Electronics > Beginners

Battery Discharging Circuit

(1/2) > >>

vincentleest:
I'm working on a project that requires me to discharge a battery from 12V to 10V with a constant current of 14A.

I read about this thread and took batee's design and modified it for a bit.
https://www.eevblog.com/forum/index.php?topic=1219.msg15720

Q6,7,8  are IRF3910
Q5 is ZXT450

I am not very confident about my changes, please point out my mistakes and give me some comments.

Thank you!

qno:
you need to replace the resistors of 1 k by a resistor of 0.15 ohms.

You need to cool the IRF3910 on a very large heatsink that can dissipate 170 Watt.

vincentleest:

--- Quote from: qno on October 30, 2010, 08:26:14 pm ---you need to replace the resistors of 1 k by a resistor of 0.15 ohms.

You need to cool the IRF3910 on a very large heatsink that can dissipate 170 Watt.

--- End quote ---

did you mean the 3 parallel resistors connected to the transistors? or some other resistors?

and thank you for you help =)

qno:
The 3 1k in the sources of the Fets.

With a source resistor of 1k no more than 12 mA can flow in the FETs.

Feanor:
I would just like to double check how you got a resistor value of .15ohm. I got 0.333ohm. Where dd I go wrong?

14 amps / 3 = 4.667A in each source resistor.

The output of the 1k/820 ohm voltage divider will settle to 0.7V greater than Vout.

So the input needs to be 0.7*1820/820 = 1.553V. This voltage is the IR drop across one source resistor.

4.667A flowing through the source resistor gives us 4.667*R = 1.553V

R = 1.553/4.667 = 0.3327ohm ~=0.333ohm = 3X1 ohm resistors in parallel.

If I made a mistake please point it out.

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks WYSIWYG Editor
Powered by SMFPacks Advanced Attachments Uploader Mod