I thought we were talking about how much current is drawn when BAT +12 is connected to the board but the engine is not running. To me, that suggests plain old DC current from the input terminal to ground, which is Delta-V/R, which is zero if R is infinite.
That’s true as far as it goes, but your “R” stands for Resistance and there
are no resistors in your sensor.
That would be a valid test if you were measuring a resistance, such as a
light bulb or a coil. It will not work to measure anything electronic, such
as an LED or a phototransistor. Try measuring the resistance if your
microwave oven or a TV and you will get the same erroneous open-circuit
result. Ohmmeters don’t tell you the whole story, they just tell you if
there’s a resistor between the leads.
Mike Eck
New Jersey, USA
www.jaguarclock.com
'51 XK120 OTS, '62 3.8 MK2 MOD, '72 SIII E-Type 2+2
1 Like
Ed,
Most semiconductor circuits need voltage before they conduct and draw current. So a resistance test (that doesn’t supply the necessary bias voltage) is not valid.
You can check the current draw by simply putting your meter in Amp mode and connecting it in series with your power lead. +ve meter lead to battery positive. -ve meter lead to + input of your hall effect board.
Be very careful not to let your meter -ve touch the chassis or any earth point (while its +ve is connected to the battery) or you will likely instantly blow the fuse in your meter (direct short across the battery). I do this all the time and have a supply of fuses to restore my meters !! On good quality meters, these are not cheap fuses!
I would expect the Hall Effect board to draw very little current (a few milliamps) but I also would not really recommend connecting directly to the battery.
- The voltage at the battery terminal is not necessarily just 12V. There can be BIG voltage fluctuations depending upon what is going on. Typically +100V and -80V are numbers I recall. However shorter transients may occur (+300V -100V) when equipment is switched or the alternator is switched off.
It is unlikely that the Hall Effect board has been designed with this in mind. This adds cost and bulk to the required electronics to survive these kinds of hits. Further reading here:
http://m.littelfuse.com/~/media/electronics_technical/application_notes/varistors/littelfuse_suppression_of_transients_in_an_automotive_environment_application_note.pdf
The hot side of a coil is also probably pretty noisy, but there should be less energy in those noise spikes than you might see direct at the battery, and hopefully (surely?) they have designed it to withstand that?
- It is always good to minimise the continuous drain from your battery when the car is not in use. Additionally, if this were an opto coupled trigger, it probably would have failed by now (because they age if left on). A Hall Effect sensor should have an unlimited lifetime … (Failure mechanism will be thermal cycles or electrical stress), but some of its supporting electronics may have operating lifetimes (such as electrolytic capacitors). No sense wasting useful operating life by leaving it powered when not required.
There is no excuse for the equipment to fail when you test it. That is just bad design. It likely results from businesses that are run by accountants, who don’t listen to their engineers. Of course it might be bad engineering, but who has ever heard of a bad engineer?
Thanks, Mark, for that very clear explanation.
A couple days ago I went ahead and moved the trigger board power source to the SW (i.e., Run) which is the second from the top pin on the input end of the ballast resistor pack. I could as well have connected it to the corresponding pin on the output side of the pack since the two are at the same potential by virtue of a direct internal connection.
Speaking of equipment failing while testing, I’m currently waiting for a replacement for the Crane optical trigger that began misbehaving after a clumsy effort to test it.
Current to trigger board is about 15ma,engine running or stopped.
Thanks, Foggyoo. I won’t bother measuring it.