What is the Wattage Calculation for a DIY Generator with Multiple Coils?

[Alkemist]

Hello, I'm building a generator and have completed initial phase. I made coil with 300 turns, which gives me 12v when rotor is turned at 2000rpm, if I short the coil it gives me 16 amp.

1] Now I'm wondering what is the output wattage of the generator I built / described?
2] If I use 18, similar coils and I rotate the rotor at 2000 rpm as before, what could be the output/wattage I may get from the generator in ideal conditions?

Any help?

Thank you.

 

[Simon Bridge]

It will depend on the load - simplest is to measure it.

 

[PhysicoRaj]

If you want the maximum wattage, you will have to multiply 12v ( if you are sure it is turned at 2000 rpm only) by the maximum current through it. But shorting will not give you the maximum wattage. Because by doing so, the voltage appears on the coil of the generator. So first measure it's internal resistance, which is equal to the load you will have to use in order to measure the max watts or current. (Just applying max power transfer theorem). Hence by that theorem if R is the int. resistance of the generator coil, the max power is v²/4R which is 36/R in your case.

 

[PhysicoRaj]

If I use 18, similar coils and I rotate the rotor at 2000 rpm as before, what could be the output/wattage I may get from the generator in ideal conditions?

You mean to say 18 coils each of 300 turns? All mounted to the same shaft in series? Then the o/p voltage will increase, also the internal resistance, whose consequence is explained by V²/4R.

If you are connecting 18 coils parallel, the internal resistance decreases (which is favourable) and I think voltage remains same but power again increases, as is obvious from- V²/4R.

 

[Alkemist]

Thank you Simon and Raj.

I wanted to get an Idea of the output before I make another coils, as it is very hard to wind them by hand. Took almost 4-5 hours for one coil and my hands were cursing me.

The resistance for 300 turns coil shows me 0.8 ohm. Also does the output wattage depend on the way I connect them [parallel or series]?

 

[Alkemist]

It will depend on the load - simplest is to measure it.

Can't we calculate something like max wattage a generator can produce based on open voltage and short circuit current output?

 

[Simon Bridge]

You can - but the results will likely be misleading.
Much the same way as you get a misleading idea of output power from a battery from the short-circuit current.
The short-circuit current depends on the internal impedance from all that wire in coils, ... you could work out what that power is based on $P=I^2R_{int}$ or something but that would be the power dissipated as internal losses. You need to know the power that can be delivered to a load.

Try it and see - use different loads and use $P=VI$ and see what happens.

You could get a ballpark figure off faraday's law with the rpm of what the generator could, in principle, supply ... but it really depends on the power that is supplied to whatever turns the handle.

 

[Alkemist]

You can - but the results will likely be misleading.
Much the same way as you get a misleading idea of output power from a battery from the short-circuit current.
The short-circuit current depends on the internal impedance from all that wire in coils, ... you could work out what that power is based on $P=I^2R_{int}$ or something but that would be the power dissipated as internal losses. You need to know the power that can be delivered to a load.

Try it and see - use different loads and use $P=VI$ and see what happens.

You could get a ballpark figure off faraday's law with the rpm of what the generator could, in principle, supply ... but it really depends on the power that is supplied to whatever turns the handle.

Thank you. Now I get it. Only way to know is to build it and test it throughly.

 

[PhysicoRaj]

The resistance for 300 turns coil shows me 0.8 ohm.

Rotating this 300 turn coil at 2000 rpm generates 12v (as you have stated) so V=12v; the load you have to connect in order to draw maximum power is same as the internal resistance so R=0.8Ω; then the max wattage is
P_m=V²/4R
=(12)²/4*0.8
=45W.

You also told about connecting 18 similar coils. Then if in series, resistance will be 18 times, that is 18*0.8=14.4Ω but o/p voltage will change, Iam not sure if that will also become 18 times so if you connect 18 coils measure the voltage.
If they are connected parallel, R=0.8/18=0.0444..Ω. But then Iam not sure about the voltage which you will have to measure.
Any way the formula V²/4R where R is the effective internal resistance will serve you through out.

 

[PhysicoRaj]

Also does the output wattage depend on the way I connect them [parallel or series]?

