- Sun Apr 26, 2015 8:56 pm
#181519
If I use a linear method, such as a resistor:
12v to 5v
.00027A(270uA)
That's a 25k resistor; .00189W.
The issue is battery life, which is measured in Amp-Hours. The PIR circuit is still drawing 270uA, where is the loss?
Car-Batt: 12v, 40AH
(Might want to reduce that rating by aroubd half if I am afraid of over-discharge, although I'd probably be using a Deep-Cycle(would have more AH); also, I could use a L-Ion or have the L-ion as a backup-batt by wiring it in parallel with the CarBatt)
---------------
It also suddenly struck me that the input voltage may vary from 10v - 16v, not a constant 12v. My concern then would be for protection of the PIR.. (operating voltage(Vdd) - 3.0VDCmin-6.0VDCmax)
10 - 5 = 5v
5v/.00027 = 19k Ohm resistor
12 - 5 = 7v
7v/.00027 = 26k Ohm resistor
16 - 5 = 11v
11v/.00027 = 41k Ohm resistor
Come to think of it, I don't know if the camera can tolerate such voltage fluctuation either. "12V" cigaret-lighter accessories for vehicles.. I am assuming they must have some kind of regulation built-in then. From what I understand, the voltage across the battery can vary 10v - 16v depending on the condition of the battery and if the vehicle is running.
What should I do, use two voltage regulators: One between the source and the main-line (left) AND one between the main-line and the PIR (right)?
---------------
Thanks for the links. I was searching earlier and found this one too: http://www.adafruit.com/products/1385
Resource: https://www.dimensionengineering.com/in ... regulators
[BTW, I don't think Window's calculator program works at all. 10/45000 = 2.2? I don't think so! ]
Sincerely,
IdahoMan
dlotton wrote:A buck converter will be most efficient... probably in the 90+% range. A linear converter will be in the 40% efficiency range. The trade-off is that buck converter will be more expensive.
Take a look here...
http://www.adafruit.com/products/1065
[edit]
These are 3A output. Probably way overkill, but cheaper...
http://www.amazon.com/Retailstore-LM259 ... +regulator
If I use a linear method, such as a resistor:
12v to 5v
.00027A(270uA)
That's a 25k resistor; .00189W.
The issue is battery life, which is measured in Amp-Hours. The PIR circuit is still drawing 270uA, where is the loss?
Car-Batt: 12v, 40AH
(Might want to reduce that rating by aroubd half if I am afraid of over-discharge, although I'd probably be using a Deep-Cycle(would have more AH); also, I could use a L-Ion or have the L-ion as a backup-batt by wiring it in parallel with the CarBatt)
---------------
It also suddenly struck me that the input voltage may vary from 10v - 16v, not a constant 12v. My concern then would be for protection of the PIR.. (operating voltage(Vdd) - 3.0VDCmin-6.0VDCmax)
10 - 5 = 5v
5v/.00027 = 19k Ohm resistor
12 - 5 = 7v
7v/.00027 = 26k Ohm resistor
16 - 5 = 11v
11v/.00027 = 41k Ohm resistor
Come to think of it, I don't know if the camera can tolerate such voltage fluctuation either. "12V" cigaret-lighter accessories for vehicles.. I am assuming they must have some kind of regulation built-in then. From what I understand, the voltage across the battery can vary 10v - 16v depending on the condition of the battery and if the vehicle is running.
What should I do, use two voltage regulators: One between the source and the main-line (left) AND one between the main-line and the PIR (right)?
---------------
Thanks for the links. I was searching earlier and found this one too: http://www.adafruit.com/products/1385
Resource: https://www.dimensionengineering.com/in ... regulators
[BTW, I don't think Window's calculator program works at all. 10/45000 = 2.2? I don't think so! ]
Sincerely,
IdahoMan