Budget alternator derating? how can I do it?

[Compass790]

Originally Posted by lateral

Post #19
https://www.cruisersforum.com/forums...-is-98527.html

Thanks for the link.

Ahh yes so that does fit with my speculation the reducing the field current should affect the alternator output. I mite try a tiny bulb in the field wire. But the fly in the ointment was there was no voltage regulator in the circuit in that example.
Edging closer to the bike spoke idea

[lateral]

Yip, but smaller bulb would be worth a try?
You run two alternators off a 7hp? No wonder it won't drive the prop.
1hp /25A or some such. I think your expectations are high.
On or off, charging/motoring, is way more practical with 7 horses.


The hitachi alts have a thermister that derates the alt as soon as the engine/alt heats up.
So I pulled my Hitachi.

[Wotname]

Originally Posted by Compass790

I've never tread carefully & ya cant teach an old dog new tricks!

Yes I got that you say I cant control the output current but I still havent received an explanation as to why reducing the current to the rotor doesnt change the stators output. Or maybe I have got an explanation & i didnt grasp it

Let's try again.

The stator output is in VOLTS.
Reducing the rotor current reduces the stator VOLTS.
So yes, the reducing the rotor current DOES change the stator output but that output is measured in VOLTS, not AMPS.

So the regulator tries to keep the output VOLTS constant.

Next question...

[Wotname]

Let's see if this helps -

The stator is producing 14.4 volts and the only thing that it providing power to is charging the battery and let's say the charging current is 15 amps at the moment.

Now you turn on some lights (2A) and a pump (3A) so the current required for the lights and pumps is 5 amps. Now the stator will be providing 20 (15 + 2 + 3) amps and the volts remain at 14.4V. Turn the lights off and now the stator current will be 18 amps (15 + 3) and the volts remain at 14.4V.

As the battery gets charged, it's acceptance current will reduce to say 10 amps. Now the stator current will be 13 amps (being 10A for the battery and 3A for the pump). Turn the lights back on and the stator current goes to 15A (10 for the battery, 2 for the lights and 3 for the pump). The voltage remains at 14.4V because that is what the regulator does. It tweaks the rotor current all the time to keep the stator voltage at the pre-set magic number.

If you want to decrease the battery charging current, you have to decrease the stator VOLTAGE.

[stormalong]

Adding to the two posts above:

In order for currrent to flow there must be a voltage difference between the source (alternator in this case) and the load (Battery).

Suppose that your battery is sitting at 12.0 volts. In order for current to flow into the battery the alternator must be supplying a voltage higher than 12.0 volts. The greater the voltage difference the more current will flow. The voltage regulator controls the voltage that the alternator produces. There are different types of voltage regulators. Most, if not all internal voltage regulators output a fixed voltage, so as the battery accepts current from the alternator and it's voltage rises the voltage difference is less and the current is reduced.

Smart regulators monitor the battery voltage and charge the batteries in multiple steps. As the voltage rises the program in the voltage regulator decides when to cut back on the charge voltage. That is when it switches (in steps) from bulk charging to float, where the alternator output only supplies the voltage needed for external loads and puts very little current into the battery.

[RaymondR]

The regulator only sets the maximum voltage output of the alternator.

The regulator will set the current output to either achieve the voltage limit or achieve the rated current output for the alternator.

The answer is to increase the back voltage seen by the regulator at the required current output.

In DC circuits Ohms law, E=IR where E is volts, I is current and R is resistance, prevails.

The alternator is set up to have a maximum voltage of 14.4

The I, current, required is 10 Amps.

Therefor E/I or 14.4/10, or 1.44 Ohms is the resistance required to produce 10 Amps through the circuit.

Batteries have an internal resistance which tends to change as they charge and is why the current tapers off when the regulator sees the 14.4 volts as the battery becomes fully charged. So in order to limit the current in the circuit you must make the regulator see 14.4 volts at it's output terminal.

Increasing R in the charging circuit by inserting more resistance of the circuit is the simplest way to do this.