I think connecting in series will give you more voltage but the drawback is that the internal resistance increases hence power decreases, while in parallel, the voltage remains same but resistance decreases so power increases.
So you will have to be sure for what purpose you are using the gen. at 12v..?

 

[PhysicoRaj]

but it really depends on the power that is supplied to whatever turns the handle.

Yes. All these calculations are based on the fact that the generator operates at 2000 rpm.

 

[Simon Bridge]

Thank you. Now I get it. Only way to know is to build it and test it throughly.

You can tell an engineer: "build me a generator to output yay much power to my load at this much turn rate." and expect to get one. The engineer may have some extra questions for you about the use of the generator.

You can buy generators that are rated at a particular power output - the rating will be assuming some standard use for that kind of generator. The manual will warn you not to expect the rated power for all uses.

When you build your own generator by putting bits together to see what happens, the question becomes quite open ended. You can use Faraday's Law to get you a $V_{peak}$ and thus a $V_{rms}$ for the open circuit case - then figure out the current for different loads/usages.
That will give you an idea of what to expect from your generator.

At some point you will need to measure things to work out the detailed performance.

 

[Alkemist]

You can tell an engineer: "build me a generator to output yay much power to my load at this much turn rate." and expect to get one. The engineer may have some extra questions for you about the use of the generator.

Here that engineer is myself. I'm building 100% cog less generator, which one day will be used as wind generator. I was able to get around 20v out of that coil after reducing the air gap. Though wind won't be able to generate that much voltage as the revolutions will be pretty low.

 

[Simon Bridge]

<Puzzled> these are odd questions for an engineer to ask in a general community.
Just to be clear: in NZ the word "engineer" is legally reserved.

An engineer of the sort who would reasonably be commissioned to build a generator would be a specialist electrical engineer... I think it would help us to help you if we knew your education better, just so we don't short change you.

 

[Alkemist]

I think connecting in series will give you more voltage but the drawback is that the internal resistance increases hence power decreases, while in parallel, the voltage remains same but resistance decreases so power increases.
So you will have to be sure for what purpose you are using the gen. at 12v..?

For simplicity let's assume it's single phase generator. All coils are in phase and in series. This gives me resistance of 0.8 * 18 = 14.4 ohm. Current = 16 Amp

Now internal Power dissipated = 16*16*14.4 = 3686.4 watt when the generator coil is shorted. [Am I right?]

Now let's do parallel arrangement of all coils. V = 12, R = 0.044 Ohm I = 16 Amp in each coil. Which makes total current of 16*18 = 288 amp if shorted without load. This gives 288*288*0.044 = 3649.536 watt [again is this right?]

I'm taking current in calculation, as I know what is the amount of current when I short it.

Both calculations are little off from each other. 3686.4 watt != 3649.536.

But I think that is the power generated by either Parallel or series arrangement of the coils.

Can we consider this as the max power output this generator can deliver?

 

[Alkemist]

<Puzzled> these are odd questions for an engineer to ask in a general community.
Just to be clear: in NZ the word "engineer" is legally reserved.

An engineer of the sort who would reasonably be commissioned to build a generator would be a specialist electrical engineer... I think it would help us to help you if we knew your education better, just so we don't short change you.

I'm Software Engineer, but not Electrical Engineer, though I think I've more curiosity of Electrical Engineering [The Technology, which runs the world].

 

[PhysicoRaj]

For simplicity let's assume it's single phase generator. All coils are in phase and in series. This gives me resistance of 0.8 * 18 = 14.4 ohm. Current = 16 Amp

Now internal Power dissipated = 16*16*14.4 = 3686.4 watt when the generator coil is shorted. [Am I right?]

Now let's do parallel arrangement of all coils. V = 12, R = 0.044 Ohm I = 16 Amp in each coil. Which makes total current of 16*18 = 288 amp if shorted without load. This gives 288*288*0.044 = 3649.536 watt [again is this right?]

I'm taking current in calculation, as I know what is the amount of current when I short it.

Both calculations are little off from each other. 3686.4 watt != 3649.536.

But I think that is the power generated by either Parallel or series arrangement of the coils.

Can we consider this as the max power output this generator can deliver?

