# Does a generator produce voltage or current?

I know that this question might sound very stupid for +90% of you, but my brain doesn't really get the electricity generation.

The thing that I don't understand is what does a generator produce, voltage or current when it spins? Sometimes I see that a generator can produce x amount of potential and y amount of current.

I'm starting from the basics of electricity. A spinning/moving magnet makes the electrons move in one direction, generating a potential due to the change in kinetic energy, right? So if one would connect a multi-meter at both terminals, one would read the voltage. I did this and it's easy to comprehend. Now things start to confuse me.

Sometimes, one can see on the specifications that a motor can deliver a certain current and this is what I don't understand. The current is related to the potential through Ohm's law, right? In other words, the current is dependent only on the resistance of a load (or wire) and the potential. Also, the current flows only if there is a voltage gradient/drop. Why is there the current specified on generators like this one?

What confuses me is this intrinsic combination of voltage-current on the generator. Can anyone help me to understand this? Thank you and sorry for dropping the IQ of this group.

• does a hydroponic pump produce pressure or flow? Sep 3, 2017 at 16:14
• Pressure. So if one adds the flow value on the specs, it means that that value is the maximum flow the pump can reach, right? Sep 3, 2017 at 16:17
• Some related links: electronics.stackexchange.com/questions/15444/… And [understanding alternator ratings][1]. The latter makes it very clear that the numbers refer to specific measurements under test conditions; actual values will vary in real circumstances. [1]: lifewire.com/understanding-alternator-output-ratings-534785 Sep 3, 2017 at 16:37
• @Physther So how come when I don't put a hose on the output to contain the water, I get lots of flow but no pressure? Sep 4, 2017 at 5:35

Internally, a generator produces voltage proportional to its speed.

However, that doesn't mean a constant voltage always comes out of a generator run at a fixed speed. The windings and other parts of the generator have some electrical resistance. To a first approximation, you can think of a generator as a voltage source proportional to speed, with a fixed resistance in series.

When current flows as a result of a load on the generator, two things happen:

1. The current causes a voltage drop across the generator's internal resistance. The voltage you get out of the whole generator is its internal voltage minus this voltage drop.

2. the current thru the generator causes a torque opposing the shaft rotation. This is how generators don't magically create power. The input power is torque times rotation speed. The output power is volts times amps. As the amps go up, the output power goes up. However, this causes the torque on the shaft to go up, so the input power goes up too. The input power is always at least as much as the output power. Physics doesn't give you a free lunch, and generators obey the laws of physics.

For example, let's say a particular generator when spun at 60 Hz (3600 RPM) produces 50 V RMS with no load. No load means the current is 0, so the voltage drop across the internal resistance is zero. Therefore, at no load you get to see the actuall internal voltage the generator generates.

Let's further say that the internal resistance is 2 Ω. If you connect a 75 Ω load to the generator, the internal voltage source now sees the internal resistance and your load in series as its load. That means the load on the internal voltage source is 77 Ω. The current is (50 V)/(77 Ω) = 649 mA. The drop across the unavoidable internal resistance is (649 mA)(2 Ω) = 1.3 V. You therefore get 48.7 V at 649 mA.

The total electrical power the generator is producing is (50 V)(649 mA) = 32.5 W. Of that (1.3 V)(649 mA) = 844 mW ends up heating the generator and you actually get (48.7 V)(649 mA) = 31.6 W out. The backwards torque on the shaft created by the 649 mA times the shaft rotation rate is 32.5 W plus whatever it takes to overcome the mechanical friction.

If you were to short the output of the generator, the maximum possible current would flow. However, you don't get any power out since the voltage is zero. All the power produced by the generator goes to heating its internal resistance. The resistance on the shaft would be high. If you managed to continue turning the generator at the same speed anyway, you'd be putting a lot of mechanical power in, and all of it would turn into heat in the generator. For many real generators, this would overheat and destroy them in a relatively short time.

