Hello Guys
I want to make a converter from a rechargable 12v-4.5 amp input sealed lead acid battery that will drop back to 9.5v output and then drop v output as the 12v drops volts, I want to use it in place of 6-c-cell batteries 1.5v ea that go inside the deer camera, I have a converter I bought but the output stays at 9.5v, is their anything i can do to make output drop as input 12v battery drops. That would make the the camera show volts as if it were useing the 6 c-cell batteries.

Thanks for any help
Will in Al

 

How are you getting 9.5V out of 6 C cell batteries? Dry cell batteries have a voltage of 1.5V and lithium and similar 1.2V.

And just because you have 6 batteries does not mean that is the total voltage of the battery pack. You need to first determine the voltage of the pack.


You will want to use a buck regulator to drop your voltage. It will be the most efficient way to regulate the voltage. The 34063 is very old and my no means the most efficient regulator out there but since it has been around so long there is plenty of support for it.

Texas Instruments also makes some newer chips that are fairly easy to use.

 

The problem is that he wants the display on the camera to indicate the charge condition of the battery by NOT giving it a solid 9.5 volts, but he can't say at what voltage the camera display indicates full, half full, or nearly dead. I already devised a circuit that will do that but I have no voltage points to match the regulator to. At least, this time, he has determined that his battery is an SLA. From that, we can find voltage points that tell the charge state of the SLA. The next job is transferring that information to the camer input but we don't know what voltage the camera uses to alter its display.

This is a circuit that will track the battery voltage, but I have no idea if it will track correctly for the camera to display charge level of the SLA. Please have a go at this, spinnaker and others.

 

The LM317 is going to be the most simple solution but OP should be aware that there will be significant loss in heat. A buck regulator would be more complicated but more efficient.

Also after reading the OP, the OP already has a "converter" but it is 9.5 and needs to drop .5 v? I would like input from the more learned forum members but I would not think .5V should make much of a difference in this case?

The biggest concern would be correct polarity.

 

This drawing has the updated information.
Edit: Adding "State of charge voltage during discharge" curves.
Using the C5 curve, the Pb-Ac voltage will be 12.1V to 10.4V.
That changes the values I used.

 

The camera shows 100% when the new c cells 1.5v batteries are put in, I only installed them for a little while, they drop as the camera takes pictures, How would i determine what percent the camera would drop, by days or what?

 

Wait until the camera shows 50%, then take the batteries out and measure their combined voltage immediately.

 

I just checked the 6 new c cell batteries. They read 9.5v total. Is their not anything I might put in line to make volts drop?
Thanks

 

Something is very wrong when 6 new C cells show a combined voltage of 6.5 V.
You wanna check that again?
Are they alkaline cells? Carbon-zinc? NiMh?, LiPO?

and yes, you can put a resistor in series with the load (camera). That would require knowing the current the camera uses and would have the same voltage drop at all times, not depending on the state of charge of the battery.

Edit: I have to bug out for a family obligation right now.

 

Hi #12
I have part of camera percent, full 100% is 9.54,but the 12v Battery % is 100% is 12.75v, 75% is 12.45v, 50% is 12.24v, 25% is 12.06v & 11.89 or less is Discharged, Can you help from this?
Thanks

 

Sorry I ment 9.5v new c cells,They are Energizer c cell batteries

 

I'm not buying the idea that a 50% full Energizer (alkaline) C cel is half discharged at half its original voltage. 9.54/2 = 4.77

Alkaline cells are dead at about 1.1 volts but the question is: When does your camera think it's dead? Try to hook up 3 C cells to the camera and find out if it even starts up with the 4.8 volts you will get. (That is allegedly 6 cells half discharged.) Then try hooking up 5 C cells to the camera and see what the energy meter says...if it starts up at all at the 7.95 volts that will result.

The numbers you gave me on the lead-acid battery pretty much fit the 100 hour discharge curve. Is there a good chance that the battery will last 100 hours driving that camera? Knowing the current use of the camera would help with this number.

 

Here's the shape of it.

 

Something is very wrong when 6 new C cells show a combined voltage of 6.5 V.
You wanna check that again?
Are they alkaline cells? Carbon-zinc? NiMh?, LiPO?

and yes, you can put a resistor in series with the load (camera). That would require knowing the current the camera uses and would have the same voltage drop at all times, not depending on the state of charge of the battery.

Edit: I have to bug out for a family obligation right now.

The batteries are Energizer 1.5v, made a mistake on volts. Total volts are
for 6, is 9.54v. The converter I bought will adjust on the output so I will get camera and see at what point the camera starts dropping. Is that what you need to be able to tell me what to do? Thanks

 

I already did it. I just put in adjustments where the information is missing. My job is done. You can build it and adjust it to suit your camera.