Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: JPK on March 04, 2023, 11:44:14 am
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There's a thread over on the Tele forum where someone posted about using multiple OT taps at the same time. Is it possible to do this with it working properly, not altering the primary reflected impedance? I thought you could only use one at a time with a selector switch. The other thing I wonder is if doing this loads the OT more than the design VA? The picture below is not mine. Copied from Tele.
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It will "...altering the primary reflected impedance ".
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The load on each of those secondary taps reflects twice the intended impedance to the OT primary.
The combination of the 2 results in the correct primary impedance.
Aiken shows this on his q&a page, but it’s been down for a while.
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^ Yeah I didn't see that each speaker impedance was double the respective tap when I posted this thread. But I did learn something. Thanks.
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There's a thread over on the Tele forum where someone posted about using multiple OT taps at the same time. Is it possible to do this with it working properly ...
If we connect to more than 1 tap, the individual loads are all in parallel.
Then as pdf64 says:
The load on each of those secondary taps reflects twice the intended impedance to the OT primary.
The User is intentionally using a speaker 2x the tap's impedance, which reflects "2x Primary Impedance." But then "2x Impedance 1" is in-parallel with "2x Impedance 2" which results in "2x Impedance / 2" = the designed primary impedance:
The combination of the 2 results in the correct primary impedance.
The reference for this is RDH4 (https://el34world.com/charts/Schematics/files/Books/Radiotron_4th_edition.pdf) Page 201-203, with this exact case shown in Figure 5.8 on Page 203.
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Thanks guys. Always fun to learn something new.
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The reference for this is RDH4 Page 201-203, with this exact case shown in Figure 5.8 on Page 203.
Interesting :thumbsup:
Franco
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The load on each of those secondary taps reflects twice the intended impedance to the OT primary.
The combination of the 2 results in the correct primary impedance.
Aiken shows this on his q&a page, but it’s been down for a while.
This makes absolutely no sense to me considering the postulation by the op. The two paralled outputs are different impedances one is 4 ohm and one is 8ohm the fact they have speaker resistances doubled to reflect correct impedances back makes no difference.. still they are different.. and 4 ohms paralled with 8 ohms equals 5.33 ohms..
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It’s the impedances reflected back to the primary that are in parallel.
The speakers themselves are not in parallel.
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Ahh ok I get it .. however determining the Z out with Kirschoffs law and the formulas is a tad beyond my pay grade. I could probably work through it but I'll just take your word for it.
pdf you don't happen to have a copy of Aikens discussion on this do you?
Thanks
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pdf you don't happen to have a copy of Aikens discussion on this do you?
Sorry, no.
He doesn’t go in depth with it.
If it’s nagging at you, maybe the the wayback machine internet archive a try?
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This makes absolutely no sense to me ...
Ahh ok I get it .. however determining the Z out with Kirschoffs law and the formulas is a tad beyond my pay grade. I could probably work through it but I'll just take your word for it.
Let's assume a case where we have 40w applied to a transformer that has a 4kΩ primary, and 8Ω & 4Ω secondary taps.
Volts = √(Power x Impedance) = √(40w x 4kΩ) = 400v on the Primary side
The fundamental property of transformers is that if 1v-per-turn is applied to any winding, all windings also have 1v-per-turn. So the Turns Ratio tells use the ratio of voltages among the windings. For 4kΩ : 8Ω, 4Ω that's
8Ω Tap ---> √(4kΩ / 8Ω) = 22.36 : 1
4Ω Tap ---> √(4kΩ / 4Ω) = 31.62 : 1
So when we apply 400v to the primary, it will be stepped down to secondary voltages in proportion to the turns ratios:
8Ω Tap ---> 400v / 22.36 = 17.89 volts
4Ω Tap ---> 400v / 31.62 = 12.65 volts
Assume we have a speaker attached each tap. They will attempt to draw:
8Ω Tap ---> 17.89v / 8Ω = 2.236A
4Ω Tap ---> 12.65v / 4Ω = 3.163A
Going from Primary-to-Secondary, the Turns Ratio stepped-down Volts (because the secondary has fewer turns) and stepped-up Current. Now going back the other way from Secondary-to-Primary, Current is stepped down by the same ratio. Let's find out how much current each load attempts to draw on the primary side:
8Ω Tap ---> 2.236A / 22.36 = 0.1A = 100mA
4Ω Tap ---> 3.163A / 31.62 = 0.1A = 100mA
If we only have one load on one tap, then either attempts to pull 0.1A through 4kΩ. how much power is that?
4kΩ x 0.1A = 400v
400v x 0.1A = 40 watts
Now recall we started out assuming 40w applied to our 4kΩ primary, which was how we knew the applied voltage. Now assume Applied Primary Volts remains unchanged at 400v, and the Applied Primary Power remains unchanged at 40w. We see for the voltages that will be delivered to the secondary taps that if both are loaded, each tries to pull 0.1A (when referred to the primary) for 0.2A total.
Let's find the Primary Impedance implied when 400v results in 0.2A of current:
Volts / Current = Impedance
400v / 0.2A = 2kΩ
This shows that if we have a 8Ω load on the 8Ω ta at the same time that we have a 4Ω load on the 4Ω tap, the Primary Impedance becomes 2kΩ rather than our original 4kΩ. This is the same result as though we used two 8Ω loads in parallel on the 8Ω tap, for a total load of 4Ω load.
Therefore we conclude that loading multiple taps is equivalent to parallel loads on a single tap. And so if we want to reflect the original-design Primary Impedance, we need to use a load of double the tap's marked impedance (for 2 taps loaded), or a load of 3x the tap's marked impedance (for 3 taps loaded), etc.
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Thanks HPB So I could actually use a 32 ohm speaker on a 16 ohm tap combined with a 16 ohm speaker on an 8 ohm tap.
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... So I could actually use a 32 ohm speaker on a 16 ohm tap combined with a 16 ohm speaker on an 8 ohm tap.
Exactly!
Or since 32Ω speakers are short on the ground, a pair of 16Ω speakers in series for 32Ω total, applied to the 16Ω tap. Then a 3rd 16Ω seamer on the 8Ω tap.
3 x 12" 16Ω Celestion version of a tweed Bandmaster?
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Cool.. I do have 4each 32ohm 10" speakers . can anyone guess what amp they came out of?
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I do have 4each 32ohm 10" speakers . can anyone guess what amp they came out of?
Not "Amp" but "cabinet."
Ampeg SVTs used 8x10 cabs with 32Ω speakers all wired in parallel for a 4Ω load. It could be that your speakers came from such a cab.
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Bassman 10?
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Yes Bassman 10 a Combo Amp weirdest one I have seen in the bass amp realm
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I gotta take some time and study post #11. Thanks HBP!