? On using two taps on OT simulataneously

[hasserl]

A question came up on another amp board I was visiting today, and I thought I would bring it here to see if I could get a better answer than what I am getting over there. No argument going on, I'm just not getting a very technical answer to the question.

If an amp is set up with multiple impedance speaker jacks, say 4, 8 & 16 ohms, each one connected to the corresponding tap on the output trans, what is the effect of running loads on two of the taps simultaneously? I.e., say a 4 ohm load on the 4 ohm tap and a 16 ohm load on the 16 ohm tap.

The typical response is no, you can't do that, but without a good explanation of why not. Since I like to understand why not, I wanted to press for a better answer. It seems off hand that the each load would present the correct load back to the primary, perhaps with the result being similar to running parallel loads on a single tap with the total primary load being halved. Is this completely wrong? Can anyone explain it?

Thanks
Hasse

[Morris Slutsky]

A question came up on another amp board I was visiting today, and I thought I would bring it here to see if I could get a better answer than what I am getting over there. No argument going on, I'm just not getting a very technical answer to the question.

If an amp is set up with multiple impedance speaker jacks, say 4, 8 & 16 ohms, each one connected to the corresponding tap on the output trans, what is the effect of running loads on two of the taps simultaneously? I.e., say a 4 ohm load on the 4 ohm tap and a 16 ohm load on the 16 ohm tap.

The typical response is no, you can't do that, but without a good explanation of why not. Since I like to understand why not, I wanted to press for a better answer. It seems off hand that the each load would present the correct load back to the primary, perhaps with the result being similar to running parallel loads on a single tap with the total primary load being halved. Is this completely wrong? Can anyone explain it?

Thanks
Hasse

Yeah, I'd guess that it'd be just like that. A transformer being a linear device and all that, the loads would just add. So if you had a 4 on the 4 and a 16 on the 16, you'd have a 50% load which some amps can handle and some amps maybe shouldn't, and you won't get maximum output power, but yeah that oughta work. If you had an 8 ohm speaker on the 4 ohm tap and a 16 ohm speaker on the 8 ohm tap, the load ought to be just perfectly right in theory. It sure seems like it should be like that . . .

Basically yeah, for each one take (load/tap) to give you the fraction of impedance and combine them like parallel resistors so if you worked out:

load fraction = 100 * 1 / ( (tap1/load1) + (tap2/load2) + (tap3/load3) ) that would probably be right.

So if you had a 16 ohm on the 4 and a 16 on the 8 for example load = 100 * 1 / ( 4/16 + 8/16) = 100 * 1 / (3/4) = 100 * 4 / 3 = 133% impedance

something like that. Of course your power would be split funny too, that's the other issue, although this impedance isn't TOO far off one of your speakers would be getting twice the power of the other one, that might be why people suggest that this shouldn't be done.

[Wilder Amplification]

The "proper" way of doing this (and a very neat way to run mismatching cabs like 16 and 8 ohm cabs simultaneously) is to run a 16 ohm cab on the 8 ohm tap, then run an 8 ohm cab on the 4 ohm tap. Placing the 16 ohm cab on the 8 ohm tap doubles the reflected plate-plate impedance, then placing the 8 ohm cab on the 4 ohm tap cuts it back in 1/2 to the stock value. This gives you a matched impedance while both cabs see the same output power.

Doing it the way the OP described would give the same effect as mismatching down with one cab, (i.e. running an 8 ohm cab on the 16 ohm tap) which cuts the reflected impedance in 1/2. It is for this reason why it's not a good idea to do it the way the OP described, but to move the cabs down a tap.

If you think of an output transformer as just a power transformer for audio frequencies (a push-pull output section is just a full wave center tap rectifier circuit turned upside down and backwards) this becomes easy to see. Example...a 50 watt amp putting out full clean output power puts out 20 volts on the 8 ohm tap while putting out 14.14 volts on the 4 ohm tap.

(20^2) / 16 = 25 watts

(14.14^2) / 8 ohms = 25 watts

25 watts x 2 cabs = 50 watts total power

Here you can see that both cabs while being of a different impedance will see the same output power since the two taps put out different voltages. Since the total output power is the same, the tubes are seeing the same load as they would see with a single cab hooked up to its matching tap.

[hasserl]

OK, thanks for the replies. It seems that both of you are thinking along the same lines as I. Would like to see any more answers that anyone would like to give, particularly if they don't agree with what's been posted.

[mooreamps]

The "proper" way of doing this (and a very neat way to run mismatching cabs like 16 and 8 ohm cabs simultaneously) is to run a 16 ohm cab on the 8 ohm tap, then run an 8 ohm cab on the 4 ohm tap. Placing the 16 ohm cab on the 8 ohm tap doubles the reflected plate-plate impedance, then placing the 8 ohm cab on the 4 ohm tap cuts it back in 1/2 to the stock value. This gives you a matched impedance while both cabs see the same output power.

I just don't know about this one. I have to think of some way to verify this. In my stereo tube keyboard amp, I have the 8 ohm woofer driven from the 8 ohm tap, and the 4 ohm ribbon tweeter driven from the 4 ohm tap. I'm still not seeing how this way would alter the reflected impedance back to the power tubes.

