? 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.

[mooreamps]

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.

I'm looking at it now. Tonight, I will post the schematic as shown in Figure 5.7 of the RDH4, with loud speakers in place of R2 and R3 ;
and then I'll draw up a schematic of a vacuum tube output transformer, with loud speakers connected to the 4 and 8 secondary taps. and point out the differences. You will see how not all the current from one load is flowing into the other load. Maybe it doesn't make any difference, but I need to look at this more closely.


-g

[hasserl]

The current for both speakers flows through the first half of the winding.

The same would be true if you had multiple speakers connected to the one 4 ohm tap, i.e. 4 - 16 ohm speakers connected in parallel. The important thing here is the total impedance, not the number of speakers. If the total impedance is matched correctly for the primary, it really shouldn't matter if just one tap is used, or two or all three, correct?

[Merlinb]

You will see how not all the current from one load is flowing into the other load.

I don't think Cbarrow meant that the current in one load actually flows into the other load. He simply meant that the current in a portion of the secondary coil is the sum of the two speaker currents; these then split off to each speaker of course. (This is assuming a single, tapped secondary coil)

Admittedly, this is probably not a serious problem in practice.

[g-one]

"If the total impedance is matched correctly for the primary, it really shouldn't matter if just one tap is used, or two or all three, correct?". I believe you are correct. For maximum power transfer you will need to use speakers double the tap impedances as has been mentioned.
Just got me thinking about an amp I was testing and the impedances didn't make sense as I wasn't getting max power transfer. I now realize the amp was designed with built-in speakers on both taps but had been made into a head version. I was testing with only one tap connected now I know why I had the problem. Sort of the reverse of using both taps on a unit designed for single tap use.
In the real world, how many times is max power transfer ignored? All those old Fenders with an external spkr. jack wired in parallel with no impedance switch. Max power with internal spkr. connected. With an ext. spkr. connected in parallel you are down on power but I guess the added spkr. more than makes up for it.

[g-one]

"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."
Power draw? We can calculate power based on current (or voltage) and impedance. Power draw is used only for convenience in AC line circuits where we specify a guaranteed line voltage. Try plugging a 50W 115V light bulb into modern ac line. The bulb will blow because it exceeded 50 watts.
In an amplifier we can not exceed the capability of the power supply (power transformer). I have a 200watt 8 ohm speaker on the 8 ohm tap of my 50watt amp. Do I have a 200watt power draw?
I agree with the rest of what you have said though. In some amps we draw bias current from the PT high voltage winding (bias tap). This is extra current through that section of the winding. Though negligible for bias, it is similar to what you said about extra current through the 4 ohm winding.

[mooreamps]

oh boy..... no ; if you have a power amp rated at 50 watts ; with 2 speakers in parallel ; then it's 25 watts for each speaker.

by the way, I just happen to have a small 50 light bulb on my desk as I an typing this message. I'm not seeing it "blow up"...

-g

[mooreamps]

I don't think Cbarrow meant that the current in one load actually flows into the other load. He simply meant that the current in a portion of the secondary coil is the sum of the two speaker currents; these then split off to each speaker of course. (This is assuming a single, tapped secondary coil)

Admittedly, this is probably not a serious problem in practice.

That's fair. I still want to look at this more closely. I do appreciate the link to the RDH4. I am also seeing Fig. 5-9 that more closely represents two separate loads on two separate output taps. I still want to take some measurements ; in circuit ; with different speaker loads on varies output taps ; and see if they concur with the statements presented in the book.

-g

[g-one]

a) That was a rhetorical question.
b) You have the specified 115V lightbulb plugged into more than 120VAC?
That's why most light bulbs are now 130VAC. W=IxE or IsquaredR or Esquared over R.

[Wilder Amplification]

That's fair. I still want to look at this more closely.

It's really simple. In the method I describe you have two cabs both consuming 25 watts from the secondary at full power output, which means that the OT primary is also seeing the exact same voltage/current ratio (which is where impedance is calculated from anyway) as it would see with a single cab hooked up to its corresponding secondary tap. What more is there to look at?

[Old Tele man]

...FWIW, Randall Aiken has this question answered in one of his Technical Questions postings for those wishing to read up on *what & why*

[mooreamps]

What more is there to look at?

I want to see how changes in the output load ; matching the output impedance and mis-matching the output impedance ; affects the reflected impedance of the primary coil of the output transformer. It could be measured from using a series resistor from one of the plates of the power tubes.

-g

[Wilder Amplification]

I want to see how changes in the output load ; matching the output impedance and mis-matching the output impedance ; affects the reflected impedance of the primary coil of the output transformer. It could be measured from using a series resistor from one of the plates of the power tubes.

-g

Um...you're an amp builder and you don't already know this??? Yet on countless other posts you've accused others (me included) of "not knowing how this stuff works" yet you have no idea how changes in output load affects reflected primary impedance???

