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Hoffman Amps Forum image Author Topic: Ohm Selector Question  (Read 5710 times)

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Offline 69SG

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Ohm Selector Question
« on: October 30, 2010, 09:40:11 am »
I built a 50 watt JCM800 type Head. 4>8>16 ohm selector.
Sometimes I use a 1>12" 16 ohm cabinet.
Sometimes I use a 2>12"  8 ohm  cabinet.
Tell me if this is correct for using both cabs plugged in
8+16=24ohms divided by 2=12 ohms
Which selector setting is safer for the transformer
8 or 16 ohms
Thanx
WDB

Offline FYL

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Re: Ohm Selector Question
« Reply #1 on: October 30, 2010, 11:20:20 am »
Speakers in series: Rt = R1 + R2 ... + Rn
Speakers in parallel: 1/Rt = 1/R1 + 1/R2 ... + 1/Rn, which resolves as Rt = (R1 * R2) / (R1 + R2)

For two speakers in //, 8R and 8R => 4R, 8R and 16R => 5.333R

Closest tap is 4R. As the actual impedance is anything but constant, you may obtain better results in the low and low/mids using 8R.

Offline PRR

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Re: Ohm Selector Question
« Reply #2 on: October 30, 2010, 04:43:46 pm »
> Tell me if this is correct -- 8+16= .... =12 ohms

Resistors in parallel, the total is always _less_ than any individual resistor.

Try some combinations you know.

8||8 is 8 your way, but we (should) know that two 8s in parallel is 4.

8 in parallel with 8,000, say a speaker with a NFB resistor or line-out tap: your way says 4,004 ohms. But we "know" that the amp will still feel that 8 ohm load, plus a teeny added load in the 8K, the real answer must be mighty close to 8.

Yeah, 8||16 works out to an odd number.

Also the 8 is probably taking 2/3rd of the amp power, the 16 only 1/3rd. Unless the 16-ohm unit does something much better than the 8, or you can cover some small area the 8 isn't hitting, you must wonder if it is worth hooking the 16 at all.

> safer for the transformer

Bah, no difference in "safer". Don't run NO load. If you have a choice, don't run 32 on the 4 tap.

5 ohms is lovely safe fit for 4 or 8. There may be some tonal difference depending on the speaker's actual impedance curve and the amp's characteristics.

Offline 69SG

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Re: Ohm Selector Question
« Reply #3 on: October 30, 2010, 08:47:47 pm »

Let me go at this from a different direction.
If I have 3>16 ohm speakers paralleled.
what would be the prefered ohm setting on the output

Offline Shrapnel

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Re: Ohm Selector Question
« Reply #4 on: October 30, 2010, 11:56:02 pm »
Lets see... 3-16R loads in parallel with each other.

1/((1/16)+(1/16)+(1/16)) = 1/(3(1/16))= 1/(3/16) = 16/3 = 5.333333R
(Side note, the final 16/3 may look like a shortcut: Speaker R/Quantity in parallel BUT that only works when the impedance is the same for each speaker.)

4 Ohm tap is closest and slightly higher impedance, 8 ohm tap is fine for a little lower impedance on the output tubes.

Hmmm.... FYL already answered this a bit back as far as impedance load Using only two values though.

PRR and FYL already came to the same conclusion I state as well as far as hook-up.
-Later!

"All the great speakers were bad speakers at first" - Ralph Waldo Emerson

Offline 69SG

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Re: Ohm Selector Question
« Reply #5 on: October 31, 2010, 08:19:35 am »
yes the simplicity of it .
16 divided by 3
 :embarrassed:
« Last Edit: October 31, 2010, 08:22:21 am by 69SG »

Offline FYL

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Re: Ohm Selector Question
« Reply #6 on: October 31, 2010, 09:23:10 am »
Quote
16 divided by 3

Does it work with Divided by 13 amps?

 :angel

Offline Shrapnel

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Re: Ohm Selector Question
« Reply #7 on: November 01, 2010, 03:29:32 am »
Quote
16 divided by 3

Does it work with Divided by 13 amps?

 :angel

ROFL - You tell me  :icon_clown:
-Later!

"All the great speakers were bad speakers at first" - Ralph Waldo Emerson

 


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