alternator field current

[markxengineerin]

I've been experimenting with a homemade alternator controller, and found this relationship for my Hamilton Ferris small frame alternator (at idle).


I am interested to know how it looks for other alternators- Balmar, large frame, 24 and 48V systems, etc. Trying to get a feel for what's out there.

So, if you know your Alternator's peak field amps at max output and feel like sharing...

[SailingHarmonie]

Certainly goes to demonstrate just of the reasons why field current alternators are so inefficient. Even before considering any other loses, you throw 7% of your power production away just magnetizing the field coils...

PMA's rule!

[team karst]

Originally Posted by SailingHarmonie

Certainly goes to demonstrate just of the reasons why field current alternators are so inefficient. Even before considering any other loses, you throw 7% of your power production away just magnetizing the field coils...

PMA's rule!

2 or 3 Amps at what Voltage?? not seeing field power shown...

[s/v Jedi]

Originally Posted by team karst

2 or 3 Amps at what Voltage?? not seeing field power shown...

Exactly

[HeywoodJ]

At least in the past (just have not checked their website recently) all Balmar 12/24V alternators had a 2.4Ω rotor. At 14.4V that works out to a 6A maximum field current. Their 80-series 50, 65, and 100A alternators all listed 6A as full field current. The ratio of field to output thus varied by at least a factor of two just across that family of alternators.

One crucial variable missing in your data set is the alternator RPM. You can full-field an alternator at zero RPM and you will get zero output. At 1000RPM you can full-field and only get moderate output, at 5000RPM you might get full rated output.

Alternator fields are fed (simplified) from the positive to the negative of the alternator, so field power dissipated is field current * output voltage. More or less.

[Dennisb2]

On a earlier boat I had a Prestolite 85 amp large frame alternator controlled by a adjustable rheostat which allow you to dial in the alternator output current to control battery charging (early 3 step charger circa 1990) The field to the alternator was a steady DC voltage which varied depending upon the resistance setting on the rheostat. RPM would also change the output.
The chart shows as I see it a plus width modulated signal to the alternator which I assume would be a pulsed DC signal. I am assuming that the alternator RPM was constant and higher than the engine rpm.
My concern would be the generation of a EMI signal. It would be interesting to see an O-scope signal of both the field input and the alternator output with the pulsed field signal.
My experience has been with a steady DC voltage to the field on the alternator.
My goal was to minimize signal noise on the DC source to all of the DC electronics on the boat.

[Stu Jackson]

Originally Posted by HeywoodJ

>>>


One crucial variable missing in your data set is the alternator RPM. You can full-field an alternator at zero RPM and you will get zero output. At 1000RPM you can full-field and only get moderate output, at 5000RPM you might get full rated output.

>>>

My thoughts exactly when reading the OP.

[s/v Jedi]

No, the field current comes from the regulator and is … regulated. It means the voltage isn’t fixed and you can’t say 6A x 14.4V = 85W loss.

[PippaB]

Originally Posted by s/v Jedi

No, the field current comes from the regulator and is … regulated. It means the voltage isn’t fixed and you can’t say 6A x 14.4V = 85W loss.

You can actually say that, using Ohm's Law. The rotor resistance is relatively fixed (the impedance does change slightly based on magnetic flux). If 6A is flowing through a 2.4Ω resistance then the voltage drop must be 14.4V. E=IR. At partial current then yes, the regulator represents part of the "resistance" in the circuit. Depending on how the regulator is designed it may dissipate the remaining power (old school) or it may be more efficient than that. But, you can still calculate the power dissipated by knowing the field current and rotor resistance/impedance.

Even from the provided data set you can get pretty close, although not knowing the exact voltage means you will only have an approximation, but the current at 100% PWM cycle is going to be provided at full input voltage, from which you can estimate impedance and work backwards. 4A at 14.4V is a 3.6Ω, at 12V a 3Ω rotor. Give or take a bit easy enough to calculate field power for the rest of the data set.

[s/v Jedi]

Originally Posted by PippaB

You can actually say that, using Ohm's Law. The rotor resistance is relatively fixed (the impedance does change slightly based on magnetic flux). If 6A is flowing through a 2.4Ω resistance then the voltage drop must be 14.4V. E=IR. At partial current then yes, the regulator represents part of the "resistance" in the circuit. Depending on how the regulator is designed it may dissipate the remaining power (old school) or it may be more efficient than that. But, you can still calculate the power dissipated by knowing the field current and rotor resistance/impedance.

Even from the provided data set you can get pretty close, although not knowing the exact voltage means you will only have an approximation, but the current at 100% PWM cycle is going to be provided at full input voltage, from which you can estimate impedance and work backwards. 4A at 14.4V is a 3.6Ω, at 12V a 3Ω rotor. Give or take a bit easy enough to calculate field power for the rest of the data set.

So the regulator doesn’t use PWM?

[smac999]

Originally Posted by s/v Jedi

No, the field current comes from the regulator and is … regulated. It means the voltage isn’t fixed and you can’t say 6A x 14.4V = 85W loss.

in bulk it would be fixed. at the current battery voltage (minus v loss from battery to regulator input). field current would drop in absorb and float.

[s/v Jedi]

Originally Posted by smac999

in bulk it would be fixed. at the current battery voltage (minus v loss from battery to regulator input). field current would drop in absorb and float.

It can’t be fixed to drop later. Fixed is fixed. Regulated is regulated.

How is the field current reduced?

Does everyone understand that the field winding is about a regulated current, not a regulated voltage.

I think the regulators use PWM.

[donradcliffe]

Modern regulators use PWM. Older ones used a power transistor. Even older ones used coils and contacts that would actually buzz when regulating.

[donradcliffe]

You can make your own energency regulator with incandescent light bulbs. A 25 watt nav light bulb will allow about 2 amps of field current, while a 10 watt bulb gives 1.2 amps. You are the regulator--when the battery voltage gets above 14 volts, switch the 25 watt bulb for the 10 watt bulb. If you need more power, use both bulbs in parallel.

[markxengineerin]

My concern would be the generation of a EMI signal. It would be interesting to see an O-scope signal of both the field input and the alternator output with the pulsed field signal.

I have done that- when the field current goes from zero to whatever it is with 12V applied, there is a small step in the battery voltage measured on my 12V bus bar. Small like 0.1 or 0.2V. I don't think it's enough to impact any boat electronics, which should all have some kind of signal conditioning on the inputs. I did not notice a big difference in the voltage output fluctuation when I changed the PWM frequency from 100hz all the way to 10khz, so I don't think it matters much.

That being said, I think it's possible to eliminate any step changes in the field current by using a very very fast PWM (hundreds of khz to Mhz) and having inductors and capacitors on the output. The chip I've chosen to use for my DIY regulator has these already and is usually used for buck converters, where this may matter more. I will be testing soon, and going to make another thread for feature requests, as I'd like to make this into an open source type product.

Short feature list for now:
Fully open source
Field current up to 12 amps
12V, 24V, 48V system all supported w/ no changes
Infinitely flexible charge cycles , alarms, settings, communication with other hardware, as you program it yourself or use common examples provided.
Current sensor and alternator temp sensor included
Cost target <$200