Noob: resistor to conserve battery power

[kurni]

My electronic knowledge is hopeless; I've scoured the forum and came to a conclusion, but I'm hoping for some confirmation through this thread. I want to know if resistor actually conserves battery power by wasting some power as heat.

Assume Q5
Vf=3.2V @ 0.35A
Vf=3.6V @ 1.30A

Assume Vin is measured under load

If Vin=3.6V without any resistor
P=3.6*1.3W
P=4.68W is consumed without any resistor

To reduce Vin to 3.2V (hence Vf = 0.35A) I need to put a resistor
0.4V=0.35A*R
R=1.143 Ohm
Hence the power consumed by the LED is 3.2*0.35W = 1.12W
The power consumed by the resistor is 0.4*0.35W = 0.14W
Total power consumption is 1.26W instead of 4.68W
This setup cuts down power usage by (4.68-1.26)/4.68 = 73% (approximately 4x battery life assuming there is no voltage sag)
Power loss of this setup is 0.14/1.26 = 11% (becomes heat)

Question 1: am I right in using 0.35A to get the value of R?
Question 2: does this setup prolong the battery life by 4x with only 11% power loss (it's a hypothetical battery without any voltage sag )

Please help
Kurni

 

[Mr Happy]

Question 1: am I right in using 0.35A to get the value of R?

Yes, exactly right.

Question 2: does this setup prolong the battery life by 4x with only 11% power loss (it's a hypothetical battery without any voltage sag )

Indeed it does. OK, I have not checked your calculations, but the principles are correct. If you put a resistor in series with the load, the current is reduced and therefore the power output from the battery is reduced.

I good mental way of validating such questions is to consider the extreme case. If you put a 1,000,000 ohm resistor in the circuit, how much power would be drawn from the battery?

This validation technique only truly works for linear systems, but it is often reasonable to assume so if you don't know for sure.

 

[CM]

The math looks correct. In the real world, things are not so ideal but you probably already know that.

 

[kurni]

Thanks guys, this is just a hypothetical example to satisfy my curiousity.

Cheers,
Kurni

 

[mpf]

No so hypothetical. This the principle of the pulse width modulator controlled linear current regulator. http://www.candlepowerforums.com/vb/showthread.php?p=2570799
The FET adds the resistance and the system depends on the non-linear led voltage versus current curve to maintain good efficiency even at low levels of current.
I use this type of regulator for all my high powered torches.
matthew

 

[Chodes]

You electronics knowledge is definitely beyond hopeless Kurni.
I've worked with qualified engineers that would struggle with what you worked out.

 

[CM]

...I've worked with qualified engineers that would struggle with what you worked out.

They're fake if they can't do Circuits 101.

 

[kurni]

Matthew, your basic tutorial to me is way too advanced I was tempted to ask a question about how boost circuit can work efficiently (e.g. MaxFlex) but resisted because I suspected that the answer would go woosh above my head.

I attempted an engineering undergrad but dropped out after 1 semester; eventually got an MBA instead

 

[BobVA]

For the specifics of the calculations, I'd say "maybe"

My caution would be that the formulas you're using are for linear devices, which an LED (or any diode) is definitely not. I'm not familiar with that device, so I can't say for sure that the resistor value you calculated would reduce the series current to the value you want.

You're spot on with the basic concept, though. The total power consumption would be reduced, at the expense of wasting some power in the added resistor.

This power waste (and consequent heat generation) can be a problem in high current circuits, but a simple load resistor works fine in low power applications (e.g. a 10mA LED used as a panel indicator) and you can't beat the simplicity / cost.


Nice work on your analysis!

Regards,
Bob

PS - +1E6 to Mr. Happy's post - that's a very useful analysis tool when you are thinking about this kind of problem.