Question about tube amplifier.

[cornelflex]

Hello

What happens if I do a short in the output transformer secondary to tube amplifiers?
The transformer output or final tubes burned?
See picture:

[Wavebourn]

No. Especially, when R29 and R30 present, and no negative feedback by voltage applied.

But if you open it your output transformer may be damaged by a spark. Also, if it is a tube of a 6L6 family, spark may appear between pins 2 (filament) and 3 (anode) causing more damages.

[cornelflex]

Originally posted by Wavebourn
No. Especially, when R29 and R30 present, and no negative feedback by voltage applied.

I understand.
Thanks for the answer!
Today I tried to short a radio receiver with tubes (EL84-end tube) at maximum volume after 30 minutes of short, perfectly functioning device.
Mathematically how to interpret it this phenomenon (if you can)?

Ps. I do not speak good English and I can not express well ( used tranlator)!

[Robert McLean]

Mathematically how to interpret it this phenomenon (if you can)?

Consider the AC equivalent circuit of a push-pull amplifier.
You will see this shown sometimes in 2 different ways, both of which amount to the same thing, either 2 equivalent tubes in seris with the load reflected throught the transformer, or 1 equivalent tube with twice the mu, 1/2 the Rp and into 1/4 the load. They amount to the same thing. I show the 2 tube version.

Normally Rl is the load reflected through the transformer, ie multiplied by the square of the turns ratio.

mu is mu of the tube
rl is plate resistance of tube.

So the AC current drawn is 2 * mu * Vg /( 2 * rp +Rl )

and normally Rl is 3000 or 5000 or whatever ohms.
rp even for triodes is 1000 ohms or more, pentodes in the 10s of K ohms or higher,

If the output is shorted then Rl is reduce to just the winding resistance of the transformer, say a few hundred ohms. But even if it is reduced all the way to 0, you still have 2 x rp as the resistance in the circuit.

so the current drawn goes up to at the most 2* mu * Vg/(2 * rp )

which might be a few times higher than it was before, but it is not dangerously high.

[Miles Prower]

Originally posted by cornelflex
I understand.
Thanks for the answer!
Today I tried to short a radio receiver with tubes (EL84-end tube) at maximum volume after 30 minutes of short, perfectly functioning device.
Mathematically how to interpret it this phenomenon (if you can)?

Tubes are high voltage, low current devices. However, speakers need high currents and low voltages. That's what the OPT does: it works as a current step up transformer, not a voltage step down transformer. 40W into 8.0 ohms needs 2.24Arms. If the finals can source 177mArms, then you need a 12.66 current step up turns ratio. The finals are going to pull that 177mA through the primary regardless of what's connected to the secondary.

If it's an 8.0 ohm load, you have: (2.24)(8)= 17.92Vrms.

A four ohm load will give: (2.24)(4)= 8.96Vrms.

If it's a 100 ohm load, then: (2.24)(100)= 224Vrms.

A dead short across the secondary will give: (2.24)(0)= 0

An open secondary will give: (2.24)(inf)= inf! Of course, it will never see infinite voltage since something's going to poof before that happens.

This is why current transformers must never go unloaded. You can get some tremendous voltages, which are dangerous, and you can ruin the transformer.

If VTs worked as voltage sources, what you'd see upon unloading the secondary is the voltage staying the same and the plate current dropping to near zero. That's not what happens. The plate current doesn't change, but the voltage shoots way up, and something just might go

[cornelflex]

Originally posted by Robert McLean


Normally Rl is the load reflected through the transformer, ie multiplied by the square of the turns ratio.
So the AC current drawn is 2 * mu * Vg /( 2 * rp +Rl )
If the output is shorted then Rl is reduce to just the winding resistance of the transformer, say a few hundred ohms. But even if it is reduced all the way to 0, you still have 2 x rp as the resistance in the circuit.

Dear Robert

Thanks for the detailed answer.
Now is clear.
I documented and little is what you said!
Best regards.

Cornelflex

[Johan Potgieter]

Originally posted by Miles Prower
That's what the OPT does: it works as a current step up transformer, not a voltage step down transformer.
.................................................. ........................................

An open secondary will give: (2.24)(inf)= inf! Of course, it will never see infinite voltage since something's going to poof before that happens.

This is why current transformers must never go unloaded. You can get some tremendous voltages, which are dangerous, and you can ruin the transformer.

If VTs worked as voltage sources, what you'd see upon unloading the secondary is the voltage staying the same and the plate current dropping to near zero. That's not what happens. The plate current doesn't change, but the voltage shoots way up, and something just might go

Somewhat off-topic,
Miles,

Darn - how do I disagree with you without actually disagreeing with you?? (I know: It is 3:10 in the am here and I should have been in bed, that's how! )

If we come down to practice, the saving grace is that output transformers are not current (or voltage) transformers all the way. Firstly they are only close to that when fed by pentodes, which also have a finite ra. Triodes make them largely voltage transformers.

But to me the main thing is that when unloaded one has the limitation of the h.t. rail. Tubes will 'bottom' at the +Vin cycle, 'shorting' the transformer then. The topmost swing (losses ignored) will then also be limited to an equivalent upward swing in p.p. circuits. In that case no damage will occur - the voltage swing will simply be the normal swing, barring some on-off effects. (In SE topology this will differ depending on a number of matters.)

As said, a correction without disagreeing with your analysis in general, and not to befuddle others. But it is not quite academical, otherwise damage would occur.

Now to bed.