't initiate.

 

I think something with the start input (pin 8) could unlock the key to the load problem.

Not sure quite how though.

 

*Update* I've got it working... there was a surface mount diode on the pcb right next to the relay, which i didn't see, which had been causing it not to work for some strange reason... anyway it works!

Now the next problem, without sufficient load, the circuit won't initiate.

That diode protects the IC from the relay's inductive spike.

The relay coil is essentially an inductor. When it is switched off, a high voltage spike is produced. The diode shorts out the negative spike quickly, protecting the IC.

Without it you will damage the IC. It might work for days, weeks, months or even years properly. But then it could fail, leaving you without any control over the indicators (that is, all indicators would fail.)

Tom

 

I've put my own diode between the voltage out on the voltage divider and pin 7. So it should all be ok.

I don't think I'm going to worry too much about the load problem... It seems to cope reasonably well with LEDs.

 

I've put my own diode between the voltage out on the voltage divider and pin 7. So it should all be ok.

I don't think I'm going to worry too much about the load problem... It seems to cope reasonably well with LEDs.

How are you powering the LEDs? Please do not say you are only using a resistor.

 

huh?

The LEDs are bulb replacement types, they sort themselves out.
Just changed the original incandescent bulbs for LED bulbs with bayonet fitting.

 

It'll work ... at 12V.

But when the battery is charging it can go as high as 13.8V and when the battery is discharged about 11.5V the regulation will go out of the window. And your LEDs will be burnt by the load dumps quickly. In short, I do not recommend those "bulb replacement" types.