Thread: Mosfet Heating Circuits View Single Post

12-05-2011, 02:26 AM
 Harvey Gold Member Join Date: Jul 2009 Posts: 1,137
Quote:
 Originally Posted by Guruji Yesterday I did two resistors in parallel and as Harvey told me the amperage remained the same. Interesting I will add more and see how much heat will I can get from more resistors.

Hi Guruji,

Are you referring to this post?
Mosfet Heating Circuits

If so, I wasn't indicating that parallel resistance does not increase current although I can see how my post may have been taken that way.

Instead, I was stressing that we can add more resistors in parallel provided the resistance in higher in each so that the resulting current stays below the MOSFET amperage rating.

The total resistance formula for parallel resistors is:

1/((1/r1) + (1/r2) + ... (1/rN))

So, lets say r1 = 5 Ohms, r2 = 7 Ohms and r3 = 10 Ohms:
1/((1/5) + (1/7) + (1/10)) = 2.26 Ohms

Clearly, the current will be more than any one of them by themselves because the formula for current is Volts / Ohms = Amps and the lower the Ohms the higher the amps.

So, if we keep our amps to 6A, the voltage would have to be no more than 13.56V with that setup. But what if we want to switch 900V instead?

Now we could put 66 sets of those 3 parallel resistors in SERIES (900V / 13.56V = ~66) and the result would be about 6A for the entire SERIES / Parallel Matrix of 198 Resistors.

900V
\ \ \
r1 r2 r3 (set 1)
r1 r2 r3 (set 2)
. . .
r1 r2 r3 (set 66)
/ / /
0V

Each parallel set is 2.26 Ohms and that in series 66 times = 149 Ohms

So the 900V across the entire matrix of 149 Ohms = 6.04 A

And the power would be 900 x 6.04 = 5.4KW

Cheers
__________________
"Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor
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