Solving parasitic draw (1 Viewer)

[george_tlc]

That is a common failure resulting in a lot more draw than you might expect. The "timer" is fed from the dome/door light circuit. When it is energized it charges a capacitor . When the circuit is de-powered, the capacitor discharges into the ignition ring light until it is done. This is what provides the timed illumination. The most common failure mode results in continuous power to the ignition ring light. I have replaced several of these over the years in different '80s. I still remember how surprised I was on the first one when I realized that all the draw was from that itty bitty little light!

Mark...

It's not actually quite like that... The capacitor is used as part of the timing, the transistor (and leaking zener) are what control power to the ring. Read my write up in the link further up and it explains a little about the circuit and the actual failure (the zener).

cheers,
george.

 

[Mark W]

It's not actually quite like that... The capacitor is used as part of the timing, the transistor (and leaking zener) are what control power to the ring. Read my write up in the link further up and it explains a little about the circuit and the actual failure (the zener).

cheers,
george.

I will!

Electronics are one step away from black magic... eye of newt and salamander tail so far as I am concerned. But it never hurts to try and understand it better.

Mark...

 

[landtank]

My understanding of zener diodes is they only pass a certain voltage. In this case it might be 12v. Since a charged battery is usually around 12.3v and has a charging voltage or 14.2v. Passing only 12v would ensure that the rings brightness is constant.

But I could be wrong.

 

[george_tlc]

My understanding of zener diodes is they only pass a certain voltage. In this case it might be 12v. Since a charged battery is usually around 12.3v and has a charging voltage or 14.2v. Passing only 12v would ensure that the rings brightness is constant.

But I could be wrong.

Not quite right

A zener has a break down voltage. It will break down (reverse biased) and conduct when you reach that voltage (it's not instant on/off...). Typically there would be a resistor in series to limit current and the voltage across the zener would then limit to the zener voltage (again, it's not a precision device).

The zener in the ring light circuit is intended to protect the transistor (spikes etc) that is providing the current path to the ring light. Nothing in the design provides 'constant brightness'. It's just a 12V nominal bulb and lights up brighter with 14V than 12V. Of course in this application, the ring light normally turns off with the engine running, so it normally would only see 'resting' battery voltage.

cheers,
george.

 

[jonheld]

You could test the voltage drop across the fuses. --this may be an easy way to isolate:

Another fuse volt drop chart: https://www.motor-talk.de/forum/aktion/Attachment.html?attachmentId=721311

I'm extremely sceptical about this method.
In order to accurately measure current, we will "assume" a fully charged battery at 12.6 VDC. There now has to be a known resistance across the fuse. And in order for either one of those charts to be accurate, the resistance has to be the same for all fuses of the same ratings, and battery voltage has to be the same for all vehicles in question, which is highly doubtful.

I just measured 6 fuses; 5, 10, 15, 20, 30, and 40 amp with my calibrated Fluke meter set to the lowest scale and zeroed out to take into account the resistance of the leads, and got a reading of 0.00 ohms across each one.
Please explain how one would measure current draw like this, as it makes no sense to me and seems to violate Ohm's law.

 

[landcrshr]

I'm extremely sceptical about this method.
In order to accurately measure current, we will "assume" a fully charged battery at 12.6 VDC. There now has to be a known resistance across the fuse. And in order for either one of those charts to be accurate, the resistance has to be the same for all fuses of the same ratings, and battery voltage has to be the same for all vehicles in question, which is highly doubtful.

I just measured 6 fuses; 5, 10, 15, 20, 30, and 40 amp with my calibrated Fluke meter set to the lowest scale and zeroed out to take into account the resistance of the leads, and got a reading of 0.00 ohms across each one.
Please explain how one would measure current draw like this, as it makes no sense to me and seems to violate Ohm's law.

Battery voltage is irrelevant, all that matters is the voltage drop across the fuse. But yes, the resistance of each fuse would have to be known. Doing the math on some of those figures in the chart, it would suggest the resistances would be in the thousandths of an Ohm. Your Fluke meter would have to be at least 10x more sensitive.

 

[jonheld]

Battery voltage is irrelevant, all that matters is the voltage drop across the fuse. But yes, the resistance of each fuse would have to be known. Doing the math on some of those figures in the chart, it would suggest the resistances would be in the thousandths of an Ohm. Your Fluke meter would have to be at least 10x more sensitive.

I guess I should have done the math before posting. Thanks.