So I have a turbine (hydro, but this question should be the same regardless wind/water so hopefully some of you wind gurus can help....) which has 3-phase output (10A) to a DC bridge rectifier. A Tristar diversion load controller dumps excess power to prevent batteries overcharging.
I'm increasing turbine output from 10A to around 20A DC (measured after rectifier) by adding a second water jet - the turbine is rated for a max 50A output at 48v so no problem there.
I don't fully understand how turbines work electrically and so how this will affect the DC output voltage of the rectifier. I'm guessing as the current from the turbine doubles the voltage will remain pretty much the same, hence doubling the output (P=VI)?
My concern at this time is if the voltage output of the rectifier will be too much for the batteries. How do I measure this DC rectifier output?
If the rectifier is connected to the batteries, then it will output the same voltage. (Well a bit higher, but it can't go much above it.)
It sounds a bit like your rectifier is rated for 10A only? To give the batteries a reasonable live expectancy, I would not charge a battery with less than 20Ah capacity. Above that it should be fine.
So if I understand this correctly the rectifier voltage output is decided by the batteries so over-voltage should not be an issue. I'm intrigued to know how this works (how does the battery know to draw slightly more than the battery rating (48v) to enable it to charge?). Does this mean for example, if the batteries are at 49v the rectifier will output maybe 51v, and if the batteries were at 52v the rectifier would output maybe 54v?
Also intrigued as to why charging at a low current 10A is not good for the batteries? I have a new rectifier to install (150A 1600v) as I'm unsure what the existing rectifier is rated at as it has no markings on it, so I thought best to go with a bigger one and a large heatsink to match.
Ignoring the rectifier for a moment and pretending the turbine generates DC output:
The generator has a maximum voltage, called "open circuit voltage". It is determined by the number of windings and how fast it is spinning. At that voltage you can't get any current out of it.
The generator also has a maximum current, called "short circuit current". That is what you get if you would short the outputs and pretend the input was still spinning. In reality that would apply such a massive breaking force that it would stop almost instantly. The output voltage would be zero.
Both cases have in common that either current or voltage is zero, and so their product (power) is also zero. You want to avoid either.
What happens in reality is that the generator spins and creates a voltage V_gen. Now the battery is already charged to some other voltage V_bat. The wires running from the generator to the batteries have some resistance, so do generator and battery. I will call that overall resistance R_total. With that the resulting current can be calculated: I = Ｕ / R = (Ｕ_gen - Ｕ_bat) / R_total. As the battery voltage rises, the voltage differential drops and less current will flow, however the voltage is slightly higher. That current will act as a brake on the generator, slowing it down. As V_gen depends on the speed of the generator, it also drops and stabilizes at some point where the braking force is equal to that of the water.
Essentially it is a (stable) self-regulating system.
Thank you! That is an absolutely brilliant answer, you have no idea how long I've been trying to extract that information from supposedly knowledgeable people! I don't know if they know their stuff but just can't explain it well enough for me to grasp, or they just don't really know it very well. Regardless of their efforts, you have managed to explain it so nicely and it makes total sense to me now!
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