# Volts / Amperage / Watts / Battery usage to empty / conversions / please point in right direction

#### scatterbrainz

Joined Aug 5, 2020
4
Greetings Everyone,

I have a little project - I'm undertaking (personal knowledge if you will). I haven't ventured into circuit boards, circuits, and electronics before, I'm actually more of a programming person, but with extra time and idle hands - so why not.

I've included a little diagram of what I'd like to accomplish. Where I'm getting confused is the conversion between multiple things. When I look at small 2" speaker. It is 500mW to max 1W. So to me this indicates for the speaker to produce sound it requires a minimum of 500 milliwatts and tolerates up to 1 watt (500 mW variance). Also an LED bulbs is rated at 2W for 250 lumens. Then also is strip LED lights, with a ratting of 14.4 watts per meter so divide by 3 (average) for 4.4 watts per foot. SO if I add it all up I'm looking for 1w + 2w + 8.8w + indicators + circuit board = 12w (loose estimate).

So then I go look at batteries. I see 2aH which is 2 amp hours (? please confirm ?). So then if a wire has a tolerance of 1.0v to 1.9v and its carrying current through the circuit board to the given attachments which require X amount of watts to work? how does amp hours in battery power translate across to watts to turn the bulb on? Further how does one determine how long before the battery is dead? Assuming that draw for all items is constant?

IF someone can point in the right direction so i can discover the pieces of the puzzle - really appreciate it - hints / tips and advice also very much appreciated.

Best Wishes. .

#### jpanhalt

Joined Jan 18, 2008
10,068
A watt is power. A watt-hour is the amount of energy.

Thus, your battery is rated at 2 Ah. Let's assume its voltage is 4.2 V to begin and 3.8 V at the end. Call that an average of 4 V. So, 4 V x 2 A = 8 W. It can deliver that for 1 hour. Similarly, it can deliver 4 A for 30 minutes = 16 W for 30 minutes. And so forth.

But things are not that simple in reality. Amp-hour ratings are at a specific discharge rate, which is usually quite a bit lower than the maximum allowable discharge rate. In reality, the higher the discharge rate, the lower the A-h rating will be so the arithmetic given above must be modified to consider the Ah rating at each discharge rate.

• MrSoftware

#### MrChips

Joined Oct 2, 2009
21,634
Let's see if I can help.

Units, units, units!
You have to pay attention to units in order to get a handle on all of this.

Here are the units you are working with (symbol in brackets):

volt (V)
current, amp (A)
power, watt (W)
time, second (s)
time, hour (h)
energy, joule (J) 1J = 1Ws
energy, watt-hour (Wh)

What do these all mean?

Power = 1W = 1V x 1A

Hence if a 12V circuit draws 2A, power consumed = 12V x 2A = 24W

Suppose that a 12V battery has to supply this amount of power for 4 hours.
Energy supplied by the battery = watt x hour = 24W x 4h = 96Wh
The energy capacity of a battery is specified in amp x hour. But energy = watt x hour. So amp x hour cannot be unit of energy.
That is the way battery capacity is specified.

In this example, the battery has to supply 96Wh of energy at a constant 12V.
This is equivalent to saying the battery has to supply 2A @ 12V for 4 hours = 8Ah @12V

So when you see a battery capacity rating of 2Ah (which is often written as 2000mAh) this is saying that at a first-order approximation, the battery will supply
2A for 1 hour
1A for 2 hours
1/10 A for 20 hours

I say first-order approximation because this is not entirely true. If you attempt to take 10A then the battery will die before 2/10 hours (12 minutes) are over.

#### Dodgydave

Joined Jun 22, 2012
9,305
You're correct with Watts in total, but Watts is a product of Volts x Amps so depending on your battery voltage and the current the loads will draw the Watts will vary.

The battery will give you the Ah rating,
so ideally you need to first measure the currents in total, then you can select the best battery for your project to give you the length of time you need in Ah.

#### scatterbrainz

Joined Aug 5, 2020
4
You're correct with Watts in total, but Watts is a product of Volts x Amps so depending on your battery voltage and the current the loads will draw the Watts will vary.

The battery will give you the Ah rating,
so ideally you need to first measure the currents in total, then you can select the best battery for your project to give you the length of time you need in Ah.
So if I round up my watts per hour to 20, that means the batter would have to be 240 watts total for 12 hours to discharge at a rate of 20 watts per hour. Is this correct? but 240 watts isn't a measure of batteries (that I know of) > but then 240 watt battery discharging at a rate of 4 volts (is this per hour)? but then it converts it to amps, but the attachments are in watts so do you have convert it backwards through to correct?

#### Dodgydave

Joined Jun 22, 2012
9,305
What voltage are you working on, you need to add up in Amps not Watts..

#### scatterbrainz

Joined Aug 5, 2020
4
What voltage are you working on, you need to add up in Amps not Watts..
This is gonna sound seriously LAME ... but I really have NO idea. This is my very first attempt at going into circuits, circuit boards, power distribution, all that kind of stuff. I don't even have the circuit board sketched out yet to be made. I sort of reverse the process. I look at say a 2w LED bulb. OK it needs 2 watts to turn on. great, but that must be DC. If I go through a circuit board, I doubt very much I'm gonna put 2 watts of current through 1 circuit on the board. But I could be wrong. So here's an example of a speaker (specifications) it does not give me anything about voltage or amperage, only wattage. So how can I tell what voltage to send the speaker if I don't know what it accepts. ? Maybe there's a calculation somewhere?

