Lead acid battery protector

[joshnz]

I'm trying to build a low cost battery protector.
this is what I have so far.


It Is meant to work like this
if power is lost run the load off battery intill battery reaches 10.6 volts.
when power returns run off power supply and charge battery.

I noticed I forgot diode to protect transistor.

I may need a second set of relay contacts to run load off when dc power supply is on.

[budm]

Re: Lead acid battery protector

It does not look right.
What is the D1 Zener voltage? If you want the battery to be disconnected from the circuit when its voltage is at 10.6V, that means when V2 (13.8) is working, then D1 will be on to turn on Q3, and when voltage is at 10.6V D1 has to be off to turn off Q3. Is that what you are trying to do. Q3 is on in normal operation, off when output voltage is at 10.6V? If that is the case then you will have to re-configure N.C. and N.O. contacts of the relay. Also the Gate of the MOSFET is tied to ground? At the same time when V2 (13.8V) will charge the battery through the body diode of the MOSFET and the R1, no protection when battery is fully charged (no trickle charge).
So how do want the Q3 to be when V2 is OK and when V2 is not present (lost power)?

[joshnz]

Re: Lead acid battery protector

I will explane later I was half asleep when I quickly put this together.
The mosfet in this setup is a very eficent diode.

[joshnz]

Re: Lead acid battery protector

Zener is 10v.
None of this has been made yet just a concept.

[joshnz]

Re: Lead acid battery protector

The .33 resistor has been chosen to to limit current to 12 amps when battery is at 10 volts to avoid over loading power supply.

When battey voltage has reached cut off voltage the relay releases connecting the power supply back then because the power supply provides more than the cut off voltage the relay pulls back in.

[joshnz]

Re: Lead acid battery protector

One issue I see is it may oscolate when power first come back.

[budm]

Re: Lead acid battery protector

"When battey voltage has reached cut off voltage the relay releases connecting the power supply back then because the power supply provides more than the cut off voltage the relay pulls back in."
OK, if you look at the circuit. When first V2 (13.8V) is connected, the current will flow throw the N.C. (Normally Closed, the bottom set of contact as shown in the diagram post 1) contact to the DC+ out terminal, at the same time this voltage is above 10.6V cutoff voltage so the Zener will conduct, turn on Q3 which in activate relay X1 which will cause the N.C. contact to go open which will disconnect the V2 from the circuit, the N.O. contact will close to switch over to battery,that is not condition you want.
And what is the MOSFET for?
You can analyze your circuit by drawing the path of the current flow, then you will see what is going on.

[joshnz]

Re: Lead acid battery protector

Mosfet Q1 is been used as a diode to stop current flow from V1 to V2 when V2 looses power.
So when the low voltage drop out activates, Current from the battery can't flow back via R1 via the NC contact of X1 that would defeat the purpose of the low voltage protection.

[ben7]

Re: Lead acid battery protector

Would this work for you?

There is a voltage divider that divides the battery voltage, and a comparator compares it with a reference voltage.

The voltage divider can be 100K resistors, to minimize the current drain on the battery. Using 100K resistors, the current drawn from the battery will be less than 10uA (about 4uA I think).

The potentiometer is to tweak the voltage divider, especially since a 100K resistor with 20% tolerance can vary in resistance quite a lot.

The voltage divider is necessary, because the comparator can't compare voltages that are greater than it's supply voltage.

When the power is on, the comparator gets power from the power supply, and when the battery voltage is OK, then the battery voltage is fed to the output. Power supply goes off, and the output voltage from the battery stays on until the battery voltage goes too low.

[budm]

Re: Lead acid battery protector

So the power from V2 will go through body Diode of the P-CH which at the same time tied directly to the battery (no protection if the battery has shorted cell?). Use low Vf diode instead of MOSFET?
How much current are we talking about here?