OT with 4, 8 16 tap...

[FunkyE9th]

If I have an OT with 4, 8 and 16 tap, and the primary is 2.5K, 5K and 10K respectively, and I have separate output jacks for each tap, can I plug in an 8 Ohm speaker into the 8 ohm jack and another 4 Ohm speaker in the 4 Ohm jack at the same time?

If so, does the tube then see a 7.5K primary?

Thanks,

-FunkyE9th

[Firestorm]

Out of my depth here. Hopefully a real transformer guy can address this. But my sense is the answer is no. Even using different taps, you'll essentially be paralleling the speaker load, so the reflected primary impedance would seem to be more in vicinity of 1.6K. If you want to do multiple speakers using multiple taps, you should look at the way a Fender 135-watt Twin or Bassman is wired. They used shorting jacks and two taps. Plug in one speaker and you use the whole secondary winding; plug in two speakers and they get put in series and you wind up connected to half the secondary winding.

[El_Martin]

Hi!

If you plug in two speakers at the appropriate outlets, you'll end up with half the reflected impedance. Mismatch. Sometimes ok, sometimes not. Depending on amplifier design.

I usually plug in two speakers at the half ohmage(?). Plug in 16 Ohms into 8 Ohm socket and another 8 Ohm into 4 Ohm socket.

If you have two speakers w same resistance, plug in parallel into half ohm socket or adjust impedance selector.

Ciao
Martin

[Dr-Joned]

What do you mean respectively? If you load the 4 ohm tap with a 4 ohm load, the resulting primary should be the same as loading the 16 ohm tap with a 16 ohm load. How are you measuring the impedance ?

[FunkyE9th]

Dr Joned,

The formula for a transformer is...

Z1/Z2 = (T1^2)/(T2^2)

Z1 is the primary impedance
Z2 is the 2ndary impedance
T1 number of turns on the primary
T2 number of turns in the 2ndary

So if you change the tap on the 2ndary, you change the number of turns, therefore you change the impedance of the primary. At least that is how I understand it. If I'm wrong please correct me, that way I'll learn.

For example here's a spec of a hammond 125ESE...

Specifications
Audio Watts: 15
Primary Impedance (Ohms): 2,500 to 10,000
Secondary Impedance (Ohms): 4 to 32
Max. DC Bias (mA): 80
Mounting centers: 4" u-bracket
Weight: 3.0 lbs.

[FunkyE9th]

Hi!

If you plug in two speakers at the appropriate outlets, you'll end up with half the reflected impedance. Mismatch. Sometimes ok, sometimes not. Depending on amplifier design.

I usually plug in two speakers at the half ohmage(?). Plug in 16 Ohms into 8 Ohm socket and another 8 Ohm into 4 Ohm socket.

If you have two speakers w same resistance, plug in parallel into half ohm socket or adjust impedance selector.

Ciao
Martin

Can you please explain how you get half the reflected impedance. I don't get it?

[Phil_S]

Let's assume you've got three taps on one secondary winding and you've wired each tap to a separate jack. All three jacks share the "0" or common wire that goes to ground.

If you plug in a 16 ohm load and an 8 ohm load, they are in parallel. Ohm's Law says that the load is now 8 and 4 or a total of 12 ohms. (I'm not going into Ohm's Law, as someone else already has done that.) So, plug the 16 ohm speaker into the 8 ohm jack, and the 8 ohm speaker into the 4 ohm jack. In this way, you can use two cabinets and "show" the appropriate load on the secondary.

If you only have 8 and 4 ohm speakers, you are out of luck as far as I can tell, unless you resort to something a bit more sophisticated.

[Dr-Joned]

Ok, lets try this. I've attached the schematic for the transformer you mentioned. They are basically showing you the possible combinations of loads you can use. I also added a modded picture to show and explain the loads.

If you put a 4 ohm speaker on the Common(0 w) and 4 ohm tap, you will reflect a 10000 ohm load on the primary side (Plates). The same goes for an 8 ohm speaker on the 8 ohm tap, a resulting 10000 ohm primary.

Now getting to your question, if you connected a 4 ohm speaker to the 4 ohm tap AND a 8 ohm speaker to the 8 ohm tap, you would be halfing the reflected load on the primary side. The load on the primary would be 5000 ohms now. This could be very dangerous for some amps if they are are not designed to be loaded that heavy.

[FunkyE9th]

Thank you! Now I realized what I did wrong. What I did was calculate the turns ratio for an 8 Ohm spkr and a 5K primary (25:1) and made the mistake of applying the same turns ratio to 16 and 4. Duh... the turns ratio is changing. That's why I arrived at the numbers I had on my 1st post. It makes more sense that the tube always the sees the same primary load if you plug in the correct impedance at the correct tap. Again duh...

So now it also makes sense that the tube sees half if I plug 4 Ohm Spkr into the 4 Ohm tap and an 8 ohm spkr into the 8 ohm tap at the same time. Both speakers will reflect the same load (Zr) on the primary and the reflected loads are in parallel, so Zr || Zr = .5Zr. Make sense now.

Again thank you for taking the time.

-FunkyE9th

[Structo]

If you plug in a 16 ohm load and an 8 ohm load, they are in parallel. Ohm's Law says that the load is now 8 and 4 or a total of 12 ohms. (I'm .

I'm a little confused by this as from what I remember, Ohms Law for a parallel circuit means the total will be less than the lowest value of the pair.
So 8 and 4 in parallel would become 2.66 ohms.
1/Rt = 1/R1+1/R2

Am I correct in this?

[Phil_S]

Structo: Your calculation is correct and mine is not. But that's not the answer to the question. I mixed things I should not have. See Martin L, above. He says same as me about plugging 16 into 8 and 8 into 4. From a practical perspective, that's what you need to know. You can't use an 8 and a 4 with this OT. (Well, you can, but it isn't appropriate.)