Mismatching the impedance up by doubling the load (16 ohm load on 8 ohm tap) doubles the reflected impedance. Mismatching down by halving the load (4 ohm cab on 8 ohm tap) halves the reflected impedance.

Example, you have an impedance ratio of 425 on the 8 ohm tap, which translates to a 3.4K reflected impedance with an 8 ohm load connected to the 8 ohm tap. Place a 4 ohm cab on the 8 ohm tap -

4 x 425 = 1.7K

Now place a 16 ohm cab on the 8 ohm tap -

16 x 425 = 6.8K

So..with just a 16 ohm cab on the 8 ohm tap alone, our reflected impedance doubles. Now place an 8 ohm cab on the 4 ohm tap in conjunction with it and this cuts our doubled reflected impedance in 1/2, which brings us back to the proper reflected impedance.

[mooreamps]

Oh I get the about putting the 16 ohm cab on the 8 ohm tap part !

Want I want to do is measure : a 16 ohm cab on the 16 ohm tap with an 8 ohm cab on the 8 ohm tap.

-g

[Wilder Amplification]

And that will halve the reflected primary impedance.

Think about it. Let's say you have a 16 ohm cab on the 16 ohm tap on a 50 watt amp. You will have 50 watts going to the one cab.

Now add an 8 ohm cab to the 8 ohm tap. The 8 ohm tap will apply 20 volts across the 8 ohm load, which adds another device trying to consume 50 watts.

So you now have two cabs both consuming 50 watts from the secondary, which makes total secondary power = 100 watts. Well power out = power in which means that primary Z gets halved in order to be able to consume double the output power.

Of course in all actuality the power supply will sag way down and your tubes would redplate long before you got 100 watts out, but the point is to illustrate that the primary Z gets halved when you do this.

When you do it the way I suggested to do it, putting the 16 ohm cab on the 8 ohm tap doubles the primary Z. Once you add the 8 ohm cab to the 4 ohm tap, it halves the primary Z, which puts the primary Z in the same place it would be with just a single cab hooked up to its corresponding tap while both cabs consume 25 watts each.

[mooreamps]

And that will halve the reflected primary impedance.

Think about it. Let's say you have a 16 ohm cab on the 16 ohm tap on a 50 watt amp. You will have 50 watts going to the one cab.

Now add an 8 ohm cab to the 8 ohm tap. The 8 ohm tap will apply 20 volts across the 8 ohm load, which adds another device trying to consume 50 watts.

With a plate current sensing resistor ; I'm going to measure it and see.

-g

[Enzo]

I guess the question then, Gary, is this: If you have a 16 ohm speaker on the 16 ohm tap and the amp running at rated output. Does adding a second speaker - say an 8 ohm speaker on the 8 ohm tap - change anything on the primary side? If it does, then the two loads have added. If it doesn;t change anything, then we get free power to one of the speakers.


And we know that if we connect a 8 phm speaker to the 8 ohm tap of a 50w amp, we get 50w in the speaker, and if we instead connect two 16 ohm speakers in parallel to that tap, we still get 50w, but 25w in each speaker. SO, starting anew, if we connect a 16 ohm cab to the 16 ohm tap and get 50w, and then an 8 ohm cab is added to the 8 ohm tap, then either the added speaker gets 50 free watts and the 16 still gets 50 watts. Or the two speakers now split the 50w between them for the same 25w each. If that split happens, then by not touching the 16 ohm speaker, we can change the power to it by plugging the 8 ohm speaker into the 8 ohm tap or not. And the only way that can happen is by the load on the tubes changing.

[mooreamps]

Basically, yes. Is it the same as driving two 16 ohm speakers from the 8 ohm tap. and to answer an eariler question, yes there is a simple cross-over network in that stereo tube keyboard amp, but it was only a non-pol series cap in line with the ribbon twitter.

-g

[R.G.]

Even the most casual reading of how transformers work and what they do shows that loads on multiple taps/windings appear summed in parallel according to the impedance and turns ratios at the primary. It used to be that you had to go find a text book to find this. With the internet, it's the work of a few seconds to find the explanation, worked examples, whatever you need to learn. Assuming you want to learn.

Arguing against this, or even arguing that it's mysterious, is the same thing as admitting that you not only don't understand, but also that you either don't know how to find the information, or possibly don't want to learn.

I don't know exactly why, but this brought to mind the two rules of pigs.
1. Never try to teach a pig to sing. It wastes your time and annoys the pig.
2. Never mud-wrestle a pig. Not only can you not win, but after a while you realize the pig enjoys it.

[mooreamps]

Assuming you want to learn.

My business is not learning electrical theory. My business is inventing it.

-g

[Merlinb]

My business is not learning electrical theory. My business is inventing it.

Talk about shooting yourself in the foot...