I have a lipo battery 3.7V, that Ive externally confirmed can supply 2.5 amps (using a 10w resistor) (3.7 * 2.5 = 9.25 watts).
However, when connected to 12V Step-Up Voltage Regulator U3V50F12 https://www.pololu.com/product/2568
to get 12 volts, I can only draw about .3 amps * 12v = 3.6 watts
Ive read the specs, and realize theres some efficiency lost, but shouldn’t I be able to get at least double the output current (70% efficiency)?
Its rated for 5A input.
That regulator is rated for an input current of 5A. When converting from a lower voltage to a higher voltage, the output current it can deliver will be lower. You can look at the graphs at the bottom of the regulators product page to see the efficiency of the regulator at different input and output voltages and also to get a rough idea of the maximum current it might be able to output. For converting from about 3.7V to 12V, I would expect to be able to get several hundred milliamps to an amp of output current.
From the details you gave about your system it is unclear if your output current is expected. What load are you using with the regulator? If it is the same resistor from before, what is its resistance value? It sounds like in your calculations you are just using the nominal voltage of your battery. Could you measure its actual value and make sure it is fully charged? How are you measuring the current?
The load is an 8 watt , 12v, led bulb.
Powering it directly from a 12v power adapter I see it consuming about .7 amps.
However when feeding a 4 volt lipo through the regulator, the input current is only .3 to .4 amps.
The battery is fully charged.
Looking at the graphs again, it looks like the regulator won’t allow any where close to the required input current of 2 amps?
If that is the case could I chain two regulators together with an intermediate voltage? The current I’m drawing now seems way to low. Should be closer to eight watts. Getting about
a quarter of that.
More specific info:
Battery input: 3.4v @ 0.6A (2.04 W)
(battery is connected to a battery charger circuit, and the step up regulator, through led bulb:)
load: (Bulb) 9.51v @ 0.23A (2.18 W) (*** When I connect the bulb directly to a 12v power source, its draws 0.7 amps)
Confused why Im getting more power out, then in (faulty meter? charger influence?)
But at any rate, seems like, directly I can get 2.5 amps out of the battery outside the circuit.
So not sure why Im only drawing 0.6 amps above.
I think its because stepping 3.4v to 12v is just not feasible while drawing 2 amps.
Anyone know if I can connect the step up regulators in series?
If I could drive two: (3.3v to 6v) regulators, and chain them together, I think Id get much greater efficiency.
But I suspect they aren’t designed to be connected this way, if they’re both being fed the same input source.
To answer one of my own questions, does NOT look possible to chain to step-up regulators in series when the have the same input.
Just confirmed the input and output grounds are tied together.
[quote]Confused why Im getting more power out, then in[/quote]I am too, since that violates a fundamental law of physics (Conservation of Energy). Obviously, you are not measuring these quantities correctly.
Incandescent bulbs are a bad choice to use as loads, because the resistance of the filament when cold is far lower than when hot, by as much as a factor of ten. If the bulb is not glowing, you could be overloading the regulator in an oscillatory fashion.
You aren’t getting anywhere with these experiments, so tell us what you really want to do with regulator/battery combination.
Not claiming to have invented free energy ;). I am a little skeptical of my meter, the lower amperage settings have since died long ago, leaving only the 10Amp in functional.
but I think the unusual measurement may be some oscillation or other effect from the battery-charger + regulator.
I will re-measure everything tonight with the battery charger removed.
Also, its not an incandescent bulb, its an LED bulb. (12v / 8 watt).
It draws a full .7 amps @ 12v from a power supply.
Ultimately, Im trying to power it from a 3.7v lipo battery, using the step-up converter to generate the 12v.
The lipo battery claims it can deliver 13amps for brief periods, and I have confirmed that I can draw 2.5amps into a 10 watt resistor.
What Im finding, and I need to confirm, is that the step-up regulator ability to draw current drops quite quickly when the input voltage is low.
It seems to draw twice the current when using a 5V / 5A adapter as input.
Its hinted at in the graphs, as Im in the extreme region of the 3.3v -> 12v efficiency chart @ 700 ma.
The graph just truncates, which may be a clue that I should move to the middle.
If Im correct in thinking it just doesn’t draw much current at 3.7V, Im thinking of adding an intermediate step up regulator (6v) into the chain.
I know the efficiencies multiply, but even then, .8 * .8 is much higher than what Im seeing.
[quote]the lower amperage settings have since died long ago[/quote]The internal fuse is blown. Open the meter and replace it (keep a spare handy).
I agree that you should try to simplify your system by removing the battery charger before doing any more testing. It sounds like you are testing with some kind of LED light bulb; if that is the case, I recommend removing that too. If you can use the power resistor that you mentioned before or have access to several low value resistors that you could place in parallel, that would be much simpler to consider the measurements of. (If you use small through hole resistors, make sure you put enough of them in parallel that you don’t exceed their power rating, which is usually 0.25W.)
Once you have your system set up with just a supply, the regulator, and the resistive load, you should measure the input and output voltages and currents and calculate the efficiency to make sure it is close to what is expected. That will show if the regulator is working properly.
As for connecting your 3.7V input to a 5V regulator and then connecting the output of that 5V regulator to the input of a 12V regulator, that is possible, but will not result in a higher current output. For regulators it is the power in (Vin * Iin) times the efficiency that equals the power out (Vout * Iout), so when converting from a low voltage to a high voltage there will be a drop in the maximum output current. For example, if you use two regulators and were able to get 2A out of the first at 5V the power into the second would be 2A * 5V = 10W. Then assuming the second 12V regulator is 80% efficient you could write the equation Pin * eff = Pout which means 10W * 0.8 = 12V * Iout, so Iout is about 670mA. That is very similar to the maximum value shown in the efficiency graphs for using the #2568 regulator to convert from 3.7V to 12V.
Also, please keep in mind that the efficiency graphs for our regulators are made to show the efficiency all the way up to the thermal shutdown point, so you should not expect to be able to draw the maximum currents shown on those graphs for more than a few seconds without the regulator over heating.
Better late to this party than to never get here at all.
How to put this…
The minimum cut off for that regulator is 2.9 volts. Your battery is 3.7 volts. You are .8 volts above the cutoff. When you connect the high power LED to the regulator, the voltage drops quickly. You get below 2.9 and it’s game over.
As the only thing the regulator does in this case is step up the voltage, not the current, you need a power source that is at least twice or better than the minimum cutoff voltage and be able to deliver the rated current. If you want to pull 5 Amps out, you have to have 5 Amps going in to it.
I would recommend if you’re building backup lighting or just a flashlight using a Cree style LED, you get a higher voltage and current battery. An 11.1 or a 14.8 5000 mah 25c LiPo will do you right nicely. LiPo’s are compact and will last through a lot of cycles before they wear out. Use a liPo charger. They sell the pololu.com/product/2588 I have one and would recommend it to any one. Good bang for the buck.