paralleled SE output tubes

[Colas LeGrippa]

this question certainly has been asked rather twice than once: in a SE amp, adding a power tube in parallel will ask for doubling or dividing the speaker load by two ? and what is the math behind the answer ? ( 2 * x = 2x    )

thanks

Colas

[tubeswell]

2 x parallel tubes = 1/2 the load resistance of 1 tube (all other things like; tube type, bias voltage, plate voltage and plate current per tube etc being equal)

[Colas LeGrippa]

As I can see, in calculating the reflected load impedance, we consider output tubes only as being resistors.. putting two resistors in parallel will divide the total load by 2.  I was not quite sure if this formula would apply to tubes too. I have never considered the tolerances of any kind in building amps, I was going rather with my ears. And it was giving pretty much good results, I must admit. But now that I have experienced weird things like getting much more power out of an EL84 than a 6L6 ( weird because a certain lack of overall understanding),  I just want to do things differently in order to avoid, by example, spending a week on a troubleshooting that appeared to be just a defective a/c cord...................

Regards

Colas

[stratele52]

Yes Colas Legrippa, tubes are "like" resistor. It is easy to have information on push pull amp. Two tubes push pull need a primary impedance output transformer twice  than  the four output tubes transformer.