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Recently I purchased two Samsung 30Q 18650 Batteries, and I want to use them to power a project, but I am confused on how, say, USB Battery packs, output a constant 5v 2.1 Amp by using an 18650 battery.

To get enough capacity and at a decently high amperage, I know I am going to have to put the batteries in parallel.

However, once I get them in parallel, how do I tell the two batteries to output 5V 4.5A that I need constantly? Do I use a converter as shown in the picture below? enter image description here

I see that the module can boost my output to 5V, but how would I tell the batteries that I want to also take 4.5A out of it? Especially in Parallel? Is there a converter where I can adjust both Voltage and Amperage?

In addition, what about protecting the batteries? Is there a board that also protects the batteries from damage due to, say, too much discharge?

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  • \$\begingroup\$ "how would I tell the batteries" ...whisper \$\endgroup\$ – Maple Aug 6 '18 at 4:36
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First thing is, you are grossly underestimating the max current you can draw from those batteries. If they match the ones I looked up, you can pull 15A @ 2.5-3.6V.

You're hoping to get 4.5A@5V out, and assuming your voltage conversion is about 80% efficient (devices like the one you are showing often are closer to 90% efficient when you run them close to max capacity.)

4.5A * 5V / 80% = 28.125W input are necessary.

At the end of it's discharge cycle your battery will only be at 2.5V, which means you would need

28.125W / 2.5V = 11.25A input, and your single batteries are capable of 15A.

Charging lithium ion batteries is a bit complicated, especially charging a series or parallel set of batteries without separating them, so if you're either using a single battery or you're taking them out to charge them, your project will be a bit simpler.

We're given Ohm's law to understand the relationship between voltage, current and resistance in a DC circuit.

\$I=E/R\$

Where \$I\$ is current in (A)mps, \$E\$ is voltage in (V)olts and \$R\$ is resistance in ohms(\$\Omega\$).

We're given Watt's Law for the relationship between voltage, current and power, in (W)atts, expressed as \$P\$.

\$P=EI\$

These relationships are fixed and predictable. The device you're looking at is a step up (boost mode) constant voltage switching regulator. Once you set it to 5v output, it will draw current as necessary in order to maintain it's output voltage, and for the most part, it should be able to do so up to it's rated output current. If you change the load, say put 2 leds in parallel instead of one, that will halve the resistance, and because the device is keeping the voltage the same it will have to output twice as much current.

So with that device, you should be able to get a steady 5V out of it up to about 5A, but you don't need to draw the 5A.

As far as the batteries, if you check the datasheets on them, they appear to be rated at 15A continuous current, they're 3.6V at full charge and 2.5V just before they're considered depleted. If you want, you can take the worst case, right before the battery is depleted, and apply Watt's law,

\$P=EI\$ where \$E\$ is 2.5V and \$I\$ is 15A, the maximum continuous current from the datasheet. \$P=\$ 37.5W max roughly per battery

You know you want to be able to draw up to 4.5A at 5V, so using the same equation, a maximum of 22.5W output

Efficiency for a voltage converter is the relationship between input and output power, namely

\$Efficiency=\frac{Power Out}{Power In}\$

And because Watt's law let you calculate your worst case input and output power, you can figure out how efficient your converter has to be to work off of 1 (or a particular set of) batteries.

\$Efficiency=\frac{22.5}{37.5}=60\%\$

Your device is marked at 94% efficiency, which is probably only accurate at full load, but nonetheless it should be OK even with only 1 almost depleted battery.

About the device you're looking at:

These devices are rated by output current, and within their ratings, they should be able to sustain their necessary input current. I find they usually have a lower dropout than rated (lower minimum voltage than they actually say) but if you want to discharge your batteries all the way down to 2.5V, you may have to find a converter with a lower minimum voltage. If you're cutting it close, you should make sure that your circuit will shut down when the battery reaches 2.5V to prevent overdischarge. Depending on what you're using your circuit for, how long you want it to run for, etc, you still may want more than one battery. Lithium-Ion batteries are good for fast discharge, but they can become quite hot in the process.

So as long as you give it 3 or more volts, your board will try to maintain a 5V output, and if your load is less than 5A at 5V, the module will output that amount instead. The current rating on the picture is the maximum, not a constant amount of output.

When you're thinking of 2.4A on a battery pack, I think the confusion is perhaps coming from a 2.4Ah lablel. An Ah is a unit of battery capacity, and 2.4Ah means that the battery is capable of delivering 2.4A of current for 1 hour. If you only draw 1 amp of current, it will last longer than 1 hour. It's referring to the max capacity and capability of the battery. In reality you could draw much less or much more than 2.4A from the battery.

A good example is battery powered tools. I'm on my third or fourth generation of Milwaukee cordless tools now. If I take an 18V battery pack with a 2.4Ah rating and hook it up to my drill and start drilling 1/4" holes in soft metal or wood, it might last about an hour before the battery runs out. This means the drill, with that particular load, is using about 2.4A. If I pop the battery in my sawzall, or even just use a larger drill bit, the battery life will be much shorter, because these tools are doing more work and using more power.

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  • \$\begingroup\$ @K H, Thank you so much for your response; however, I have a quick question. Say that my lithium-ion battery discharges to the 2.5V at the end of the dischage cycle. Won't that mean I need a booster board that can handle around 11.25 Amps input in order to boost the voltage to 5V while the cycle is at 2.5V? This one only states 5A max. Or is it that these boards output 5A maximum, higher than what I need? In addition, how come most common battery packs only output 2.4A? \$\endgroup\$ – Omar Sumadi Aug 7 '18 at 0:15
  • \$\begingroup\$ Added to my answer in reply to this. Just keep reading at the bottom. \$\endgroup\$ – K H Aug 7 '18 at 1:02
  • \$\begingroup\$ @K H, you talk about protecting the battery against over-discharging below 2.5V and protecting against short circuits. I know there are boards that protect against this sort of thing: how would I go about combining one of these 5V boost boards with a protection circuit? Should I be able to find one all-integrated into one board instead of buying a protection board and boost board separate? How would I go about combining the two if not? In addition, so, this board should be able to take in near 12A I need because the output will be 5A @2.5V, and it's rated based on output? \$\endgroup\$ – Omar Sumadi Aug 8 '18 at 3:35
  • \$\begingroup\$ @K H, in addition, if my load is wanting to draw 5V 5A cconstantly until battery depletion, this board-battery combo will supply that constantly given there is battery power? Just wanted to confirm. \$\endgroup\$ – Omar Sumadi Aug 8 '18 at 3:37
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The load determines the current drawn from a constant voltage source.

If your load is designed to draw 2 amps from a 5 volt source, it will draw 2 amps regardless of the source's capability, as long as the source can supply at least 2 amps at 5 volts.

If you attempt to draw 4 amps from a source that can only supply 2 amps, something bad will happen - the source voltage will drop to a point where it can supply the current the load demands (but if it falls too low, the load may not work correctly.) Or components in the source will overheat, and release the "magic smoke" that all electronics requires to work.

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I am confused on how, say, USB Battery packs, output a constant 5v 2.1 Amp by using an 18650 battery.

They use a circuit to convert the 3.2-4.2V from the battery up to 5V.

Also they don't output a constant 2.1 amps.

how do I tell the two batteries to output 5V 4.5A that I need constantly?

You can't.

Do I use a converter as shown in the picture below?

You can use a converter to change the voltage.

The converter doesn't tell the batteries anything though. You can use soap to make soapy water. Does the soap tell the tap to make soapy water? No, the soap makes soapy water. And the converter makes high(er) voltage electricity.

I see that the module can boost my output to 5V, but how would I tell the batteries that I want to also take 4.5A out of it? Especially in Parallel?

You don't.

Is there a converter where I can adjust both Voltage and Amperage?

No such thing exists. You can set the voltage, or the current/amperage, but not both.

As a power supply designer, you get to choose one. The other one will depend on the one you chose, and the load resistance.

In addition, what about protecting the batteries? Is there a board that also protects the batteries from damage due to, say, too much discharge?

Yes, that is a good idea. If you are careful you might not need one, but it can save you from accidentally damaging your batteries.

It's a good idea, but you don't need one.

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