# Battery Replacement Project and voltage divider physics

Discussion in 'The Projects Forum' started by Jeffery Vahrenkamp, Jun 16, 2016.

1. ### Jeffery Vahrenkamp Thread Starter New Member

Jun 16, 2016
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0
Hi,

A little background before my questions:
I working on replacing a lithium ion battery I have in my camera with a wire that hooks it up to a much large capacitance car battery. Normally I would just drop the voltage down to 8.4v with a buck converter and call it a day (I've done this and it worked on an older camera) but my newer one has a pesky feature where it checks for an "Authentic" battery. It does so by doing a hand shake with a chip in the battery. I have a battery I have dismanteld and it's a little trick. There is a chip on top of a two lithium ion cell battery which has 3 terminals. Terminal 1 and 3 are simply hooked up to the group of cell 1 and the + of cell 2, with the cells in a serial configuration to give the ~ 8v of power. Terminal 2 is hooked up at the junction point of the two batteries. According to people who have studied the battery, the + terminal(1) and the middle terminal(2) power the chip with 4.2V and open a mosfet controller and also handle the hand shake.

Here is my dilemma, and the root of my question:
I have a 12V battery. I need 0V, 4V, 8V outputs. The 8V output should be able to put out 2.5 amps, using two buck coverts will be too bulky, and I"m not sure what would happen connecting the two grounds from the two buck converters.

My thought was to create a voltage divider with the setup (12V+ ----- (R1=5ohms) ----+Terminal (8V) ---- (R2=10ohms)-----4V terminal (4V)).
My calculations indicate this should give me 8V at the + terminal and 4V at the 4V terminal. The ground would connect to the ground terminal.

I realize this isn't as simple as it seems because 1) I don't know what the resistance of the camera or if the 4V terminal in incircuit with the camera, so I can't predict how that would affect the voltage divider. 2) I also was thinking about the power dissipated by the resistors. They are 1 Watt resistors, but from my calculations it should be producing either 12 or 9 watts of heat, which would be undesirable waste of energy in my book and not the best for keeping my camera sensor cool. My first thought was to change the value of the resistors, but this is where I get a bit confused.

For the formula V=IR, if I drop the resistance the heat dissipated by the resistor (V*I) goes up. This makes sense to me as a short circuit with just a copper wire will produce more heat than a high value resistor due to current limitation. But at the same time, my brain tells me that lower resistance should be more efficient, and produce less waste heat. This makes sense if the V is voltage drop over a single element, then as resistance goes down, so does voltage drop and dissipated heat. My brain tells me the second case is the one I will experience setting up my voltage divider.

Ideally I'd like to have as little power dissipated by the resistors as possible, and I need 2.5 amps to reach the camera. I was thinking in a perfect world I'd have 1 mm (resistor 1) and 2 mm (resistor 2) lengths of copper wire to give very tiny amounts of resistance with my V+ terminal soldered at the junction and the 4V terminal soldered at the end of the divider.

So which should I do?:
1) I high resistance voltage divider
2) low resistance voltage divider,
3) or does someone have a better alternative to doing the voltage divider.

2. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,967
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Welcome to AAC!
Personally, if I've understood you correctly, I'd forget about resistive voltage dividers, use a switch-mode buck regulator to get the 8V high current supply and run a linear regulator from that to get a low current 4V supply, with the chip having the 8V as its +ve rail and the 4V as its -ve rail.

3. ### AlbertHall Well-Known Member

Jun 4, 2014
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447
If this chip has just three terminals and all three are connected to the cells in the battery I don't understand what 'handshake' is possible. Is there another connection from the chip to the camera?

Sounds OK - except we don't know which direction the current in that middle pin of the chip flows. The linear regulator will source but not sink current.

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
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It will source current to a load, but it will get that current via the 8V reg, i.e. it will sink current too.

5. ### AlbertHall Well-Known Member

Jun 4, 2014
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What I mean is if the IC draws current from the 8V supply and tries to return it to the 4V supply, like R1 below, then the 4V regulator cannot properly regulate its output voltage.

6. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Agreed. But put a load (resistor to ground) on the IC2 output and its fine. Current through R1 is sunk by the load.

7. ### AlbertHall Well-Known Member

Jun 4, 2014
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447
Aha, now at mod state 1.
I still have no idea what that three pronged IC is up to in there.

8. ### Alec_t AAC Fanatic!

Sep 17, 2013
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..... and therein lies the rub. Whatever supply we come up with we may not fool the IC.

9. ### Jeffery Vahrenkamp Thread Starter New Member

Jun 16, 2016
5
0
Sorry about being slow getting back. This is the relevent line from the posts I have read describing what the different leads are doing

The battery internal board is powered with the following connections center tap and positive, the negative point is also connected to the board,
this is controlled by a high current mosfet IC switch, if a short was to happen across the battery terminals or pulled down the inner workings
would trigger the IC switch open disabling the battery output p- p+, and that one cell powering the board is not effected and continues to keep voltage reference checks.
That sounds to me like the chip and its MOSFET are driven by the second battery cell with the V+ terminal as the + and the middle pin as the ground, giving it ~4V power. This runs the chip which is responsible for protecting the camera and the battery cells. The Chip then connects to the camera. If the chip is powered it does a hand shake with the camera and leaves the mosfet open to deliver power from the V+ and V- terminals to the appropriate terminals on the chip. From the sounds of this, it looks like the middle terminal would need to be able to sink the power.