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I am working on a walking robot for my class in mechanical engineering. I have two 12 VDC solenoids that need to be actuated. Their resistance is 4 ohm so basically will require 6 amps @ 12V for the pair.

My battery is 6 x C cells in series, so 9V total. They have 8 Ah capacity. I want to boost the voltage to 12V using a module like this one.

Will this give me the 6A @ 12V I need?

I could also run the solenoids at 9V but I want to get as much force as possible out of them.

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    \$\begingroup\$ 8A max input current, so getting 8A output at a higher voltage would imply>100% efficiency, which isn't possible. \$\endgroup\$ Dec 6, 2023 at 16:49
  • \$\begingroup\$ @user1850479 Is 8 amps really the max though? Just because battery is rated at 8 Ah does not mean it will not discharge more amps momentarily. I really only need the 8 amps for a few seconds as my robot is turning. \$\endgroup\$ Dec 6, 2023 at 16:52
  • \$\begingroup\$ That’s significant load. Have you looked at different battery options? \$\endgroup\$
    – winny
    Dec 6, 2023 at 17:21
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    \$\begingroup\$ The Amazon listing says a max of 8A input current continuously. I wouldn't bet on higher working for something cheap like this. \$\endgroup\$ Dec 6, 2023 at 18:09
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    \$\begingroup\$ I’m thinking something other than alkaline, probably lithium ion. \$\endgroup\$
    – winny
    Dec 6, 2023 at 21:03

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if 6 C cells is 9V, then they must be alkaline, not NiCD or NiMH. 8A at 12V is equivalent to 10.6A at 9V. Call it 11A to allow for boost converter efficiency.

Check the manufacturer's data sheet for your C cells, but I would be surprised if they are capable of giving you about 11A without the cell voltage collapsing.

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No, it won't.

Even if the boost converter would be 100% efficient, 12V 6A out requires 9V at 8A.

If one battery has 150 milliohms of internal resistance, 6 batteries in series has 9V and 0.9 ohms internal resistance. If your load is a short circuit, only 10A will flow.

So even a 5A load the voltage would drop the voltage to 4.5V, and that's 22.5W and as per the maximum power theorem you can't draw any more power that that.

So if you need 72W, the batteries cannot provide any more than 22W.

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The key factor here is the internal resistance of different types of cell.

A fresh alkaline C cell has an open circuit voltage of 1.5V, and an internal resistance in the range 150mΩ to 300mΩ (according to this Energizer datasheet).

This implies that even a dead short across the cell (which will totally collapse the output voltage) will only give 5A-10A.

If you maximize the useable power by matching the load resistance to the internal resistance, you can't expect more than 2W-3W usable power per cell (with the same amount lost in internal heating).

For each solenoid the power needed is 12*12/4=36W, so 72W for the pair. Allow 90% boost efficiency, and you're looking for ~80W.

This implies that you will need somewhere between 30 & 40 fresh cells to drive the pair....

Even at 9V, the solenoids still need 9*9/4=~20W each, or 40W the pair, so 15-20 C cells.

Checking the datasheet for the large D cells, we see they have very similar internal resistance, so this doesn't help.

You should therefore consider using rechargeable batteries which typically have much lower internal resistance (NiCd, NiMH, Lead-Acid, or one of the lithium chemistries).

For instance, the equivalent Energizer NiMH C cell has an internal resistance of 11mΩ full charged, and has speced discharge curves at 5A with just 55mV drop. Even 1/2 charged the resistance is only 22mΩ.

This means you can get 15-30W of usable power out of each battery, so 6xC NiMH cells would be able to supply 90-180W of power, which is what you need. However, be aware that there will be significant internal heating of the batteries.

Also worth considering use a dedicated NiCd or LiPo battery pack designed for use in RC vehicles, as these have even lower internal resistance, which will drastically reduce the internal heating.

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