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.
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.