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By IdahoMan
#181519
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! :evil:]


Sincerely,
IdahoMan
By dlotton
#181521
IdahoMan wrote:
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?
First, do NOT use a resistor. A resistor is not a regulator. It will lot correct for fluctuations in input voltage or load current.

With a linear regulator the loss is in the regulator. In your case, for a 12V input you need a regulator that drops 7V across the regulator and supplies a 5V output. It's kind of a brute force method of regulating. The power source (your battery) outputs 12V and 270uA. Power = Voltage x Current. The 270uA is common to both the linear regulator and the PIR sensor. P = 270uA(7V+5V) So, 7/12ths of that power is dissipated in the regulator, and 5/12ths is dissipated in the PIR sensor.

A switching regulator is a different animal. It actually converts power. In the case of the buck regulator it converts a high voltage low current to a lower voltage and higher current. Power in equals power out times the efficiency factor of the regulator, which is generally pretty high for switchers.

Pin = efficiency x Pout
or
Vin x Iin = efficiency x Vout x Iout

The regulators I linked were way overkill for a 270uA load. They were rated at 1A and 3A. At a 270uA load the efficiency will likely not be in the 90% range. In fact, at that load, the efficiency might be lower than a linear regulator. You'd want to find something that's better scaled for your use.
By IdahoMan
#181523
First, do NOT use a resistor. A resistor is not a regulator. It will lot correct for fluctuations in input voltage or load current.

With a linear regulator the loss is in the regulator. In your case, for a 12V input you need a regulator that drops 7V across the regulator and supplies a 5V output. It's kind of a brute force method of regulating. The power source (your battery) outputs 12V and 270uA. Power = Voltage x Current. The 270uA is common to both the linear regulator and the PIR sensor. P = 270uA(7V+5V) So, 7/12ths of that power is dissipated in the regulator, and 5/12ths is dissipated in the PIR sensor.
Thank you. I'm sorry if this is a silly question but help me wrap my head around it: How much more drain on the battery -if any- do I get by having a linear voltage regulator added to the PIR circuit if the current is common to both reg and PIR? It's still the same amount of Amp draw, the battery life is measured in AMP-HOURS.
The regulators I linked were way overkill for a 270uA load. They were rated at 1A and 3A. At a 270uA load the efficiency will likely not be in the 90% range. In fact, at that load, the efficiency might be lower than a linear regulator. You'd want to find something that's better scaled for your use.
Thanks.
By dlotton
#181524
The extra current, just to run the regulator, is pretty minuscule if you get a good linear regulator. You're probably looking at 3-5uA.

The real inefficiency in this use case isn't the extra current to run the regulator, it's the power dissipated in the linear regulator... 270uAx7V
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