When I was charging my house batteries I would monitor the alternator output and as the voltage in the circuit increased with the increase in internal resistance of the battery, slide the two wires on the bike spoke closer together to reduce the resistance through the bike spoke to maintain a constant resistance in the circuit.

[Wotname]

Originally Posted by Compass790

..........I still havent received an explanation as to why reducing the current to the rotor doesnt change the stators output. Or maybe I have got an explanation & i didnt grasp it

Maybe you would like a slightly more technical explanation than my previous 'simple' one.

1. rotor current creates a magnetic field
2. the rotating magnetic field cuts the stator winding which induces electromotive force (EMF) in the stator windings.
3. The size of the induced EMF in the stator is directly proportional to the rotor current.
4. the EMF is measured in Volts.

So reducing the rotor current reduces the EMF (i.e. Volts) in the stator.

For more detail, wade though this link https://en.wikipedia.org/wiki/Electromotive_force

[Wotname]

Originally Posted by RaymondR

The regulator only sets the maximum voltage output of the alternator.

The regulator will set the current output to either achieve the voltage limit or achieve the rated current output for the alternator.......
.........

Err.... No. Totally incorrect. The regulator does not set the current output, it has no idea what the output current is. It does not monitor the current.

The rest of your post is basically correct.

[RaymondR]

Originally Posted by Wotname

Err.... No. Totally incorrect. The regulator does not set the current output, it has no idea what the output current is. It does not monitor the current.

The rest of your post is basically correct.

The regulator sets the current output of the alternator by varying the excitation current in accord with the output voltage of the alternator. The current output through the circuit resistance generates the back voltage the regulator senses therefor it monitors the current output.

What's going on in the battery is a little more complex than what I have stated however it presents as an increase in resistance.

Prior to Hall effect devices measuring the voltage drop across a resister (shunt) was the commonest method of measuring current.

[Compass790]

Originally Posted by lateral

Yip, but smaller bulb would be worth a try?
You run two alternators off a 7hp? No wonder it won't drive the prop.
1hp /25A or some such. I think your expectations are high.
On or off, charging/motoring, is way more practical with 7 horses.


The hitachi alts have a thermister that derates the alt as soon as the engine/alt heats up.
So I pulled my Hitachi.

I didnt say I run 2 alternators I just said it has 2 alternators. Usually we just have a belt on the 55amp Bosch & the Hitachi just sits there as a spare.
When we bought the boat I did notice the Hitachi was pretty useless for any serious work due to the thermistor & I had to get the 55 amp Bosch functioning which started my alternator education ( Clearly a work in progress lol )

Yes I know off is more practical but if I could get 8 amps charging it would be better than nothing.

[Compass790]

Originally Posted by smac999

How does the alt currently charge everything? It should not be putting out more then 10a while charging the engine battery.

Just disconnect the charge to the house bank when you need power. Ie if you have an acr. Disconnect the ground wire via switch. If you have a 1-2-all switch move it from all to eng batt.

This is easier the screwing around with a fake regulator

I'm not sure where you got the 10A limit whilst charging the battery.
We have 225 amp capacity batteries & if the voltage is 12v in the morning the alternator will be putting 27 amps @ 1500 rpm into them for quite some time.
We use the batteries for both house & starting purposes.

[Wotname]

Originally Posted by RaymondR

The regulator only sets the maximum voltage output of the alternator.

The regulator will set the current output to either achieve the voltage limit or achieve the rated current output for the alternator.....................

Originally Posted by Wotname

Err.... No. Totally incorrect. The regulator does not set the current output, it has no idea what the output current is. It does not monitor the current.

The rest of your post is basically correct.

I should have added 'the only time the regulator "sets" the current is when the alternator output reaches it rated maximum output current although it is better stated as "limits" the current to the rated maximum.

Just think of the alternator as a constant voltage, current limited power supply.

[Compass790]

Originally Posted by Wotname

Let's try again.

The stator output is in VOLTS.
Reducing the rotor current reduces the stator VOLTS.
So yes, the reducing the rotor current DOES change the stator output but that output is measured in VOLTS, not AMPS.

So the regulator tries to keep the output VOLTS constant.