Ya, I've read similar threads on this before, and like the original poster of this thread, I've not read a better explanation of why not. If I set up a power transformer doing it the way Wilder suggests ; putting a 6 volt lamp off the 12 volt tap ; and a 3 volt lamp from the 6 volt tap ; it would work. But it doesn't. I'd put the 12 volt lamp from the 12 volt tap and the 6 volt lamp from the 6 volt center tap ; and yes I'd buy that.

Let's take it one step further. 12.6 volt filaments from the 12.6 volt tap, and then 6.3 filaments from the 6.3 volt center tap ; and yes I'd but that too. So, I'm not seeing any difference between the secondary coil of a power transformer compared to the secondary coil of the output transformer. So, for an output transformer ; 16 ohm tap drives a 16 ohm loud speaker "and" the 8 ohm tap driving an 8 ohm loud speaker.

-g

[Koreth]

Doesn't doing it that way effectively put the 8 and 4 ohm loads in parallel with each other, giving a 2.66...Ω load on one side and God only knows on the other side?

[Wilder Amplification]

If I set up a power transformer doing it the way Wilder suggests ; putting a 6 volt lamp off the 12 volt tap ; and a 3 volt lamp from the 6 volt tap ; it would work. But it doesn't. I'd put the 12 volt lamp from the 12 volt tap and the 6 volt lamp from the 6 volt center tap ; and yes I'd buy that.

mooreamps as usual has it WAY backwards. Maybe...just maybe...there's a reason why he's been banned from quite a few message boards.

[mooreamps]

Doesn't doing it that way effectively put the 8 and 4 ohm loads in parallel with each other, giving a 2.66...Ω load on one side and God only knows on the other side?

I don't see how. If you draw a schematic of it, you will see there is a part of the secondary coil between the two different loads.

-g

[mooreamps]

mooreamps as usual has it WAY backwards. Maybe...just maybe...there's a reason why he's been banned from quite a few message boards.

Then post a mathematical proof it's way backwards ; or post a way to measure the change in reflected impedance.

Besides ; as a side note ; that tube keyboard amp I use still works set-up that way. Maybe we should ban it too ????



-g

[pdf64]

'12.6 volt filaments from the 12.6 volt tap, and then 6.3 filaments from the 6.3 volt center tap ; and yes I'd but that too. So, I'm not seeing any difference between the secondary coil of a power transformer compared to the secondary coil of the output transformer. So, for an output transformer ; 16 ohm tap drives a 16 ohm loud speaker "and" the 8 ohm tap driving an 8 ohm loud speaker.'

But the supplies to the primaries are very different in the case of a power transformer and output transformer. The power line will put a constant voltage on to whatever load you connect to it, until it trips out, whereas power tubes require an impedance matched load.

Re the keyboard amp, I'd assume that the tweeter has a crossover of some sort to roll off (ie impede) low frequencies, and probably the converse for the woofer. That should maintain the primary impedance at a fairly constant level, as the drivers won't both be fully on at the same freq.

[Wilder Amplification]

Take a look at my math above and you shall see it very clearly.

Stop thinking in terms of "impedance" and look at the OT as a simple power transformer and all should become clear.

[cbarrow7625]

Impedance matching aside, you still have to make sure there is not excess current draw through the transformer. Assuming we have a transformer designed the deliver 50 watts; wiring an 8 ohm, 50 W speaker to the 8 ohm tap and a 4 ohm 50 watt speaker to the 4 ohm tap will result in a net "power draw" of 100 Watts. Even worse, The current for both speakers is going to pass through the 4 ohm tap, doubling the current flowing through that portion of the winding. Sounds like a good recipe for an overheated & subsequently dead transformer.

Connecting the speakers the way Wilder describes it, and as he showed with his math, will maintain the proper 50 Watts of output power through the transformer.

I would assume that the keyboard amp you are describing has a passive crossover between the transformer & speakers. The speakers are not drawing current at the same frequencies, so connecting them to their matching impedance taps probably is the correct thing to do. For a guitar amp where both speakers are drawing current at the same frequencies I would "mismatch" in the manner wilder describes.

[mooreamps]

Take a look at my math above and you shall see it very clearly.

Stop thinking in terms of "impedance" and look at the OT as a simple power transformer and all should become clear.

I am looking at it as a simple power transformer ; that's why your set-up does not make sense to me.

and no not all the current is passing thru the 4 ohm tap.

-g

[Merlinb]

To restate the point; suppose you have a nominal 4k transformer, with 4 and 8 ohm taps.
If you hook an 8 ohm speaker to the 8-ohm tap, it reflects 4k to the primary.
If you hook a 4 ohm speaker to the 4-ohm tap, it also reflects 4k to the primary.

If you hook both up simultaneously then you would have 4k in parallel with 4k, making 2k primary impedance.

This is described in RDH4, page 202-203.

[cbarrow7625]

Assuming the transformer has a typical single wind tapped secondary to give us the different impedance taps, can you explain how it would be possible that all of the current is not passing through the 4 ohm part of the winding (not out of the 4 ohm tap, through the 4 ohm part of the winding)?

For simplicity, let's assume that the current is flowing from ground and out the 4 ohm & 8 ohm taps. The current going to the 4 ohm tap comes from ground and leaves through the 4 ohm tap. The current to the 8 ohm tap also leaves ground flows through the same bottom part of the winding as the 4 ohm tap & then continues on to the additional part of the winding leading up to the 8 ohm tap. The current for both speakers flows through the first half of the winding.