P.S. to my previous post: The statement I quoted at the beginning of that post is incorrect. The reference to 1/2 of the OPT’s (output transformer’s) winding being in use for the 8 ohm tap should have said 0.707 (i.e., 1/(square root of 2)), assuming there is a 16 ohm tap, and the reference to 1/4 of the OPT’s winding being in use for the 4 ohm tap should have said 0.5.
Tube amps having output transformers are usually designed to have approximately the same maximum power capability when a 4 ohm load is connected to the 4 ohm tap as when an 8 ohm load is connected to the 8 ohm tap, etc.
For a resistive load: Power = (Voltage squared)/Resistance
... so to provide equal power into a halved resistance voltage squared must be halved. Which means that the voltage and hence the turns ratio of the transformer must be reduced by a factor of 0.707.
Regards,
-- Al
Tube amps having output transformers are usually designed to have approximately the same maximum power capability when a 4 ohm load is connected to the 4 ohm tap as when an 8 ohm load is connected to the 8 ohm tap, etc.
For a resistive load: Power = (Voltage squared)/Resistance
... so to provide equal power into a halved resistance voltage squared must be halved. Which means that the voltage and hence the turns ratio of the transformer must be reduced by a factor of 0.707.
Regards,
-- Al