I am trying to get a 4 cell Lithium battery pack to work with a 12 volt inverter. The inverter I am using has a maximum voltage of 15 volts and a minimum voltage of 11volts. The battery that I am using to power the inverters has a nominal voltage of 14.7 volts and a voltage of 16.5 volts when fully charged.
What kind of options do I have for getting the inverter to work with the battery voltage?
The inverter will output 120 volts and 1 amp so the inverter will be drawing about 10 amps from the battery.
How can I drop the battery voltage down by a volt or two so it will be within the voltage range accepted by the inverter?
Well I did that and It worked! Hurray! I used two big 35 amp diodes and the battery voltage came down enough to run the inverter. From what I calculated the diodes are going to disapate about 2 watts of power at full load which is kind of a bummer.
Is there a way to bypass the Diodes when the battery voltage drops below 14 volts to make it run more efficently? I was thinking about using mosfet as s switch in parallel to bypass the diodes and some kind of compairator like an opamp or a attiny85. Does that seem like somthing that would work?
Is there a way to bypass the Diodes when the battery voltage drops below 14 volts to make it run more efficently? I was thinking about using mosfet as s switch in parallel to bypass the diodes and some kind of compairator like an opamp or a attiny85. Does that seem like somthing that would work?
You can use an op-amp as a comparator or a specialized IC [an comparator] which ends up driving a mosfet or relay.
Or place a resistor and zener across the input, and use the [lack of] voltage across the resistor to finally drive a relay or mosfet.
It is true this can be done with a mosfet. Once you think about doing this, you might as well bias the mosfet in a two stage process on powerup. (In this way you will no longer have need for the diodes.)
(1) Bias the mosfet a little, so that it provides a few ohms of resistance, thus dropping a couple of volts for the first five seconds,
then...
(2) Bias the mosfet a lot, so it turns full on, providing very low resistance, thus maximizing efficiency.
This two-stage startup up can be performed automatically, with a few components.