4
\$\begingroup\$

WAY back in my past I did some serious EE work, but not having used it for nearly 40 years, I'm a little lost on how I do this...

I have some 6 and 12 v cars I'm trying to restore and was lucky enough to find some very rare new switches. There won't be any more, and even good used ones are hard to find. It occurs to me I really should off-load the power from going through the switches - I'm only interested in protecting the headlight and starter solenoid / relay circuits because they're the high-power circuits and the ones most prone to trouble.

What I thought I'd do was get an IC - I do not want to use relays, solid-state only - and wire it in such that the switch is only used as a signal that throws the IC into action, with the IC passing current to the desired components.

I recall, however, that there's a need to run some (more than the minimum necessary to toggle the IC) current for "whetting" to act as a cleaning function for the switch contacts.

The headlights are rated at about 55W per bulb, so 110, plus a safety margin would be good, and there are halogen upgrade bulbs available but I'm told they require too much current for the original switches - a big reason the switches are so hard to find. Some poking around indicates these higher-output bulbs are rated at 100w per bulb, thus roughly doubling the power demand.

There are no published rating values available for the starter solenoid for my chosen type, but I'd guess that nearly all autos will have similar values for the "pull in" voltage / current on the solenoid, what will differ most is the current fed to starter motors - and that isn't involved in my goals here.

Of course, the voltage range available on 12v will vary from about 11 to 14.7 and on 6v, from about 5.5 to about 7.3, give or take a little, depending on state of charge and load - and whether a charging circuit is active or not!

It would be nice if all the various versions of headlight bulb and the starter solenoid can be accommodated with one IC (per switch) per voltage range, so in theory I'd need two per vehicle and they'd be the same type and have the same circuitry. Just a guess here, but I don't think the starter solenoid draws 100w, much less 200w, but it is just a guess.

OK, that's where I'm at.

I'm unclear if anything is needed to accommodate the wiping current - or even how to come up with any value for it, or, (presuming I have to do something special to accommodate it other than just pick the correct IC) how to wire it. For all I know, there may well be an IC that will work as-is quite nicely, with no extra circuitry required since this is perhaps a more common problem than for just little-ole-me.

I'm also unclear on how to select (and find a vendor to sell) the IC. I have a vague recollection that the wiring will be very different if I use an NPN vs PNP, but my skills are so rusty there's nothing left but the rust!

Any assistance appreciated.

\$\endgroup\$
  • 2
    \$\begingroup\$ Are you sure you don't want to use a relay? They are cheap and readily available. Automotive is really one of the more challenging environments for electronics design. \$\endgroup\$ – user28910 Dec 5 '13 at 22:16
  • \$\begingroup\$ ^^^ What he said, car manufacturers still fit relays today, and I'm fairly sure that solid-state would be cheaper these days, so presumably there's a few good reasons. \$\endgroup\$ – John U Dec 6 '13 at 9:29
  • \$\begingroup\$ Re: higher-power headlamps - the optics of an older vehicle's headlamps may not be sufficient to avoid dazzling oncoming drivers, even with dipped beams. I suppose there is no plastic around to be melted. \$\endgroup\$ – Andrew Morton May 17 '14 at 20:43
1
\$\begingroup\$

There are many 'High-side switch' IC's available for this purpose. Check the application notes for details on how to use them. But as mentioned by smashtastic, check that the vehicle is negative grounded.

The wetting current can be provided by bridging the two contacts with a capacitor (from http://en.wikipedia.org/wiki/Wetting_current).

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

Use a 8 A 12 V or 16 A 6 V relay.

The coil resistance will provide the wetting current required for the switch.

The relay will switch the load to your lamps. Typically this will be high side switching and the low side from the lamps to the car's chassis for the 0 V return. Unless your car is positive ground. Then it will be low side switching on the 0 V side.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.