This is the power dissipated in the coil of the generator, not in the load. When you connect the load, the current reduces, hence also the power. The power drawn by the load depends on the load and the output voltage. Hence shorting is not a good way to measure wattage.

The best thing is to first get an idea for what use you will put the generator to (lighting or heating etc.) and decide the output voltage. With that done you can check the wattage needs of your load and design the generator according to that. If you go the other way around, it will be trickier.

 

[Alkemist]

The best thing is to first get an idea for what use you will put the generator to (lighting or heating etc.) and decide the output voltage. With that done you can check the wattage needs of your load and design the generator according to that. If you go the other way around, it will be trickier.

Thank you for the inputs. I guess rechargeable battery would be best load for this generator. But for calculation/design let's say the load requires 25V, 100Amp and have resistance of 0.005 ohm. How should I proceed for this case?

 

[Alkemist]

You can tell an engineer: "build me a generator to output yay much power to my load at this much turn rate." and expect to get one. The engineer may have some extra questions for you about the use of the generator.

Do we have any book or online resource, which I can refer to build such generator as I mentioned in earlier post?

 

[PhysicoRaj]

But for calculation/design let's say the load requires 25V

If you can produce 25v with your gen and still have the internal resistance at 0.8Ω, then
P_max=V²/4R
P_max=195.3W
That's all right, on course. The generator is upto it's mark. All depends on the load now.. Ok let's see..

...have resistance of 0.005 ohm.

So your load has a resistance 0.005Ω. But this is very much small compared to the int resistance of your generator. So, much of the voltage appears on the generator coils leaving only a small voltage on the load.
By voltage divider rule, voltage across your 0.005Ω load is:
V_L=VR_L/(R+R_L)
V_L=0.155 volts!
The power drawn by the load is
P_L=V_L²/R_L
P_L=4.8W!
See how low the values are! Reason: load is smaller than the internal resistance.
It's very important that you match the load and int resistance. See: http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem
Your load resistance must be greater than or equal to the internal resistance (or source resistance).

 

[Alkemist]

Understood. So if I increase the resistance of the load by adding a resistance in series, it will change the calculations. Wouldn't that reduce the amount of current passing through load?

Another thing is, I was looking for some DC motors online and found one interesting one. It's configuration is :

Battery: 14 Cell / 51.8V
RPM: 150kv
Max current: 190A
Watts: 9800w
No load current: 51.8V/5.2A
Internal resistance: 0.011 ohm
Winding: 8T
Stator Pole: 24
Motor Pole: 20
Suggested ESC: 250A 14S Compatible

How can this motor draw 9800 w? I*I*R gives me 190*190*0.011=397.1 w or 52*52/0.011 gives me something unreasonable. I must be missing something in my calculations or have poor understanding of how this works [and looks like both things are correct].

 

[PhysicoRaj]

So if I increase the resistance of the load by adding a resistance in series, it will change the calculations. Wouldn't that reduce the amount of current passing through load?

Yes it will and also the power to be supplied to the load. It is simply a current limiter but not an efficient way to transfer power.

Another thing is, I was looking for some DC motors online and found one interesting one. It's configuration is :

Battery: 14 Cell / 51.8V
RPM: 150kv
Max current: 190A
Watts: 9800w
No load current: 51.8V/5.2A
Internal resistance: 0.011 ohm
Winding: 8T
Stator Pole: 24
Motor Pole: 20
Suggested ESC: 250A 14S Compatible

How can this motor draw 9800 w?

You are right. I don't get to see how they put those. Where did you find this? I think the motor along with the stator have to be considered while calculation. They have used multiple poles and the calculation and ratings might have been in a different way.

 

[Alkemist]

You are right. I don't get to see how they put those. Where did you find this? I think the motor along with the stator have to be considered while calculation. They have used multiple poles and the calculation and ratings might have been in a different way.

This is one of the smallest and highest performing machine available in the market for now. Called as Turnigy Rotomax which outputs about 150CC engine. More details are @ http://www.hobbyking.com/hobbyking/...Max_150cc_Size_Brushless_Outrunner_Motor.html

There are other types also available, which runs on Hallback Array principles, but not available commercially yet.

I can only see V*I = 98xx Watt, but still it's very very high, don't you think?

