DC/DC conversion to icncrease battery life?

[Dave01]

Afternoon All

Hope somebody can help me.

I am looking to produce a battery operated continuity tester, I want it to run from a 9v battery but with the 50 LEDs its supplying the battery life will only be around 2 hours. If I increase the voltage with a step up dc/dc converter will the battery life increase? If so how do I calculate it battery life? (No data sheet)

Thanks for your time

Kind Regards

Dave01

 

[Nigel Goodwin]

You can't get something for nothing, increasing the voltage decreases the current, plus you make a loss on the deal as well.

If you're talking just a small 9V battery, replace it with 6xAA (or larger) batteries, they will greatly out perform it.

A lot though really depends on the exact circuit, and what it's doing.

 

[dknguyen]

What Nigel is saying (in case you didn't completely understand it the first time) is that a battery can only supply a fixed amount of energy. THat energy comes as power supplied over a certain amount of time. The power may come in the form of a lot of current and a little bit of voltage, or a little bit of current and a lot of voltage. This is what a DC converter lets you control- increase current or voltage at the expense of the other. It can't increase both at the same time (which would be making power from nothing).

 

[Grossel]

.... increasing the voltage decreases the current, plus you make a loss on the deal as well.

You forgot to tell Dave01 WHAT current that decreases (not from battery).

In general - by putting an voltage increase circuit between the battery and the load, current from the battery will increase with an even greather factor than the voltage multiplier because of loss in transformer.

 

[Dave01]

Thanks

Can I explain exactly what I am trying to do, I am again hoping you will be able to help.

I am trying to use a work project (design & build a continuity tester for a modular wiring system). I am hoping I can use the same project as my uni project but it obviously needs to be much more involved/complex which is why i am trying to find a use for a DC - DC converter in the tester.

Does anyone have any ideas on how I can improve on the complexity of my project?

Again, Thanks for your time, it is much appreciated.

Kind Regards

Dave01

 

[dknguyen]

Well, to increase the voltage [current] by double, an ideal perfect DC converter would need to have double (more for real converters) the current [voltage] supplied to it by the battery. So the battery still supplies the same amount of energy for the same amount of time before dying, it is just in a different form (balanced between voltage and current). You can't make something from nothing as we've established...

but you can make something less from something even more and this is how linear converters work. Linear converters can only step down voltage. WHat they do is burn away the extra "voltage" to provide a reduced voltage. What this means is that if I need 5V@1A from the regulator, but I am supplying 9V to it, it will still draw 1A at the input and burn away the extra 4V as heat (the input and output currents are always the same, ideally in reality a little bit more is drawn to keep the regulator running). If you use a switching converter (step-down in this case) you can extend battery life because a perfect switching converter does not burn off the extra energy as heat. A perfect step-down switching converter only draws an Iin = (Vin/Vout)*Iout. It actually converts the balance of power between current and voltage at the output rather than burning off the extra. So in this case battery life would be extended.

So if your battery voltage is higher than what the circuit is using, you could increase complexity and lifetime by doing that...but not if your battery voltage is lower than your circuit voltage.

 

[crutschow]

A switching circuit such as a Joule Thief (do a Google search) can increase the amount of energy you can extract from a battery, since it can maintain a constant output power even as the battery voltage drops to a very low level.

 

[Dave01]

How will the amount of LEDs I need to use work with this circuit, they all seem to show only one LED?

This help is brilliant, Thanks!

 

[blueroomelectronics]

Why do you need 50 LEDs?

 

[crutschow]

How will the amount of LEDs I need to use work with this circuit, they all seem to show only one LED?

Forgot about you wanting to light 50 LEDs.

The Joule Thief, as designed, only operates one LED. I assume you need 50 because you are checking the continunity of 50 wires in a cable. For that you would need 50 Joule Thiefs, or a multiplexer to sequentially connect the 50 LEDs to one Joule Thief.

 

[blueroomelectronics]

Ahh now a mux should do nicely, 50:1 won't be bright but should be usable for a simple cable tester.

 

[crutschow]

You might actually want the multiplexer to go slow enough to light each bulb individually for a short period so that any short between wires would be indicated by two LEDs lighting simultaneously. Thus you could test for continuity and shorts with one test.

 

[audioguru]

Go to a battery manufacturer's website like Energizer Battery Company. Energizer Batteries & Flashlights. Energizer.com .
They have a detailed datasheet for every battery they make.

A 9V alkaline battery starts new at 9.0V. When it has a 270 ohm load (33mA) then its voltage drops to 8V in about 3 hours and drops to 6V in 20 hours.
Increase the current 10 times to 330mA and the duration is much less than 1/10th.