How to power via 24v

I can’t seem to find a 24v power supply. Ive found this
(https://uk.farnell.com/mean-well/rs-25-24/power-supply-ac-dc-24v-1-1a/dp/2816011?gclid=EAIaIQobChMIv-mNqYHW7QIVw7TtCh3lQALIEAQYBSABEgKo3fD_BwE&gross_price=true&mckv=stG9SXdgF_dc%7Cpcrid%7C459820069594%7Cplid%7C%7Ckword%7C%7Cmatch%7C%7Cslid%7C%7Cproduct%7C2816011%7Cpgrid%7C104445605501%7Cptaid%7Cpla-335119492449%7C&CMP=KNC-GUK-SHOPPING-SMEC-Whoops-Newstructure-31Aug2020-Desktop-Lo)
But I don’t think it’s what I’m thinking and what i require.

Basically, all i need is some external power (like a battery pack) that i can hook up to my 24v motor and power the arduino.

(Find electronics extremely confusing, after months of research i still do not understand volts,watts, amps, etc. I don’t get why 8 AA batteries produce 9 volts, yet a car battery can be 12 volts, there’s no way that approximately 13 AA batteries can power a car. Also, there’s the size difference. I’m clearly missing something very important and thinking it very basic.

You need to know the current rating (Amps or milliamps) for the motor. The power supply should be rated to match the required voltage with at-least enough current capability to power the motor.

Ohm's Law defines the relationship between Voltage, resistance, and current. Current = Voltage / Resistance.

Ohm's Law is a law of nature with man-made units-of-measure.

after months of research i still do not understand volts,watts, amps, etc.

When you take an electronics class, Ohm's Law is the 1st thing you learn. :wink:

Resistance is "the resistance to current flow."

There is an analogy of voltage being similar to water pressure, and electric current being similar to water flow. The main difference is if you break a water pipe the resistance goes to zero and you get water flowing out all over the place. If you cut a wire the resistance goes to infinity and current flow stops. And with no water-resistance nothing bad happens. Zero electrical-resistance across a battery or power supply is a "short" and bad things can happen.

We usually don't know the resistance of the motor, but there is a current rating. And, it takes more current at start-up and under full-load than when running freely with no load, which means there is more resistance with no-load.

Usually the voltage is (relatively) constant and the current varies with the load. for example, here in the U.S. 120VAC is always present at the power outlet but no current flows until you plug something in (and turn it on). A hair drier takes more current than a regular light bulb and if you plug-in two hair driers you'll probably draw too much current and blow a circuit breaker, and then voltage drops to zero and no current flows.

If you try to start a car with two 9V batteries, the voltage will drop to (almost) zero and the batteries will quickly drain.

Power (Watts) is calculated as Voltage x Current (Amps).

Find electronics extremely confusing, after months of research i still do not understand volts,watts, amps, etc. I don't get why 8 AA batteries produce 9 volts, yet a car battery can be 12 volts, there's no way that approximately 13 AA batteries can power a car. Also, there's the size difference. I'm clearly missing something very important and thinking it very basic.

Basic Electronics

Think of it like this:
Volts is the measurement across a battery. The battery provides current, measured in Amperes (Amps, or milliAmps for low power devices) to a load. (sometimes presented as water pressure, and flow in gallons per minute or similar).

How many amps will flow depends on the voltage and the load.
An ideal 1V battery will provide 1A of current into a 1 ohm load. That is Ohms Law; Voltage (V) = Current (I) x Resistance (R). V = IR and this can rearranged, as V/R = I, or V/I = R.
Say an ideal 9V battery was connected to a 5 ohm resistor. The current will be 9V/5ohm = 1.8A

Similarly you can divide Volts by desired current to determine a current limit resistor.
Say you wanted to limit the same battery to just 0.5V; 9V/0.5A = 18 ohm resistor.
We use this to determine current limit resistors for LEDs.
Say you had an Arduino, and you wanted to limit the current from a pin to 8mA so it wasn’t so bright.
The Voltage from the pin, across the LED, across the resistor, to Gnd will be 5V.
An LED has a current where it turns on, say 2.2V for a Red LED. This is Vf.
The 5V source will be split across the LED and the resistor. If the LED needs 2.2V, that means there will be 2.8V across the resistor. The current thru the LED and the resistor will be the same - there is only path for the current, so could it be otherwise.
So, resistor value needed = 2.8V/8ma (0.008A) = 350 ohm.
You may see this as (Vs - Vf/current = resistor. (5V - 2.2V)8mA = 350 ohm.
Or say you had a resistor and wanted to know how much current it would allow to flow:
(5V - 2.2)/200 ohm = 14mA (0.014A).

Capacity is how many mA can be produced by a battery over a time period.
An ideal 9V battery with 400mAH capacity could produce 0.4A for one hour into a load (like a 22.5 ohm resistor).
However, small battery internal chemistry limits the creation of electrons * to a smaller amount; if more are asked to be created than the battery can supply, the voltage drops. So a small square 9V battery will maintain 9V for smaller currents, like 10-20mA, and trying to draw more will not work - the battery may overheat.
A car battery on the other hand has different battery capacity and can create many more electrons, often expressed as Cold Cranking Capacity - this can be many hundreds of amps short term to turn the engine over while you start the car, or stretched out over an hour or two while you sit in the car and listen to music with engine off.