 

[Alkemist]

I tried to hook up 13W and 7W bulbs to the generator. I know my coil shows 12 V in open circuit situation and 16 Amp when short circuited, but when I hooked up these bulbs each time the meter showed me nothing for current and voltage. So I wonder why can't the generator lit up the bulb not to full strength but just a dim / sparkle?

I don't know the resistance of the bulbs, but they sure say that 120V, 13W and 120V, 7W.

 

[PhysicoRaj]

I can only see V*I = 98xx Watt

You can't take into account their value of internal resistance of 0.011Ω. Because their rating of 9800W=52*5.2 does not match with it. 52/5.2=10Ω.

but still it's very very high, don't you think?

Yes. Approx.10kW

 

[PhysicoRaj]

I tried to hook up 13W and 7W bulbs to the generator. I know my coil shows 12 V in open circuit situation and 16 Amp when short circuited, but when I hooked up these bulbs each time the meter showed me nothing for current and voltage. So I wonder why can't the generator lit up the bulb not to full strength but just a dim / sparkle?
I don't know the resistance of the bulbs, but they sure say that 120V, 13W and 120V, 7W.

The rated wattages are for the rated voltages (rms). 13W @ 120V means it has a resistance of 120*120/13=1.1kΩ
1.1kΩ against 12V generator means a power of 12*12/11077.6923=0.13W!
By similar calculations you get to know the 7W bulb draws a power of 0.07W!
This will not even heat the filament let alone light it!
Try bulbs with ratings closer to your generator output voltage. You can throw in a 120V - 12V transformer (whose watt rating is more than the wattage of the bulbs) and get a decent output to power your bulbs.
By the way, what are you using to drive the gen?

 

[Alkemist]

The rated wattages are for the rated voltages (rms). 13W @ 120V means it has a resistance of 120*120/13=1.1kΩ
1.1kΩ against 12V generator means a power of 12*12/11077.6923=0.13W!
By similar calculations you get to know the 7W bulb draws a power of 0.07W!
This will not even heat the filament let alone light it!
Try bulbs with ratings closer to your generator output voltage. You can throw in a 120V - 12V transformer (whose watt rating is more than the wattage of the bulbs) and get a decent output to power your bulbs.
By the way, what are you using to drive the gen?

To turn it, for now I'm using muscle power for 30-60 sec duration, will set up another motor/belt for that, when I wind more coils.

I don't think if I use 120V - 12V transformer, it will help lit the bulb. If it is not lighting the bulb by 12v o/p of generator, it won't light if I just step up that volt to 120v. It will never work. If it does, it will be Magic!

 

[PhysicoRaj]

I don't think if I use 120V - 12V transformer, it will help lit the bulb. If it is not lighting the bulb by 12v o/p of generator, it won't light if I just step up that volt to 120v. It will never work. If it does, it will be Magic!

Who told you! It will!
I once had to do a project on this in my school and I had powered a 240V 25W incandescent bulb from a 12V 66Ahr car battery through a rotary converter and a similar transformer of 15W rating. That's why it's rightly called a 'Power Transformer'!
You generate a max of 45W (as previously calculated) So you can light up six 7W 'or' three 13W bulbs, by a 12-120 trans of 50W rating!

 

[Alkemist]

I don't know how it worked. For me it didn't. May be you used Battery, that's why. But if you use generator of 12V, you won't be able to lit that bulb for sure. I have done that experiment. I had a bicycle dynamo which have me 20V o/p. I stepped up it to 240V and it didn't lit the bulb.
I think, using battery as power source and generator as source makes lot of difference.
Inverter works different way. Also you can draw more or less power out of battery but then it discharges accordingly, whereas generator doesn't behave that way.

 

[Alkemist]

Who told you! It will!
I once had to do a project on this in my school and I had powered a 240V 25W incandescent bulb from a 12V 66Ahr car battery through a rotary converter and a similar transformer of 15W rating. That's why it's rightly called a 'Power Transformer'!
You generate a max of 45W (as previously calculated) So you can light up six 7W 'or' three 13W bulbs, by a 12-120 trans of 50W rating!

If you get chance, try the same experience with small 12v generator as source and not DC battery, you will get a surprise.

