Battery charging with PMGs

[powerbuoy]

Hi,

I read some of the threads on battery charging and permanent magnet generators. Just did not find the answer to the following:

Let's say my PMG has a no load voltage of 50V at 800RPM. I connect it to a battery bank

and the voltage will be pulled down by the battery to 12 or 13V. At this point in time the charging current is 2Amps. What happens to the current if the RPM increases to 1000? The battery should still clamp the voltage down to 12V and the generators internal resistance is constant ... the amps should not change or do they? I figure that something has to give but am not quite grasping it.

Thx,

Powerbuoy

[Flux]

If you have 50v from the generator and a 12v battery you have a voltage difference of 38v and if you have 2 amps the generator will have a resistance of 38/2= 19 ohms.

Now if you run at 1000 rpm you have 50 x 1000/ 800 v = 62.5v

You now have a voltage difference of 62.5 -12v or 50.5v.

From this and the resistance found previously current 50.5/19 = 2.66A

Flux

[Opera House]

The real issue is that you are creating more heat in the generator than power into the battery.  You should use a switching regulator to match the generators best operating point to the battery voltage.

[powerbuoy]

The battery is fixed at 12V, ... the generator (single phase winding) has 12V across its terminals as well since these are paralleled (or screwed onto) with the battery. Where would I measure the 50.5 or38V differential? It must be internally in the generator ...do I have to see the single phase coils as an RI network (resistive - inductive)? Where does the voltage division take place?

Powerbuoy

[Flux]

I thought that by pmg you meant a permanent magnet dc generator. The internal emf will be the same as the no load terminal voltage. The difference between the emf and the battery voltage is forcing the current through the circuit resistance.

Now you tell me the thing is a single phase alternator ( I assume you include a rectifier before you connect it to a battery otherwise you will have fun)

If the alternator has inductance as well as resistance you will have to use the impedance value instead of resistance.

To be honest, if it has significant inductance you will never be able to do simple sums to find the current. If the reactance is high enough the current may remain constant with speed as the inductive reactance rises with frequency ( and  speed).

If you are already confused by the simple resistive case then it is beyond me to explain it simply here.

Flux

[powerbuoy]

Flux:

It is actually a permanent magnet alternator. The unit has 18 poles, is single phase and was running on a lathe at 300RPM for a while. Open circuit volts are 36VAC, shorted

Amps are 4.8  Rectification to DC is no problem, I guess I'll just use a bridge rectifier and start pumping.

I believe that the case is not that simple at all. True, the question is simple, the answer might not. The permanent magnet alternator is basically a coil which can be divided into a resistive and inductive portion. This "RL" portion is now being connected in parallel with the battery. I start running the lathe. The generated voltage E is now being dictated by the flux density x conductor length x magnet speed x the number of conductors. This simple formula tells me exactly what I have to assume in terms of voltage at a given RPM. If I bring the stator impedance into the equation I will also be able to figure out what the inductive current should be. I double the RPM -> I'll double the voltage ... no problem here. However, once I connect to a battery I'll have a natural voltage sink in the circuit which will keep the generated voltage at constant. Therefore, one variable (E) just became a constant. The generator resistance (R) should stay constant as well. Therefore my question: Will there be an amp rise if I raise the RPM? Or will the ampacity stay pretty much the same but I'll see higher losses on the generator due to the inductive reactance?

Thx again,

Powerbuoy

[powerbuoy]

opera:

I would like to keep it simple and leave the electronics out. But, if you have a suggestion for a switching regulator manufacturer I'll be more then happy to look it up.

Thx,

Powerbuoy

[oztules]

Powerbouy,

 the answer to your original question is yes

If you increase the rpm from 800 to 1000 rpm, more power will be generated. ie W increases with rpm.

Yes the internal resistance will remain roughly the same (not worrying about temp at this stage) so assume for now R= constant

Yes the batt voltage will remain essentially the same (use this as an assumption not real, but for this case we will assume it remains the same) so assume E= constant:

So if more power is generated, and the losses are dived up proportionatly up amongst the original loss areas evenly (yet another assumption for these purposes), then

If W=ExI then as W increases with rpm increase, so to will I. So yes the current into the batteries will increase as rpm increases, and voltage will remain stable.

Now the assumptions...... For small changes in rpm , they hold good. As higher differentials from the original settings become apparent, these assumptions will start to fall apart. At the extremes for any particular system, the accumulated loses will lead to over temp, over voltage charging, over current charging, blown up components or any combination of the above.

Within the design parameters of any single system, the yes holds good.... more rpm= more amps.

