I currently have a set up where I have attached a permanant magnet dc motor to a bicycle. The motor is attached through a diode to a 12v regulator and then to a 12v to 240v inverter. Everything works fine and I can power up my devices.

But I have been told that if I connect my 12V lead acid gel battery to the inverter as well via a diode, this will act as a back up source, so if I stop pedalling, the battery will power the inverter via a process called forward/reverse biasing.

Please can I get some advice on this and whether it will work?

Thanks

 

Don't know anything about bias in this case, but it woulde only be beneficial to let your bicycle-generator charge a battery, and run the inverter directly from the battery. Then you can take a break and still have your devices powered. Or pedal just for fun/charge.

How much Amps does the dc-motor give at normal pedalling?

 

Without getting complicated, and using only a diode, you could try connecting the battery pre-regulator per below. When the motor isn't running, the battery will power the regulator and the inverter.

MTR --> DIODE --* REG --> INV
BAT --> DIODE --*

However, the motor must supply a higher voltage than the battery to run the regulator, and the battery must have enough headroom to run the 12V regulator (at least 12.5V if the regulator is an LDO with 500mV drop out). If a fully charged battery is 13.6V, then this would work until the regulator drop out voltage is reached (probably in the +12.5 to +13.0V range). So, this arrangement will only work until the battery is down to 13V or so. Not very long, nor efficient use of the battery.

If you can't lower the regulator output voltage (to say, 9V) to better match battery performance (hence lower input voltage to inverter), it may lead to trying to connect the battery post-regulator.

MTR --> DIODE --> REG --> DIODE --* INV
...........................BAT --> DIODE --*

Now the regulator must supply a higher voltage than the battery. If a fully charged battery is >12V (say 13.6V), and joined to the regulator 12V, you could pedal all day but it would always be running from the battery -- not the motor voltage.

So, the regulator output voltage needs to be increased to greater than that of the battery and compensate for diode drop.

Batt - diode drop = Vbin
Reg - diode drop = Vrin
Vbin < Vrin

If the Vbin is 13.0V (13.6 - 0.6 = 13V), set the regulator to a higher voltage (at least 14.0V) where Vrin = 14V - 0.6V = 13.4V.

Now the question is, what voltage can the motor supply to the regulator? Can it supply at least 15V, to stay above the drop out of the regulator?

The critical points to be balanced by circuit design, are the fully charged battery voltage, the output voltage from the motor, and the regulator Vout and drop out.