 

[PhysicoRaj]

It all depends on how much power the source is ready to give. Your generator ousts 45W and this is more than sufficient to light a 13W bulb if you provide it with the correct rated voltage. Check if the bulb's not burnt out!
You are still turning it by hand, at 2000rpm, may be you are using a set of gears,... I don't know but you need to turn the generator so that it is able to give out the required power at the stated voltage. It can't be a mistake from the 'transformer addition' side.

 

[Alkemist]

It all depends on how much power the source is ready to give. Your generator ousts 45W and this is more than sufficient to light a 13W bulb if you provide it with the correct rated voltage. Check if the bulb's not burnt out!
You are still turning it by hand, at 2000rpm, may be you are using a set of gears,... I don't know but you need to turn the generator so that it is able to give out the required power at the stated voltage. It can't be a mistake from the 'transformer addition' side.

In case of battery 12 V battery, can you say what's maximum o/p it can give? 12V Battery can give give o/p of 1000W or may be more, but that would be for short duration. If you draw only 10W, it will last longer.
In case of 12V Generator it cannot give more o/p like 1000W for the same load which we used for battery test, by adding inverter to it.

And yes, the bulb is not burnt. I'm using it at night time. I've gears in place, which ratio of almost 1:60.

 

[PhysicoRaj]

In case of battery 12 V battery, can you say what's maximum o/p it can give? 12V Battery can give give o/p of 1000W or may be more, but that would be for short duration. If you draw only 10W, it will last longer.

Batteries are not rated for power but in the amount of charge they contain. I used a 66Ah battery which is nothing but 66Ah*12V=792Wh. You are right that they can give large powers but for short duration. But in case of a generator, you are continuously supplying that power by mechanical means. A generator doesn't have stored energy does it? As long as you are putting 45 joules a second into the gen, it will give out that 45 joules in a second to the load. The transformer only steps the voltage so that the load is able to receive this power. Conservation of energy, you see.
Here's a quote by Simon:

... but it really depends on the power that is supplied to whatever turns the handle.

That's the point.

Theoretically, everything is right. Practically, there must be something faulty with the apparatus or procedure in your experiment, which I can't see.

 

[Alkemist]

I understand that, but I got little confused in between because of Battery and Generator conversation we had and your Bettery/inverter experiment.

So to make sure we are on same page, I'd like conclude this fact considering we use same loads for both cases and ideal sources / no internal resistances:
Battery Case: 12V 66Ah battery can supply 12x66=792w for an hour or it can supply 1584w for half hour or 3168w for 15 min. This holds true even if we use inverter.
Generator Case: 12V 16A generator can only provide 192W of power continuously. Even if you step up or step down, or use inverter [on DC o/p] it will still deliver same wattage.

Also : Consider that a lamp of 13w needs 120V and I use my 12V generator as source for it. If 12V cannot lit the bulb, then it won't lit, even if I step up my generator to 120V.

 

[PhysicoRaj]

If 12V cannot lit the bulb, then it won't lit, even if I step up my generator to 120V.

No, that's not true. Power consumed is a function of both the voltage and the load resistance. P=V²/R

What I am telling you is the generator supplies the same wattage regardless of the voltage, but not that the load will consume the same wattage if you change the voltage across the load. The generator supplies that much power which the load 'asks'. And the load 'asks' more if you apply more volts. The rating on the bulb means the same. 13W@120V means - "This bulb 'asks' for 13W of power from the source if applied with a voltage of 120V."

As for your bulb not lighting, I have thought of that. You measured the output volt of the gen with a multimeter? Was that the rms value? Or the peak? If 12v was peak value, then the stepped up 120v is also peak and not rms. And the bulb is rated for 120V rms! 120V peak means you get a voltage of 85V rms! It is simply like applying 85v dc to the bulb. Wait, I think I have calculated the resistance of the bulb in a previous post...
Here it is - 1.1k for 13W
Ok..
P=V²/R
P=85*85/1100
P=6.5W
Maybe that's why it's not lighting.

Therefore, make sure what type of voltage you have measured. If it's the peak, then you will have to change the o/p of gen from 12v to 17v. That will give you 120V_rms on the bulb and it should be doing:thumbs: