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The more I searched throughout the internet the more confused I became.

Imagine I have 13 batteries, each of them 3.7V, 20A, 3000mAh and I want to power 1000 W 48V DC electric motor.

If I connect the batteries in series I will have 48V but only 3000 mAh. If I connect them in parallel I will have 39 000mAh, but only 3.7V

However there are a lot of current being unused and I figured I could use some step-up DC-DC voltage regulator, but I am unable to see how they work and what kind of it will I need. I assume they will decrease the current, hence decreasing the capacity in order to increase the energy.

What voltage regulator is best for this, how much of the amps/mAhs will it affect?

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    \$\begingroup\$ I think you are mixing units. mAh is not mA, and Wh is not W. If you put batteries in parallel there is no current being unused. \$\endgroup\$ – Wesley Lee Mar 27 '18 at 11:56
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    \$\begingroup\$ I see no mixing of units in this question. 20A refers to the rated max continuous discharge current, while 3000 mAh is the nominal capacity. \$\endgroup\$ – Dampmaskin Mar 27 '18 at 12:21
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You can convert between different combinations of voltage × current, but you can never make more voltage × current than you started with. Physics can be inconvenient like that.

Before even thinking about voltage and current, look at the total available power. If the batteries can't deliver the power, then no amount of conversion can help.

You say each battery can output 20 A at 3.7 V. That means they can each put out (20 A)(3.7 V) = 74 W. You have 13 such batteries, so the total available power is 962 W. You want to run something that takes 1000 W. There is a obvious problem here.

Note also the short time the batteries can sustain the 74 W per cell. Each battery is rated for 3 Ah. At 20 A, they would be completely drained in 9 minutes in theory. In practice, the capacity is usually lower at high current. You need to look at the battery datasheet carefully.

You can simply try to connect all 13 cells in series with the motor. That produces the right voltage with 20 A capacity. Battery current capacity is squishy, and this can probably get the motor going, especially if it is not initially loaded.

However, continuing for minutes will damage the batteries. If you continue long enough for the weakest cell to be empty, it will be seriously abused. That could include pyrotechnics, depending on the current the remaining batteries can still push through.

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  • \$\begingroup\$ So I need 14 batteries. That will make 1000W and 48V and 3000mAh. Seems like I will dry them very easily \$\endgroup\$ – Edenia Mar 27 '18 at 12:04
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    \$\begingroup\$ @Ed: It's not that simple with real batteries. \$\endgroup\$ – Olin Lathrop Mar 27 '18 at 12:06
  • \$\begingroup\$ Those batteries are LG HG2 li-ions. Very expensive \$\endgroup\$ – Edenia Mar 27 '18 at 12:07
  • \$\begingroup\$ 1000W is a lot for any battery. Are you making an electric motorcycle? Anyway, harnessing that much power from batteries in a safe and efficient manner takes some careful engineering. \$\endgroup\$ – Dampmaskin Mar 27 '18 at 12:18
  • \$\begingroup\$ I think those exact batteries are very good for this (Yes, electric bicycle), if something drains 3 amps for an hour, one battery can last for around 6 hours. If it is 14 of them, I can afford 42 amps for 6 hours at 3.7V \$\endgroup\$ – Edenia Mar 27 '18 at 12:22
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As Olin said, you need to carefully consider your power source. To answer the question in your title:

A DC-DC converter is any power system which takes a DC input and provides a DC output to the load. For example, a voltage divider is a type of DC-DC converter, as it is capable of providing a dc output to the load which is different from the source. (However, the voltage divider tends to be horribly inefficient, as it just sits there eating power.)

A step-up converter is a type of DC-DC converter and is the same as a boost converter. They are called this because they step-up/boost the voltage from the source to the output. These tend to be much more efficient than a voltage divider. In the ideal case we assume that the input power is equivalent to the output power, but you can realistically expect around 90% efficiency for most scenarios.

If you are assembling this circuit, I highly recommend that you take safety precautions and test for short circuits at low voltages before powering it up to higher voltages. If you try to power the circuit without a load, the current will blow up the capacitor. Furthermore, if your switching device has a duty cycle that approaches 1, you are in danger of short circuiting your power supply, so you should limit the duty cycle somehow.

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