SE transformer as plate choke

[RayP]

I seem to remember years ago somebody writing about using a typical SE transformer as a plate choke. If so, I have three questions about the inductance.

1 - would the inductance in the primary be sufficient if the secondary is not connected?

2 - would the inductance in the primary change if the secondary is connected to a power resistor such as 8 ohms?

3 - would the inductance in the primary change if the secondary is connected to a power resistor such as 100 ohms?

ray

[Paul Joppa]

1 - would the inductance in the primary be sufficient if the secondary is not connected?

It will remain what it was.

2 - would the inductance in the primary change if the secondary is connected to a power resistor such as 8 ohms?

The inductance will remain the same, but the impedance will drop drastically - the combination looks like a resistor equal to the primary impedance.

3 - would the inductance in the primary change if the secondary is connected to a power resistor such as 100 ohms?

Same as 2 - but the resistance will be larger, and the frequency band over which  the impedance is resistive will be smaller

[RayP]

OK, thanks.

ray

[2wo]

How about an ungaped PP transformers. Will it saturate.?

[Paul Joppa]

Yes it will. A large one that I measured some years ago could take 2 or 3mA only, before saturation.

Incidentally, in the early days of the SET revival some people would use a P-P output transformer in parafeed to avoid the saturation issue. I was disappointed when I tried it with the same transformer as above (from a Stereo 70) - I think it's because of the way P-P transformers are wound differently from SE transformers. Great bass but the treble just wasn't right.

[RayP]

I am still a bit puzzled by answer #2 where the secondary is connected to an 8 ohm resistor. This would be a typical non-parafeed SE amp feeding a loudspeaker. If the impedance drops dramatically, how does it load the tube? I must be missing something which I presume is a formula for how the impedance is calculated from the inductance.

Paul, could you show how that would work with a transformer with a 40 H primary such as the Magnequest DS-050.

ray

[Grainger49]

The load on the secondary is reflected back to the primary of the transformer.  Open is an infinite load, 8 Ohms is a very low load, 100 isn't that much larger.

It is one of those transformer things like turns ratios.

[RayP]

Perhaps if describe what I am pondering which is whether I could design a spud amp using an 829B which has two plates but only one cathode.

(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fi87.photobucket.com%2Falbums%2Fk125%2Frperry13%2F829Bspud_zps4e7e88e2.jpg&hash=c8e7d0900ab62133fce99d121ad14dce36c2d84b)

Pretend the two triodes U1 and U2 are the two plates of the 829B. U2 is acting as the power tube loaded by L3 and R5 as the typical output transformer loaded with a speaker.

U1 is the driver triode loaded by L2 and capacitor connected by C1 to the grid of U2.

Since both U1 and U2 are sharing a single cathode they have to work together. I was wondering if I could use the same kind of output transformer as the load for U1. Would the circuit work better with an 8 ohm resistor as R4?

ray

[Paul Birkeland]

The 829B will be a bit of a difficult tube for a spud amp.  It's really the common screen connection, not the cathode connection, that will end up being bothersome.  Triode strapping both tubes, gain will be really problematic without an interstage between halves. 

[RayP]

I wondered if that screen would be a problem and I was not planning to build such an amp. It was just a thought bubble.

However to get back to the original question and presume we have a imaginary dual triode with a common cathode, would an R4 of 8 ohms be the best solution of the three ( R4 not there, R4 = 8 ohms, R4 = 100 ohms).

ray

[Paul Birkeland]

Loading the transformer with a resistor would decrease the gain and increase distortion.  Not having the resistor there is by far the best option. 

[RayP]

OK thanks.

ray

[Paul Joppa]

I am still a bit puzzled by answer #2 where the secondary is connected to an 8 ohm resistor. This would be a typical non-parafeed SE amp feeding a loudspeaker. If the impedance drops dramatically, how does it load the tube? I must be missing something which I presume is a formula for how the impedance is calculated from the inductance.

Paul, could you show how that would work with a transformer with a 40 H primary such as the Magnequest DS-050.

ray

The impedance across the primary terminals would be the primary winding inductance in parallel with the reflected load resistance (5K ohms). At very low frequencies, the inductive impedance is much lower and dominated; at high frequencies the load resistance dominates. So the impedance transitions from inductive to resistive; and the impedance magnitude is always less than 5000 ohms.

The transition happens when the inductive impedance equals the load resistance, about 20Hz in this case.

Without the loading resistor, the impedance is just that of the inductance; at 1kHz that's around 250,000 ohms.

[RayP]

Thanks for the detailed answer. I keep learning.

ray

[Paul Birkeland]

I thought it'd also be worth mentioning that you can't actually have both halves of the 829B triode strapped unless they are paralleled.  If you redraw your schematic to be two pentodes, you'll notice that you get the choice of either connecting G2 to the plate of the driver stage, or the output stage.  Modulated G2 voltage that's out of phase on the driver stage should add some fun complications!