The above discussion was for basic generators. Apparently some are interested in discussing automobile alternators. Those are nowadays complete units with rectifiers, regulators, and feedback, to which the above doesn't apply. It does apply to the bare generator inside a car alternator, but you don't have access to that directly, and its workings are actively modified and obscured by the controller.

• This is a great answer! Thank you so much! It makes so much sense now. Sep 3, 2017 at 18:31
• -1/2 pt , technical errors on term " generator" and -1/2pt for non-relevant examples and no mention of field current controlled current source output. Sep 3, 2017 at 19:03

Sometimes, one can see on the specifications that a motor can deliver a certain current and this is what I don't understand. The current is related to the potential through Ohm's law, right?

Not in the way you think.

Yes, if you connect a lower resistance load the current will increase if the generator's output voltage remains constant. This is shown by $I = \frac {V}{R}$.

However, there are a few limits including details such as the maximum current carrying rating of the windings. The motor can supply more and more current but the winding temperature will increase as you do and at some point you will burn them out. A rule of thumb is that every 10°C rise in running temperature halves the winding life.

In other words, the current is dependent only on the resistance of a load (or wire) and the potential.

Yes, as explained above.

Also, the current flows only if there is a voltage gradient/drop.

Yes. A difference in voltage or "potential" between two points will cause current to flow. It's a bit like two water tanks connected by a hose. Water will flow until the levels are the same. Current will drop off as the height difference decreases.

Why is there the current specified on generators like this one?

It is the design limit for the output.

Update: I missed the "this one" link in the question.

That looks like a car alternator so the situation is a bit more complicated. To charge a car battery will require > 14 V so the output is a nominal 12 V but actually more like 14 V. If you monitor the voltage on a car battery while someone revs the engine from idle to medium you might see the voltage rise. (You will also see the headlights brighten noticeably.) That alternator has a regulator on it which will adjust the field winding current to maintain the voltage close to the specified output - probably 14 V or so.

• Thank you very much! So basically, when I look at a generator that has its current specified on it, it means that that's the maximum amount I can ever achieve with it, right? And that is only because of its winding/construction and it is independent of what is connected to it. Did I understand it correctly? Sep 3, 2017 at 16:13
• Yes. You can safely run any load up to the rated current. Some generators will allow a short overload - maybe to start a large electric motor which draws a high current on startup - for a short time. This will be detailed in the machine specifications. Sep 3, 2017 at 16:19
• And a last question. About that alternator... The specifications say 12V, 160A. I imagine that the voltage would go to 12V, and then the only thing that will change would be the current, reaching a maximum of 160A. Is that correct or will the current reach 160A only when the voltage is 12V? In other words, can there be different currents for the same voltage drop? Sep 3, 2017 at 16:24
• @Physther A thinner wire could have been wound inside your example alternator. Many more turns would fit the same internal winding space. The alternator would be rated for higher voltage, and lower current. Sep 3, 2017 at 16:25

In vehicles, there is a difference between generators and alternators.

First, what is shown in the photo, is not a (DC) "generator" rather it is an AC alternator which includes a diode bridge (3 phase, 6 diodes) to rectify into DC.

Let's make sure you understand this first.

Originally generators were used with commutating windings with brushes making contact with rotor windings to alternate the AC voltage back to DC.

Due to brush arccing failures in the late 60's these were redesigned with Alternators that use copper slip rings to provide DC to the rotor instead, which we call self-excited since it can use the rectified output voltage to excite the rotor current so that the rotating flux can effectly increase the output AC current amplified by the work done to rotate the pulley and regulated by the output voltage to 14.2V with a regulator that has a fixed low voltage reference.

Next we know that motors/generators and alternators produce a "no load" EMF voltage proportional to RPM, but in this case the regulator controls the rotor (rotating DC) "Field current" to also regulate voltage as RPM changes. However below a certain RPM, no amount of field current can raise the voltage, this only gives current gain.