Operahouse is right in as much if you wish to get sexy with it and improve performance outside the useful operation range for a set system, PWM would offer better power return on your system.....at what cost?...depends how big.

There is no voltage divider as you pondered up above,  So If we have to keep the voltage constant..............hmmmm....

If you think of the start point as that where the alternator voltage is the same as the battery I=0   E= batt. As we speed up and increase the power, E stays the same..... how's this?

If W=IxE and E=IxR then W=Ix(IxR) so as we increase the power (W), and R is treated as a constant, then the I must increase. so the power can increase even without worrying what happened with the E.

In the real world it is a very complex relationship, and many other variables come into it, but I think that answers your quandry in simple terms, and without taking into account things which would muddy the waters so to speak....inductance, heat,ambient temp,wire losses rectifier losses etc. etc.

Is this what you were after?

...........oztules

[powerbuoy]

Right on ... I am seeing more light now. All these inputs are appreciated!!!

[Flux]

You are confusing emf with terminal volts. Emf is as you say the voltage induced in the coils by the magnets and this does not change when you add load. The terminal voltage does fall to the battery voltage.

It is the difference between the emf and the battery volts that is the effective voltage driving current through the internal impedance of the machine.

If it is an axial machine with no iron core then the thing will be resistive.

If you have an iron core there will be some inductive reactance. If the inductive reactance is greater than the resistance it's output will fall below the straight line with speed and it will eventually limit to constant current.

If it is an air gap machine the output will rise linearly with speed, with iron in the stator anything could happen.

Flux

[oztules]

Oh dear me

I'm hoping this is a difference of perspective, otherwise we have the idiot v's the flux, and that leaves me not just down the hole. but at the very far end of the hole... so here I go.

I was trying to avoid dealing with the emf in the coils, as I felt it added ideas which did not need to be introduced. I was trying to describe increase in rpm giving an increase in power.

The original question was ""At this point in time the charging current is 2Amps. What happens to the current if the RPM increases to 1000? The battery should still clamp the voltage down to 12V and the generators internal resistance is constant ... the amps should not change or do they? I figure that something has to give but am not quite grasping it.""

soooo.....

I had defined the E as the battery voltage ie :

"Yes the batt voltage will remain essentially the same (use this as an assumption not real, but for this case we will assume it remains the same) so assume E= constant"

I didn't want to involve the induced voltage in the coils, as it is not necessary to understand it in order to measure the Watts out. I have assumed E= terminal voltage, as if we measure I out and terminal voltage(in this case defined as  E), we can calculate power at a given rpm.

By assuming E=constant, we no longer needed to look at its potential variance(coil EMF), but rather the measurable under load value....which we nailed as fixed for this discussion.

I fixed R  because I wanted to simplify the model. I didn't want to introduce inductance and a host of other possible contributors to the discussion, as their relevancy to it ceased to exist once we defined R as fixed for this purpose ie:

"Yes the internal resistance will remain roughly the same (not worrying about temp at this stage) so assume for now R= constant"

So given the assumptions that were laid down, working E out of the equation was the best way of avoiding having to discuss forcing volts etc, which in the end don't matter to the alternator user, (yes they matter to the alternator maker for sure), he seemed primarily concerned with output changes corresponding with rpm input changes. This by necessity would seem to dictate that we deal with what is directly measured at the output.

then there was:

""There is no voltage divider as you pondered up above,  So If we have to keep the voltage constant..............hmmm""

It's here that I think I made it confusing...

"" If you think of the start point as that where the alternator voltage is the same as the battery I=0   E= batt. As we speed up and increase the power, E stays the same..... how's this?""

Here E was defined as Battery volts, not EMF in the coil, and fixed it's value at E=batt.

So this still works I think

   "If W=IxE and E=IxR then W=Ix(IxR) so as we increase the power (W), and R is treated as a constant, then the I must increase. so the power can increase even without worrying what happened with the E."

this avoids having to talk about how the alternator works, but instead shows how to explain what we measure at the output, and that we don't have to agonise over what happened in the coils, what happened with the voltage, or how it got there.

I don't think this absolves my sins completley, but hopefully shows where I thought I was coming from.

One feels a rouge asteroid could come my way soon

...........oztules

[Flux]

Oztules

Don't worry that was not directed at you. For some reason replies to things here usually end up under the post you reply to but at times they come several posts later or even at the end.

It was powerbuoy that seemed not to get the concept that emf remains unchanged but the terminal voltage falls due to internal volt drop.

It was a little confusing to call battery volts E but you did define it.

Flux