Next question...

hang in there Wottie!
It takes a 12lb hammer to pound an idea into a metal basher lol

So I'm down to an adjustable voltage regulator or an ss bike spoke.
Actually it will be a 2.5mm 316L welding rod not a bike spoke if I try that as thats what I have handy.

[RaymondR]

You could have 1,000 volts of EMF at the alternator's output and until current begins to flow there would be no output. If you then put a "zero resistance" dead short across the alternator's output the voltage would then drop to zero but the amps go up to a maximum value which depends upon the speed of the alternator and the stator's resistance and the high ampage across the stator resistance will generate sufficient heat to burn it out.

I will say it again; the regulator controls the maximum voltage of the alternator, the charging circuit resistance ie. the sum of the interconnection wiring resistance, stator wiring resistance and battery resistance controls the current flow and whilst doing so causes a voltage increase at the alternator output terminal which the regulator senses to limit the output current from the alternator. Increasing this circuit resistance will result in a decreased current output from the alternator which in this case is what is required.

If the regulator had an external voltage sensing terminal (some do and it is there to bypass the wiring and allow the regulator to sense the battery terminal voltage) increasing the circuit resistance will not work and some means of increasing the apparent battery terminal voltage would be required in order to decrease the alternator output ampage and
placing some value of voltage in series in this circuit would do so. The converse to what is required in this instance ie. reducing the voltage being sensed at the regulator to increase the alternator current output can be achieved by inserting diodes in this circuit. The roughly 0.6 volt forward biased voltage drip "fools" the regulator into causing a higher current output from the alternator. This also I have done at times to speed up battery charging (Not highly recommended though as it has an adverse effect on battery life)

E=IR is a law of our physics related to electrical circuit dynamics. The factors, ie. EMF, current flow and resistance, are interdependent and any change in one will cause a corresponding change in one or both of the others.

[Compass790]

Originally Posted by RaymondR

You could have 1,000 volts of EMF at the alternator's output and until current begins to flow there would be no output. If you then put a "zero resistance" dead short across the alternator's output the voltage would then drop to zero but the amps go up to a maximum value which depends upon the speed of the alternator and the stator's resistance and the high ampage across the stator resistance will generate sufficient heat to burn it out.

I will say it again; the regulator controls the maximum voltage of the alternator, the charging circuit resistance ie. the sum of the interconnection wiring resistance, stator wiring resistance and battery resistance controls the current flow and whilst doing so causes a voltage increase at the alternator output terminal which the regulator senses to limit the output current from the alternator. Increasing this circuit resistance will result in a decreased current output from the alternator which in this case is what is required.

If the regulator had an external voltage sensing terminal (some do and it is there to bypass the wiring and allow the regulator to sense the battery terminal voltage) increasing the circuit resistance will not work and some means of increasing the apparent battery terminal voltage would be required in order to decrease the alternator output ampage and
placing some value of voltage in series in this circuit would do so. The converse to what is required in this instance ie. reducing the voltage being sensed at the regulator to increase the alternator current output can be achieved by inserting diodes in this circuit. The roughly 0.6 volt forward biased voltage drip "fools" the regulator into causing a higher current output from the alternator. This also I have done at times to speed up battery charging (Not highly recommended though as it has an adverse effect on battery life)

E=IR is a law of our physics related to electrical circuit dynamics. The factors, ie. EMF, current flow and resistance, are interdependent and any change in one will cause a corresponding change in one or both of the others.

Thanks for the tutorial RaymondR

Yes we too used the diode in the field wire trick but have since just put a 14.8v VR in as generally we cant use the alternator driving anyway unless the batteries are near full


Cupla questions for ya.

Would a 2.5mm ss rod be ok instead of a bike spoke to use as an adjustable resistor?

Alternatively could I use one of these in the field wire to dial up the voltage to fool the regulator to reduce the alternator output? Ok I realise the 2amp current limit might be a bit low but I could have a 2 pole switch & a wire with no voltage converter in it. The budget would stretch to this

https://www.aliexpress.com/item/3227...B&gclsrc=aw.ds

Feel free to shoot me down in flames ( you too Wotname )
I have many a baldrick plan.