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4 pi series/ parallel resistor headache [message #35064] Sun, 10 February 2002 08:16 Go to next message
tdc is currently offline  tdc
Messages: 34
Registered: May 2009
Baron
I am building the the 16 ohm 40 watt resistors all from 16 ohm 12 watt parts. I did read all the Gilmore does the Kama Sutra posts below and remain a little confused. But then the Kama Sutra lost me after the 40th or so position. Maybe a drawing would help the math challenged?? Do you combine two 16 ohms resistors in parallel and then hook the ends of the two parallel sets in series or do you hook two 16 ohms in series and then fuse the ends to make a parallel??
same question here [message #35065 is a reply to message #35064] Sun, 10 February 2002 08:50 Go to previous messageGo to next message
trancemitr is currently offline  trancemitr
Messages: 74
Registered: May 2009
Viscount
I was just looking at all the parts sitting on the floor wondering the same thing. I'm actually drawing up a picture right now to send to Wayne to (hopefully) get some confirmation. Wayne, I'll have a picture off to you shortly. Please watch your e-mail. Thanks.

Kevin

here we go, since it's wayne's day off [message #35066 is a reply to message #35064] Sun, 10 February 2002 09:21 Go to previous messageGo to next message
Sam P. is currently offline  Sam P.
Messages: 307
Registered: May 2009
Grand Master
Ok, take TWO of your 16 ohm resistors, and lay them side by side. Twist the very ends of their leads together, just a turn or two. It helps to hold the two wires near the ends, and use a second pair of needlenose to gently twist the exposed ends together. Solder the ends.

Take a second pair, and repeat the above steps.

Now take the first pair, and place it end to end with the second pair. One end of each pair will get soldered together. Measure the two free ends left...eureka, a 16 ohm,48 watt resistor.

Yes, feel free to connect them as two 16 ohm resistors in series first, then you are effectively now combining two 32 ohm ones in parallel to get back to 16 ohms. Same result either way. Sam

maybe I need to mention some guidelines?
1) two resistors in series will add, ie two 10 ohm resistors in series equals 20 ohms.
2) two resistors in parallel(sidexside) will be less, oversimplified, two 10 ohms in parallel will equal 5 ohms. easy to figure AS LONG AS the two R's are the same.

Re: here we go, since it's wayne's day off [message #35067 is a reply to message #35066] Sun, 10 February 2002 10:05 Go to previous messageGo to next message
Adam is currently offline  Adam
Messages: 419
Registered: May 2009
Illuminati (1st Degree)
Just connect all four leads on one side of the resistors together and connect all four leads on the other side together. It's just four resistors in parallel. The result is one collective resistor with four times the power handling and 1/4 the rating of a single resistor.

Adam

Re: here we go, since it's wayne's day off [message #35068 is a reply to message #35067] Sun, 10 February 2002 10:52 Go to previous messageGo to next message
tdc is currently offline  tdc
Messages: 34
Registered: May 2009
Baron
I quess either of the two ideas works. I understood the fact that the resistance came out but was wondering if the power capacity is the same either way. Sounds like it is. Thanks to you both. tdc
four 64 ohm resistors in parallel [message #35069 is a reply to message #35067] Sun, 10 February 2002 10:56 Go to previous messageGo to next message
Sam P. is currently offline  Sam P.
Messages: 307
Registered: May 2009
Grand Master
will also equal 16 ohms, Yep, if you are using regular wirewound power resistors, vs. the NON INDUCTIVE mills brand he mentioned. Building a 16 ohm part with the values available means they need to use the four 16 ohm ones, arranged as specified previously.

resistors in series
Rt = R1 + R2 + R3+...
resistors in parallel
1/Rt = 1/R1 + 1/R2 +...

Hope the formulas make this clearer. Sam


Us dummies need to stick together! [message #35070 is a reply to message #35065] Sun, 10 February 2002 13:56 Go to previous messageGo to next message
BillEpstein is currently offline  BillEpstein
Messages: 886
Registered: May 2009
Illuminati (2nd Degree)
My network only looks a little like Waynes, but it works. The key to making it go was measuring the Ohms (Ohmage?) as I went along. Then making the hook-up to the PXBJR4XXII.6 crossover was a little tougher.

Take 4 resistors and make a 4 pack so they're parallel, touching and all 4 leads pointing straight out at each end of the group. Put a wire tie around all 4. Now twist together ANY 2 pair of leads at one end of the 4 pack. Go to the other end and twist together the OPPOSITE pairs of leads from those you did first. If you twisted A to B and C to D at one end, you twist A to C and B to D at the other. You should be able to measure 16 ohms + or - 10% at any point ON THE SAME END.

From end to end you should measure 12 ohms. And no, I don't know why except E=IR, which means that the hypotenuse can be squared.

Make another bundle the same way EXCEPT leave the twists at the A-B, C-D end twisted in such a way as they stick out parallel to the bundle.Set both bundles on the window sill to cool.

Now we get to the dangerous part. Attach a wire to each connection A-B and C-D, on one of the bundles. Make'm long enough so one can reach from the tweeter + to the resting place of the crossover and the other long enough to reach the tweeter, errr horn.

At the other end of this same bundle attach one lead from the capacitor thingy to each of the twisted pairs A-C and B-D, and solder. That would be one lead to one pair only. That guy Faraday, what a monster! His Farads were so big we only use 47 TRILLIONTHS of one for this job! And of course, as we all know, his cousin, Abner Faraday invented Baseball. How cool is that?

Now you take the other bundle, you haven't soldered anything yet, right?, and solder the A-C, B-D ends. The A-B pair of leads becomes a sorta kinda wire you put through the tweeter + of the X-over and the C-B leads go through the - terminal. Put the wire from the C-D of the first block through the + terminal. Put a wire that only goes to the horn negative through the Tweeter - .

Measure 16 Ohms from the + X-over terminal to the end of the wire running to the horn +.
Measure 0 ohms from the - X-over terminal to the end of the wire that goes to the - horn connection.
Measure 16 ohms from + to - tweeter terminals. (Your meter may jump a bit 'cause of the capacitor)

Now you can solder all the terminal connections.

Solder wires to the = and - woofer terminals of your Eminence PXBII-1K6 X-over and you're done.
You know not what awaits you, unless you're Tom Brennan and he didn't read this far.

BTW, if there's anything wrong with this construction you only have yourselves to blame for following an admitted moron over the precipice and you can certainly disregard anything I've had to say about the Truth and Beauty of Pi Speakers.


by george, I think he's got it!(nt) [message #35071 is a reply to message #35070] Sun, 10 February 2002 14:03 Go to previous messageGo to next message
Sam P. is currently offline  Sam P.
Messages: 307
Registered: May 2009
Grand Master
nt
Re: four 64 ohm resistors in parallel [message #35072 is a reply to message #35069] Sun, 10 February 2002 14:23 Go to previous messageGo to next message
Adam is currently offline  Adam
Messages: 419
Registered: May 2009
Illuminati (1st Degree)
One problem is wirewound resistors are inductive. Don't use them in speaker crossovers. The other problem is no one makes 64 ohm resistors. Might be better to do a series/parallel network instead.

Adam

other problem is no one makes 64 ohm [message #35073 is a reply to message #35072] Sun, 10 February 2002 14:35 Go to previous messageGo to previous message
Sam P. is currently offline  Sam P.
Messages: 307
Registered: May 2009
Grand Master
resistors, I refered to them w/o looking at a chart. 68 is a real value, but four of them gives 17 ohms when you are done! Sam

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