? On using two taps on OT simulataneously

[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.

[g1]

"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.

[g1]

"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

[g1]

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