Automatic Charging Relay

Hey

For a class project, my group is constructing a small VAWT power generator. We want it to provide power to a street light via a rechargeable battery. We also want it to be able to switch the light over to the grid when the battery dies (if the wind is too low for too long) and back to the battery when it is charged again. Basically we want it to reduce but not necessarily eliminate external power consumption.

I’m in charge of the part of the circuit that will do this. From my research the best thing I can find is an automatic charging relay, but only if it works in reverse: remove battery from circuit to light when voltage drops below a certain value and connect it when the voltage exceeds a different value.

Is there any way of modifying an ACR to do this or any other way of doing this? Thank you.

A popular approach to this challenge is to always run the load from the battery and keep both charge inputs (solar and mains) connected at all times (no need for a relay).

If you want to maximize use of solar power, you would choose a set point for mains charging that is just above the safe drain level for your battery (for SLA this is about 11.7V). That is whenever battery/load voltage drops below 11.7V your circuit will be supplied from mains.

A series diode on the mains charger lead going to the battery will keep it out of the circuit when solar power is available. Mains charging should provide a fixed voltage output at 11.7V plus the diode voltage drop (0.7V).

How would I implement a set point? Would the series diode on the mains do it? Do you have any suggestions for part numbers?

Also, just to clarify, VAWT is a Vertical Axis Wind Turbine. The set up could be the same for VAWT and solar, but your suggestion mentioned solar and I just want to make sure.

And so I know I understand, your suggestion is: generator > battery > wire > load (light) main power > diode ^

Am I understanding this correctly?

Sorry for the solar/wind mixup, but the solution is the same (substitute with "green power" if you like).

Wiring is simple:

Generator +/- to battery +/- respectively Load +/- to battery +/- respectively Mains charger +/- to battery +/- respectively

The generator obviusly needs some sort of regulator on output appropriate for the battery, but this was outside the scope of your question.

The series diode goes from mains charge + to battery +. To regulate voltage output from the mains charger you could use an LM317 regulator and follow the guidelines in the datasheet to select approriate resistors (or fixed plus variable resistor) for the voltage set point required.

Current output from the mains charger would need to match the load current requirement plus a safety margin.

For a commercial product one might consider a range of possible optimizations to this design with respect to power efficiency. Examples would include a switch mode mains charger, substitute series diodes with FET's etc. This would add complexity to the design and probably is not something you would build as a student.

An "ambitious student" however might identify at least some of these and possibly include the math (to show the power savings potential) in the project documentation.

A member of my group has an Arduino Uno microcontroller and if we use that we will have less parts to wait for. Are you familiar with this model? The way I understand it is that the wiring won't change much from you're first suggestion: generator > battery main > Uno power and Uno pins > load battery > Uno pins battery > load

then on the programing side, program it so that if the voltage on the Uno-battery wires drops below 0.7V, let main voltage go to load. If Uno-battery wires is above 11.7V, stop main voltage from going to load

If this understanding is correct, that should let the battery charge continuously off the turbine (as long as the wind is blowing, regardless of main being connected) and if lack of wind leads to the battery getting too low, the main power kicks in until the battery recharges.

Does this make sense?

circuit diagram quick

If this understanding is correct, that should let the battery charge continuously off the turbine (as long as the wind is blowing, regardless of main being connected) and if lack of wind leads to the battery getting too low, the main power kicks in until the battery recharges.

Does this make sense?

Sort of.

You would choose a cut-off point for the turbine output at just above 11.7V (assumining a 12.7V nominal SLA battery). This means that the load would partially draw from the turbine and partially discharge the battery when battery (system) voltage drops towards 11.7.

At 11.7, mains will kick in and feed the load, but not charge the battery. This is just to keep the light on and prevent the battery from discharging further.

When the generator kicks in (just above 11.7V), the output will feed the load AND charge the battery. This will maximize the use of "green power".

If you're concerned with battery longevity (an SLA battery is good for only about 500 charge/discharge cycles or less) you would choose a mains set point just below 12.7V and let the generator kick in at 12.7V. This would improve battery life, but increase mains power draw.

An SLA battery can be charged at up to 14.4V, but not for long. To keep it simple, you could limit turbine charging at 13.7V max. At this level you can keep it running forever and the battery will last for 10 years.

ok I think I understand.

Now I'm having trouble with the coding side. As of now I have this

const int Check = A0; //battery to pin A0 const int Relay = 9; //pin 9 to relay

void setup(){ pinMode(Check, INPUT); pinMode(Relay, OUTPUT); }

inside a loop I plan to read the voltage on Check and call it Result. If Result is less than 0.7 V, write a voltage to Relay to switch the relay and take the battery out of the circuit. Else if Result is greater than 11.7 V, write a voltage to Relay (or a Relay2 if the relay is built that way, to switch the relay and connect the battery to the circuit.

This seems logical to me but I might be completely off. Do I need a circuit element to convert the voltage signal to analog or does the Uno do that? I've done analog conversion for 5V before (analogRead 0-255) but how would I do this for 12V?