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how much resistance to reduce a current of 5A (24V DC) to 2A? without changing the Voltage.

in a 24V alternator i'd like to manually (and variably) reduce the field current (5A) to 1A  (in order to regulate the alternator output)  can i do it with a rheostat?  what kind of resistance must it have (rheostat specs)?
is output current linear to field current?  5A field = 75A output,  1A field = 15A output ?
thank you!

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raedy07 (author) 1 year ago

hello! all's well that ends well - i got myself a 25W/20ohm rheostat and it does d job just like i thought it might :) cheers, RJ

-max-1 year ago

If you do not change the load, you CANNOT change the current through it without changing the voltage. Remember, Volts push Amps through Ohms.

To get less amps, you need to make less volts, or increase resistance of the load. The rheostat would literally just drop away the voltage that is proportional to the current through it, so that way the load "sees" less voltage and does not draw as much current.

If you think of the rheostat and the load as one system, one thing, then it is as if you have increased the resistance of the load (series resistances add up) and so less current is drawn.

raedy07 (author)  -max-1 year ago

"If you think of the rheostat and the load as one system, one thing, then
it is as if you have increased the resistance of the load (series
resistances add up) and so less current is drawn."

that's just what i mean! but Joseph says it'll change d voltage?! well d 1 important thing is to reduce d strenght of d electromagnet (to reduce O/P), so a reduced V wouldn't matter really.....? i guess. i am lacking d theoretical background. but as no higher current will flow, field wire shouldn't be burnt

-max- raedy071 year ago
I think what you might be thinking about when asking about how to limit current without limiting voltage, is output impedance... Let me explain. Say you got a LED, LEDs tend to be very picky about the voltage across them. 0.5V too much, and you'll likely kill the LED instantly; 0.5V too little, the LED might not even glow. In addition, the exact operating voltage of LEDs is dependant on temperature, type, age, and several other factors.

So we need a "ballast" of sorts to power the LED properly. We want the current to be relatively constant, and the voltage to change dynamically with the LED. A Constant Current (CC) power supply does exactly this. It adjusts voltage across it dynamically in real-time to attempt to keep the output current constant, In the same way that a constant-voltage (CV) power supply attempts to keep the voltage across it constant with ever-changing loads and currents.

In nature, there aren't any ideal CC or CV sources for power. Realistically things tend to fall somewhere between those 2 extremes. This is where the concept of output impedance can be introduced. Imagine adding a really large resistor to the output of a CV supply. When no current is being drawn, the resistor does not produce any voltage drop, so you will see the full voltage of the CV supply. However, as you increase the current by adding more and more load, you will notice that the output voltage "sags" and that voltage sag will be related to the current. You can actually imitate a CC power supply with a CV supply and a really large resistor, this comment is already way too long, I'll let you explore that!
raedy07 (author)  -max-1 year ago

thanx a lot 4 this info!!! i'll look into those videos asap

-max- raedy071 year ago
Effectively all the rheostat does is divide the power being used between it and the load, reducing the current as well. I don't see any reason to use a large, expensive, inefficient and lossy resistor instead of some gearing to lower the RPMs of the alternator.

I am not an expert at alternators, I don't have much experiance with them, but I will guess that the open circuit voltage on the output is going to be proportional the rotational velocity of the spindle. I sort of recall someone else saying there actually is more advanced battery charging electronics built into them, or at least some rectifiers.

Yes, its ALSO a function of RPM, as well as field current

What is field current? Is that the current through the windings inside of generators, dynamos, and alternators? I am guessing that it would depend on load, more load = more field current.

raedy07 (author)  -max-1 year ago

d other way round - more field current, more O/P

The field current is the current in the armature of the alternator. It is a controllable variable, and it affects output voltage.

raedy07 (author)  -max-1 year ago

did that, d useful O/P is now 20-40A at 1300-1600 (cruising
speed)rpm. better, but not good enough, range is too narrow ;)
field current, also excitation current, varies d strenght of d el.magnet
(rotor in alternator) hence d O/P. d other factor is rpm

-max- raedy071 year ago

I recommend if you want to learn some of the basics of electronics, that you find videos on youtube about resistance, voltage, current, power, stuff like that. I have 2 videos that are 28 minutes total that cover all the essentials on my youtube channel, Power Max (same profile pic.)

Bear in mind the O/P IS changing the output volts by playing with the field current. It IS a kind of amplifier, but the gain is very low. I am surprised the alternator field is pulling 60W.

Output current depends on the LOAD, output volts depends on the field current

raedy07 (author)  steveastrouk1 year ago
the alternator when necessary charges a battery bank of 2 x 24V x 150Ah. i'd like to reduce charging current as batt. voltage reaches 28V. lets say batts are empty, there flows a max.charg.current of 75A. if i don't regulate field current, voltage will just keep rising, 30V, 32V, i don't know how high, but TOO HIGH. so how do i best do that, manually? rheostat idea no good?

The alternator's internal regulator should handle the current control for you, to limit the o/p voltage when charging finishes.

raedy07 (author)  steveastrouk1 year ago

should but i probably wouldn't b looking elsewhere if it would - it didn't, and was therefor removed. also d original rectifier. i replaced it with 3 external bridge rectifiers. here i d tropical marine environment automotive stuff doesn't last all that long and hence has to b adapted/improved :)))

Then the big voltage regulator is the way to go. Try the LT338.

http://www.linear.com/product/LT338A

PS. With a big heatsink...if this is in a "tropical marine environment", and its on a boat, could you mount the heatsink on the cooling water inlet on your boat ?

You don't use a variable resistor to regulate current that high use a variable current regulator.

First it would be a very big pot about the size of your fist and it would change the voltage.

You want something like this circuit only rated up to 5 amps.

https://www.instructables.com/id/Circuit-Testing/

raedy07 (author)  Josehf Murchison1 year ago

i see - would this speed controler do d job?

http://www.ebay.com/itm/New-10V-30V-10-100-15A-250W-DC-Motor-Speed-Control-PWM-HJ-A-1017/390818354561?_trksid=p2047675.c100011.m1850&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D1%26asc%3D20140602152332%26meid%3Dc0519e7f0338412dac65b6ad4009c178%26pid%3D100011%26rk%3D7%26rkt%3D10%26sd%3D201227091777

That is a PWM, it works like a switch being turned on and off at different speeds to control the speed of a motor.

I'm not sure what it would do as a regulator for an alternator except giving you a pulsing current.

-max-1 year ago

Also you would need a big and probably expensive rheostat to do that. The power lost in a resistor is equivalent to the voltage dropped across it, times the current through it, or V^2/R, or I^2*R, where I = current in amps, V = voltage in volts, and R = resistance in ohms.

A large adjustable voltage regulator would be a far better choice.

You fail to undestand the basic principles of generators and batteries.

As Steve pointed out: The produced current depends on your load.
No load and there will be no current.

The field current stabilises the output voltage under load, so more load and the field current goes up so the alternator can still produce the required voltage.
If you would limit this regulation the voltage would break down under load.