So RPM creates the voltage while field current automatically regulates the output current determined by the internal voltage regulator and load current (Ohms Law, Vout= Iout(load)*R(load)). So the field current naturally reduces with rising RPM at a fixed load.

However we know that the load also affects voltage by impedance ratio, so the source impedance must be lower than all expected loads except for the starter motor, which is a much lower impedance but like a generator it has a commutator with heavy copper brushes.

So the Alternator produces both voltage and current regulated by RPM and field current, so we say it generates power that depends on load. Since the available torque is high from a serpentine belt, it may be undervoltage below 500 RPM but be able supply full power at 1200 RPM or so using engine RPM scaled to alternator RPM by pulley ratios.

In short (no pun intended as that would blow the 6 diode bridge at high RPM) the alternator is a voltage regulated current controlled current source at fixed RPM that supplies all the current needed to charge the battery and other loads. It must be suitably sized for current as the battery ESR determines the max current to raise the voltage to 14.2V and the battery ESR reduces with rising CCA capacity (when new) needed to turn a big truck starter motor.

## side effects from mismatched alternator and battery

Although a vehicle, once running can operate without a battery, the alternator must have the current capacity at correct regulated voltage voltage and current to meet expected demands. The biggest load is a an undercharged battery. When batteries age each cell becomes more mismatched and the weakest cell can boil the electrolyte from over voltage from excess alternator current to reach 14.2V so instead of 14.2/6=2.366V per cell, anything 10% higher will age the battery rapdly. So installing a bigger alternator on an old battery, can result in battery failure quicker than normal. Also as diode bridges get hotter and age slightly with higher ESR a fresh battery with higher CCA rating (and lower ESR result) can increase the stress on the alternator bridge and fry it in ar least one phase so the alternator reduces its capacity from 3 phase to 2 or 1 phase and can no longer keep up to worst case load current causing the headlamps to dim at normal idle.

Lets consider a new 1000 Amp CCA rated battery that when warm has CA rating of 1200A. This means the ESR is (12.5v-7.5V)/1000A= 5 mΩ from the std test at drop to 7.5V and Ohm's Law. So what is the max current to raise the battery from 12 V (undercharged) to 14.2V ? Again from Ohm's Law, (14.2-12)V/5mΩ= 440 amps !! Fortunately the diodes can handle brief over current periods and the weak alternator could fry its diodes , while a stronger one might need more RPM to raise the voltage to 14.2 until the charge level rises.

Thus the alternator current and battery CCA are designed for each vehicle to increase longevity at the least cost by careful selection of matching diode ESR and battery cell ESR mismatch from aging. This is why they sometimes dont last long after replacing one or the other. This is aggrevated by excessive V regulator settings and poorly sized or maintained batteries. (warped plates from brief shorts), excess ambient temperature baking in the sun in Arizona. etc etc.

Sorry for the long answer, I hope this raised your IQ on Alternators. It is more complicated as the control loop is a current controlled current source (CCCS) with a Voltage Reference to get an output of 14.2V +/-0.1. This results in it becoming a regulated Voltage source. The other answers at this time, do not mention this at all.

• Thank you very much for the nice answer! Definitely raised my knowledge on Alternators. Sep 4, 2017 at 12:38
• ESR on batteries, diodes are not constant but with constraints in current, v, time, temp can be made linear constant. Batteries are just a bank of caps in series/parallel, with a no charge threshold voltage with large kF *ESR =T constants, with memory (// effects) but can be constrained for either transients or steady state to make estimates. Sep 4, 2017 at 12:44
• The votes are not an indicator of content, but the capacity to understand of the readers. Sep 4, 2017 at 12:48

A generator, operated into a short, produces the maximum current; that current inside the generator coils will act to oppose the field provides by the "magnets" and implements a feedback system by means of flux bucking, flux opposition, thus partial flux cancellation.

That partial flux cancellation limits the internal "B" field of the coils.