#### MrChips

Joined Oct 2, 2009
21,634
So if I round up my watts per hour to 20, that means the batter would have to be 240 watts total for 12 hours to discharge at a rate of 20 watts per hour. Is this correct? but 240 watts isn't a measure of batteries (that I know of) > but then 240 watt battery discharging at a rate of 4 volts (is this per hour)? but then it converts it to amps, but the attachments are in watts so do you have convert it backwards through to correct?
No.

watt refers to instantaneous power.
watt-hour is energy.

You can consume 20W for 1 hour. Energy consumed is 20 Wh, not 20W per hour.
20W for 12 hours is 240Wh.

If you battery is 4V, then the battery capacity required is 240Wh / 4V = 60Ah.

#### MrChips

Joined Oct 2, 2009
21,634
Assuming a speaker is correctly rated at 2W. (This is a huge assumption because speaker may be rated in peak music power which means nothing to me.)

Typically, a loudspeaker has an impedance of 8Ω. Now, impedance is a measure of resistance to AC.
But let us assume that this is DC resistance.

Power = volt x amp
Power = amp x amp x resistance
Power = volt x volt / resistance

Since we know power and resistance, we can calculate voltage and current.

volt x volt = power x resistance = 2W x 8Ω = 16 VxV
volt = 4V
current = 4V / 8Ω = 0.5A

To verify, power = 0.5A x 0.5A x 8Ω = 2W

#### Audioguru again

Joined Oct 21, 2019
1,733
MrChips did not see the attachment for the cheap small 3W, 4 ohms speaker. The speaker has no audio spec's.
A speaker uses an AC signal, not DC from a battery. The battery can power an amplifier or a tone generator.

3W into 4 ohms might burn up the speaker if it is fed a continuous tone but if fed music that has 3W of momentary peak power but an average power of 0.3W then it will survive. The maximum momentary power of 3W into 4 ohms is produced by (the root of 3W x 4 ohms)= 3.46V RMS at a current of (3.46V/4 ohms)= 0.865A RMS. The maximum average power of 0.3W with music is produced by (the root of 0.3W x 4 ohms)= 1.1V RMS at a current of (1.1V/4 ohms)= 0.275A RMS.

#### MrChips

Joined Oct 2, 2009
21,634
I am just showing @scatterbrainz here how to calculate basic current and voltage if you know power and resistance.

If a resistor is 8Ω and rated at 2W, the most it is rated for is 4V or 0.5A.
If this were a real install, you would derate the wattage to 1W. This brings the maximum voltage and current down to 2.8V and 0.35A.

#### scatterbrainz

Joined Aug 5, 2020
4
Audioguru / MrChips - Thanks a bunch for inputs. The speaker is just an example, i don't know if its actually an appropriate one. I know that the speaker is going to be used for either a) a beep or b) a 5 - 10 word sentence. Audibly clear from 10 feet away. (I am still researching what type / power of speaker would be used), I do know it has to be small under 4" diameter. SO based on the information provided I could take 3w / 2 -> 1.75w (since it would not likely run at peak 3w power) provides a result of 2.65v and .66A which would be 660 mA. So the connecting wire to the speaker (HOT) would need to be supplied 660mA for the speaker to function, now here's what's not straight in my head. If I was to put something into motion, it would take a larger initial amount of "energy" / "power" then to maintain it at a given velocity provided that constants are employed (like a flat surface, same friction co-efficient for entire length driven, etc.etc.etc). Thus if I ran a car off a car battery (say alternator was broken), I may get 10 - 20 minutes out of the battery before its dead, and the car shuts down. So how do you convert for example a car battery is 130CCA (I'm assuming this is Cold Cranking Amps). It is a 12v and has a capacity of 6aH / 10h > SO does this mean the TOTAL 6 amp hours is discharged over 10 or that you can discharge 6 amps PER hour for every HOUR up to 10 hours. Which would give you a total of 60 amp hours? But going back to the little speaker. if it requires 660mA is that for 1 minute, 60 minutes? If I wanted it to sound for 12 hours, how do you equate time into that equation? I hope I'm making sense. (Its alot to wrap my head around ) - thanks again everyone.

#### MrChips

Joined Oct 2, 2009
21,634
You are overthinking this.
Forget about the a 130CCA car battery.

You can determine your requirements by simple tests and experimentation.
Take your boom box and play music at the desired volume. Measure the average current required by the boom box (not the loud speaker).
Let us assume that the boom box takes ½A @ 12V.
Now, suppose you want to play music continuously for 8 hours.
Battery capacity required = ½A x 8h = 4Ah

• scatterbrainz

#### djsfantasi

Joined Apr 11, 2010
6,692
...Take your boom box and play music at the desired volume. Measure the average current required by the boom box (not the loud speaker).
...
The wattage rating of the speaker doesn’t directly relate to your battery requirements. It is used primarily to match a speaker to the amplifier.

By directly, I mean various speakers will require different voltage/current from the amplifier. Thus, indirectly affecting the power used by the amplifier.

You need to select an appropriate amplifier and speaker combination, and measure the peak power required. Use this parameter when sizing your battery.