Automotive Relays and Resistors and DIIM-DIP

[Electronics2000]

Hello All.


I am a new entrant in this forum.

I had many questions that I couldn't find answers to and one day it struck me that why didn't I ask someone online. So today I joined this forum and hopefully someone out there will have more knowledge than myself and will be able to answer my questions/help me.

Well that's my introduction and 'ice breaker' over with, now to my questions.

I want to retro-fit DIM-DIP lighting in my car.
I have got the basic circuit sorted out.
The sidelights are wired direct to the battery via the dashboard switch.
The operation of the sidelights operates the DIM-DIP circuit via a relay.

How can I make said relay operate only with the ignition on AND the sidelights on? at present the it operates via the sidelight circuit which as explained is wired permanently via the dashboard switch to the battery.
In other words requiring/insisting the DIM-DIP circuit to have two separate inputs for it to trigger one output [the illumination of the 55w headlight via a inline 1 ohm resistor].

An automotive headlight bulb is 55w. A car battery supplies approx 12 to 13 volts, 4·548 amps.
Using the above figures as inputs, next component after the 55w bulb comes a 1 ohm resistor. What will be the resultant wattage at the headlight bulb? i.e. 55w without resistor, so how much w output after a 1 ohm resistor has been applied before the voltage reaches said bulb? How did you calculate this, so I can do it for myself.

There I have set quite a big challenge for the more expert members of this forum to solve [sorry ahem! I meant, to help me with]

[Electronics2000]

Sorry forgot to add this as clarification to my message above.

"two separate inputs for it to trigger one output" the two separate inputs referred to are "vehicle ignition on AND sidelights on"

Thanks.

[Simon (Webmaster)]

Hi

Presumably the power to the dim-dip circuit comes for a permanently live feed, via the relay that is operated by the side lights. You either need to power the circuit from a 12V feed of a suitable rating that is only live when the ignition is on, or perhaps better, connect a second relay in series with the side-light one, i.e. the power to the dim-dip circuit passes through the contacts of both relays in series, both needing to be on to power the circuit. Feed the coil of the second relay from the ignition switch. Hence both will need to be live to activate the circuit.

[Electronics2000]

Hello Simon. Many thanks for your answer; whilst I was waiting for that, what you suggested did pop through my mind, I was just a bit reluctant to have a double relay in series solution, but if that's the best way then so be it! Thanks once again anyways. Much obliged!

Any ideas for/about my second question perhaps? I have re-quoted it below.

"An automotive headlight bulb is 55w. A car battery supplies approx 12 to 13 volts, 4·548 amps.
Using the above figures as inputs, the next component to come [in-series] before the 55w bulb comes a 1 ohm resistor. What will be the resultant wattage at the headlight bulb? i.e. 55w without resistor, so how much w output after a 1 ohm resistor has been applied before the voltage reaches said bulb? How did you calculate this, so I can do it for myself."

12v +ve -> {no resistor} -> 55w bulb = 55w output
12v +ve -> 1ohm resistor -> 55w bulb = ?w resulting output

[Simon (Webmaster)]

Sorry about that, missed the second part. Working as follows:

Power (W) = V * A
Voltage (V) = I * R - ohms law

Rearrange the above equations as necessary.

Current through 55W lamp = 55 / 12 = 4.58 amps
Resistance of 55W lamp = 12 / 4.58 = 2.62 ohms

Adding a 1 ohms resistor in series will give a total of 3.62 ohms

Current through lamp & resistor = 12 / 3.62 = 3.32 amps

Voltage across the lamp (the remainder being dropped across the resistor) = 3.32 * 2.62 = 8.7 volts

Wattage dissipated in lamp = 8.7 * 3.32 = 28.88 watts

Wattage dissipated in resistor = (12 - 8.7) * 3.32 = 10.97 watts.

This last figure is important - the 10W will be dissipated (effectively wasted) in the resistor as heat. You will need a resistor rated above this, and it will get HOT! A more practical solution would be a voltage regulator to reduce the voltage. This will still be dissipated as heat, but can be easily attached to a suitable heatsink. The more efficient way to reduce the light output without wasting power and creating heat is to use a PWM (pulse width modulation) controller. The only losses then will be in the switching transistor, these will be small.

[Electronics2000]

Simon, you're a star, "you're grrreatt!", I mean that very respectfully sincerely and kindly.

You're explanations have been very simple to follow yet detailed in their depth; precisely what I wanted.

Mind you I guess you wouldn't be the moderator of this forum for nothing, especially as this forum is to do with all things electronic. I take it you're a professional in your own right?

Some techy type musings in reply to your post:

#1 I did originally want a "solid-state" type solution do dim the lights, but that was before I came on here, and most people who I thought would know such a solution said I didn't exist or couldn't be done, so hence I bought the resistor. And now I have the thing in my hand, might as well use it, than discard it for, ultimately a better solution. But at least that better solution will be used next time should I require.

#2 How much in £ would a voltage regulator be roughly? Would a place like Farnell stock such an item or is it be a bespoke solution?

#3 My resistor is a 100w golden coloured aluminium type component; I take it that that should be more than ample for what I want to put it to use as?

#4 As #2 How much in £ would a PWM (pulse width modulation) controller be roughly? Would a place like Farnell stock such an item or is it be a bespoke solution?

#5 From your calculations, would I be correct in deducing that the effective light output at the lamp end would be circa 20w [55w-(28.88w+10.97w)]?