Robust delay circuit?

[Cryptin]

Hi,
I'm a mechanical engineer that recently started to look into some electronics due to a rare car I'm driving that have electrical problems...

One of the problems is a halogen bulb that melts down the housing of the lights so I'm looking to design a circuit that could replace the halogen bulb with a LED-equivalent, but does not give bulb-failure indication and not building unnecessary heat.

For this I'm planning to use a circuit that switches of the the power resistor (or some other current limiting device) after like 2s.

Since I'm a novice in electrical engineering I'm having trouble figuring out what would be the best way to switch off the resistance circuit after 2s.

Supply is pulsating like the indicators on the car, but the same bulb is also brake light, so when holding the brake pedal it's constant 12V on supply side, but during brake light no bulb failure indication is active, so I want to turn off the circuit not to build excessive heat.

As I'm a mechanical engineer I'm thinking like valves, I need a normaly closed valve, that will open when supply has been on longer than 2s and will close again directly when supply is off. So when indicators are on, current will flow through the resistor (since indicator is like 1s on 1s off 1s on...) but when brake light is on, no current needs to flow through the resistor and only the LEDs.

I feel kind of stupid not to figure it out so any help is appriciated

Best solution would be to have supply and trigger the same (meaning no constant supply to the delay circuit)

 

[kellys_eye]

If you aren't concerned for the loss of power (i.e. power saving is not an issue) then simply connect a suitable 'bleed' resistor across the LED to draw the 'difference current' between the LEDs consumption and that required to maintain the monitoring current flow.

i.e. if the LED draws 200mA and the original bulb drew 2A you need to 'bleed' (2 - 0.2 = 1.8A @12V) which, using Ohm's Law gives a resistance of

12/1.8 = 6.66Ω

power dissipation is I^2 x R = (1.8 x 1.8) x 6.66 = 21.6 watts.

A resistor of 6.66 ohms (6.8Ω being the nearest easily available value) rated at 25 watts (aluminium-bodied resistors of this type are readily available) connected across your LED (assuming the numbers are as per above - adjust to the correct values as required) will ensure the LED works AND the current monitoring will not be affected.

Current will only flow when the light is on, of course.

 

[Cryptin]

Hi and thanks for your response.

Yeah I know I can use the resistor, but I'd like to switch the resistor off after 2s, since monitoring of the current draw is only on during indication and not during brake. I'd like to do this on a small PCB that could easily be installed in the harness... So in a way I'm concerned about the loss of power, or rather the build up of heat.

As for the current limiting it self I have a few things I'm going to try:
1. resistor - will always work but builds heat.
2. Current limiting with e.g. AL8861Y-13 @ 1.5A
3. Mix of the two might be needed if ECU senses the switching of the AL8861Y

 

[kellys_eye]

The brake lights/indicators can't be on for any length of time to be a significant contributor to 'wasted energy' can they? This is taking energy-saving to ridiculous levels! In a motor vehicle you would never be able to calculate the saving in terms of mpg.... a careless 'blip' of the throttle would likely waste more 'energy' than using the brake lights/indicators for a couple of years!

If it was an e-vehicle the lights would be LED anyway.....

The idea behind the resistor is to relocate the heat - that was your primary concern for changing the type of bulb wasn't it? - to an area where it can do no damage.

As for the heat.... if the application of the signal to the resistor is intermittent then the average heat will be a lot less and the use of a 25W resistor would be overkill anyway.

I'd also query your statement of the use of a halogen bulb - in the brake/turn signal unit???? Isn't there a legal limit on the power (illumination) of such signal lamps (5/21 watts for indicator/brake respectively)? How is that enough heat to melt the housing?

From your answers:

1. the heat is negligible and, anyway, removed from the are where melting is a problem
2. that device limits the current for the LED replacement but won't 'compensate' for the missing current needed to prevent a 'blown bulb' event to be recorded.
3. see 2.

Something about your statement/calculations doesn't add up......

 

[Cryptin]

Sorry to say, but
1. Heat is not negligible, specially when you consider to place it in either the housing or free in the air.
2. Yes I'm aware of the device function, if you use it instead of the resistor it will limit the shortened current to 1.5A, so if putting it in the same place as the resistor as "shorten to GND" it will limit the current to 1.5A adding the LED current draw it would land somewhere around 1.7A.

The reason to turn off the resistor or device is not to build up heat when brake pedal is on. Circuit is still active in means of powering the LED however the "21W simulation" can be turned off when you have constant voltage of 12V for more than 2s (meaning brake is active). 21W of power builds up quite some heat, you for sure burn your fingers on a 21W halogen bulb lit for more than 2s.

For sure a resistor is simple, but still it's not an elegant design, so I'm looking to improve, not to save mpg but to save other components in the system, minimizing the heat in the housing.

 

[dorke]

Do you know how does the sensing of the lamp's current work?
Do you have a schematic ?
This circuit can be "fooled" in many ways,but details are needed.

 

[Cryptin]

Unfortunately I'm not exactly sure how the sensing of lamp work, since no documenation or schematic exist that I'm aware of. It's a GM BCM that monitors it and I've been suggested that it works with current sensing.

 

[AnalogKid]

1 capacitor
1 resistor
1 MOSFET

What is the current (at 12 V) you want to sink through the ballast resistor?

ak

 

[Harald Kapp]

Please observe your local regulations concerning this planned modification. You may not be allowed to use the car on public roads with a modified lighting system.

 

[Cryptin]

1 capacitor
1 resistor
1 MOSFET

What is the current (at 12 V) you want to sink through the ballast resistor?

ak

Current is about 1.5-2A

I was thinking about this setup first, but I was told it was inefficient and the mosfet would get very hot from switching with a RC-system.

 

[Cryptin]

Please observe your local regulations concerning this planned modification. You may not be allowed to use the car on public roads with a modified lighting system.

I am aware of the regulation, nothing is really changed except for removing the problem with hyper-flash, same light just from LED instead of halogen bulb. No functionallity is changed.

 

[Harald Kapp]

same light just from LED instead of halogen bulb. No functionallity is changed.

Anyway this modification may void the approval of your car for use on public roads. At least it would here. As far as I'm aware in Germany you would need approval for such a modification.

 

[AnalogKid]

I was thinking about this setup first, but I was told it was inefficient and the mosfet would get very hot from switching with a RC-system.

When the circuit first turns on the MOSFET is saturated and dissipates very little power. As the voltage ramp on its gate decreases, after a while it pulls out of saturation, moving through its linear active region. After the timeout period it is off and dissipates zero power. The length of time the FET is dissipating power depends on the gain of the FET, but a good guess is that it is dissipating power for less than 1 second. During that time the power increases, peaks at around 6 W, then decreases to zero. A first-pass approximation is that the average power is around 3 W per cycle.

One approach is to oversize the MOSFET and let the package deal with the heat. Anything in a TO-220 package or larger should be fine. To reduce the FET power to near zero requires a second (smaller) transistor to add positive feedback (hysteresis).

Another option is to use a power darlington instead of a FET. It has much higher gain so the base voltage range for linear conduction is much smaller than a FETs, reducing the heating time. Peak power dissipation would be the same, but the heating time period would be a smaller percentage of the overall timing period.

ak