Benchtesting an alternator

[SiWhite]

I'm having a few charging issues (as seems to be so fashionable at the mo!)

I get the same symptoms on all three of my spare alternators (produces about enough juice at startup, but soon drops off to not charging, battery light only glowing very dimly). I suspect either wiring or a past-it's-best battery.

Is it going to be possible to benchtest the alternators to make sure they're not all cream-crackered?

I'm hoping to bolt them to the aux alternator mounting point on the block, and running a positive and negative (from the terminal and alt body respectively) to an out-of-the-car battery. I'd bodge a charge light / exciter circuit too, to make sure it all works, and measure output at the alt terminals as normal.

Is there anything glaringly olbvious with this plan that'll cause me injury, death or unnecesary expense?

[David Sparkes]

There is nothing "glaringly olbvious with this plan that'll cause me injury, death or unnecesary expense", but you need to consider some sort of variable electrical load, so you can attempt to draw different amounts of current from the alternator, while monitoring the voltage. Multiple headlight bulbs, each with a seperate switch (or say three bulbs to each switch) might be easy to rig up. At least then you have some idea of the load withouit having to measure the current.

I'm not trying to be too rude, but the comment "soon drops off to not charging, battery light only glowing very dimly" makes me ask whether you really know what you are doing?

If the alternator is working, the 'battery light' (presumably the red charge or alternator warning light) shouldn't be glowing at all. It should have full battery voltage (via the ignition switch) on one terminal, and alternator output voltage on the other terminal. With a balanced voltage there is no current flow, so the filament isn't heated.

Two reasons cause a difference in voltage across the bulb, one is if the alternator isn't producing enough voltage, the other is if there is some resistance in the battery / ignition switch feed. Say this resistance takes 3 volts to force current through, the bulb will have 11 volts on one side, and 14 (alternator output) on the other. the 3 volt potential difference across the bulb will cause some current to flow, and the filament to glow sightly. As this fault is in the car wiring, EVERY alternator you fit will appear faulty.

By all means rig up an alternator test facility, but check the standard installation as well.

HTH.

[FridgeFreezer]

David,

Just to pick you up on your comment - no matter what the resistance in series with the charge light it would not cause the bulb to glow if the voltage at either end was the same, in fact resistance in series with it would discourage the bulb from glowing.

What would cause the bulb to glow is an earth fault at either end of the charge light wiring, a short causing a high load on the alt, or as I have had a poor connection from the alt to the battery +ve terminal.

[David Sparkes]

Hmmm,

I've had a high resistance ignition switch cause this 'ignition light glow', and poor battery charging.

Your comment about 'resistance not causing the lamp to glow,' is where I disagree.

What I'm suggesting is a circuit that is Alternator voltage - light - resistance - Battery voltage.

Normally the Alternator and Battery are at the same voltage, and the resistance is zero. I think we can safely ignore tenths of ohms in the wire and connectors.

If the resistance (perhaps of a faulty ignition switch) is 10 ohms, say, the voltages 'seen' by the light are unequal. It is the voltages at the light that are important, not the voltages at each end.

I think to problem is in the detail: the voltages at the alternator and battery, even in a perfect installation, will be slightly different (otherwise the battery would not be fed any charging current). Thus some current will flow through the bulb filament, which may in fact glow slightly, but this is not visible due to the red cover. Add in the high resistance ignition switch and the current flow causes a voltage drop across the switch. Now the light sees markedly different voltages at it's terminals, so more current flows (heating the filament, which gets brighter, causing it to be seen). The increased current flow increases the voltage drop across the resistance until a new balance is obtained.

Do you mind, I had to start thinking then :-)

[rtbarton]

Introducing resistance in the warning light circuit alters the trip-in voltage of the alternator, it doesn't effect the charging voltage. Altering the bulb wattage (and hence its resistance) is a way of setting the trip-in voltage.

If there is a resistance in series with a bulb the potential difference accross the bulb will decrease, but the pd accross the combined circuit will be the same. The same lower current will flow through BOTH the bulb & resistance, so the bulb will be dimmer.

[David Sparkes]

"If there is a resistance in series with a bulb the potential difference accross the bulb will decrease, "

I disagree, in this circuit.

What you say is correct if the circuit is conventional, with an negative at one end, and a positive at the other.

With this circuit you are dealing with TWO sources of voltage (positive), with a common (we hope) negative.

If the value of one positive changes, the potential difference across the bulb increases.

[rtbarton]

"If there is a resistance in series with a bulb the potential difference accross the bulb will decrease, "

I disagree, in this circuit.

What you say is correct if the circuit is conventional, with an negative at one end, and a positive at the other.

With this circuit you are dealing with TWO sources of voltage (positive), with a common (we hope) negative.

If the value of one positive changes, the potential difference across the bulb increases.

It depends which voltage changes and in which direction.

If one end is say +12 wrt earth and the other +10v wrt to earth then as far as the bulb and resistance is concerned this equates to a pd of 2v accross the pair. If one positive changes from 10 to 11 the pd accross the combination is now only 1 volt, so the current will decrease, and will be the same in both the resistance and the bulb. (NB you will only get a current as long as there aren't any reverse biased diodes in the circuit).

If the change is +10 to +9 the pd = 3v so current increases and so on.

[SiWhite]

Not entirely sure who's winning the electrickery tech competition, but rest assured you CAN bench test an alternator with a battery, some bits of wire, a bulb and a spinning engine or an impact wrench

Thanks for everyone's help and the deep tech which I don't understand in the slightest

[rtbarton]

.... you CAN bench test an alternator with a battery, some bits of wire, a bulb ........ an impact wrench

That'll sort out any loose connections

..... deep tech which I don't understand in the slightest

Me neither

I was considering rebuilding alternators at one time, just need an electric motor, some kind of frame, battery, load, ammeter and voltmeter.