Watts are just a measure of the Voltage and current being used. Power = Voltage x Current, or P = IV.
You know V = IR, so can you do some substitutions to find other things. V=IR, V/I = R
P = IV, and V=IR, so P = I x IR, or I^2R. How many watts will a 350 resistor dissipate with LED above?
P = .008A x .008A x 350 ohm = 22.4mW, so part rated for 1/10W would be okay to use.

P = IV, and I = V/R, so P = V x V/R or V^2/R. 2.8V x 2.8V/350 = 22.4mA, so there a couple ways to determine power dissipation.

8 batteries connected in series add up voltage-wise. 8 x 1.5V (AA battery) = 12V.
AA batteries have a capacity of 2.2AH (2200mAH), so the current available will be limited to that, but at a higher voltage, say for driving a small motor.
If many, many, are connected in series, you can have 36 for 48 volts for powering an electric bike.
Take that string of 32 batteries (48V), and connect many strings in parallel, and the current available adds up.
10 strings at 2.2A = 22A for one hour.
Put a lot more in series, and many strings in parallel, and you can supply more volts and more amps → more power.
That’s what electrics cars do - big battery packs of small batteries in strings, with many parallel strings.

  • electrons are not really produced - they are charged up by voltage, and their charge jumps from electron to electron, and this is what gets referred to as current. The higher the voltage, and the more electrons present, determine how much current can flow into a load. Resistors keep current from moving too much. Capacitors store up charge, and keep direct current from flowing thru - but will let variations in current flow. Inductors do the opposite - they let steady current thru, but prevent variations from flowing thru.
    As an example, capacitors and inductors are used to make filters to use with speakers. Low pass filters let the low variations into the woofers for low notes, while high pass filters keep the low notes out of the tweeters for nice crisp high notes.

How many Amps at 24V do you need?

shazhazel:
I can’t seem to find a 24v power supply. Ive found this
(https://uk.farnell.com/mean-well/rs-25-24/power-supply-ac-dc-24v-1-1a/dp/2816011?gclid=EAIaIQobChMIv-mNqYHW7QIVw7TtCh3lQALIEAQYBSABEgKo3fD_BwE&gross_price=true&mckv=stG9SXdgF_dc%7Cpcrid%7C459820069594%7Cplid%7C%7Ckword%7C%7Cmatch%7C%7Cslid%7C%7Cproduct%7C2816011%7Cpgrid%7C104445605501%7Cptaid%7Cpla-335119492449%7C&CMP=KNC-GUK-SHOPPING-SMEC-Whoops-Newstructure-31Aug2020-Desktop-Lo)
But I don’t think it’s what I’m thinking and what i require.

Basically, all i need is some external power (like a battery pack) that i can hook up to my 24v motor and power the arduino.

(Find electronics extremely confusing, after months of research i still do not understand volts,watts, amps, etc. I don’t get why 8 AA batteries produce 9 volts, yet a car battery can be 12 volts, there’s no way that approximately 13 AA batteries can power a car. Also, there’s the size difference. I’m clearly missing something very important and thinking it very basic.

Eight AA batteries if 1.5 volt/cell produce 12 volts when all in series. You mention a 24 volt power supply which is very common however you omit the load so nobody will know how much current you need? It's not just about volts. Power = Volts * Current in a simple DC circuit. You need to understand the very basics. You do not learn to swim in the deep water.
Ron

Gxuys, i VERY HIGHLY appreciate your help! The only thing ive learn about electricity in the past few months is ohms law, but ive not understood anything about it - the only time ive understood it's relevency is when you want to find out I, R or V. i didnt think there was any other use for it.

This information is very helpful, reading it once has opened my mind but i did not understand some of it (my fault, not yours) - im sure im going to have to read it all over and over again for it all to make complete sense.

I cannot thank you enough for the help and almost "spoon feeding" me.

Before trying to power a large motor with "high voltage" try some smaller steps: power a LED -> small motor -> large motor. This way you will hopefully learn the basics "safely" - the low power mistakes are much less dangerous for you and the devices. And even in case of damage it is much cheaper. Also buying the wrong part for 10 cents is no problem; buying an expensive driver only to find it is not suitable is ... expensive mistake.

Since you have stated you are new to electronics it must be assumed that you have not learned what the rest of us have learned the hard way, which is that given enough power is available (24V could qualify depending on the PS current rating) is something goes wrong , a semiconductor device , even as small as a 2N3906 NPN transistor, can explode violently with a very loud bang and propell
plastic shrapnel clear across the room at high velocity or, alternately,!with enough current, a
small 8-pin IC like an LM555, can burst into flames
and glow red, melting the plastic breadboard.
This is one of the reasons companies require
safety glasses in the labs. Just sayin'...

A solderless contact on a breadboard can easily have ~ 1 Ohm. Not a problem for Arduino electronic until you try precision analog or pass some serious current (let say 1 amp) trough it. You don’t need any wiring error to melt the BB or connecting wires. It is easy to forget.

I use electronic contact cleaner and blow it out
with a can of Air Duster.

On the other hand, the breadboard contact might be what you'd expect:
(LCR METER: UNI-T UT612)