[Guy 13]

Hi Blair and all Audio Circle members.
Why a solid state amplifier gives (Lets say) 100 watts at 8 Ohms and almost always double the watts at 4 Ohms ?
It does not seems to be the case with tube amplifiers.
Guy 13

[Niteshade]

It could work that way with a tube amplifier. For example, if a tube amplifier had a single tap, such as 4 ohms, the power would increase until you reached 4 ohms if starting from a higher impedance.

Some speakers have impedance plots that go all over the place. Sometimes it's best to plug an 8 ohm speaker into the 4 ohm tap to compensate for the low dips. Since much of the impedance spectrum is above 4 ohms, there may be less overall power (not always!), but the power output will be more consistent and the speakers will sound better.

[BobRex]

Hi Blair and all Audio Circle members.
Why a solid state amplifier gives (Lets say) 100 watts at 8 Ohms and almost always double the watts at 4 Ohms ?
It does not seems to be the case with tube amplifiers.
Guy 13

Ohm's Law is why.....  Let's do some math (and this is simplified, so don't piss and moan that the numbers aren't exact)

According to Ohm's Law, power can be calcualted as the square of the voltage divided by the resistance.

P=(E*E)/R

Now let's say you have an 8 ohm load and 100 watts.  Determining the voltage gives a rounded value of 38 volts.

And plugging the numbers into the calculation as such P = (28 * 28)/8 gives you 100 watts (okay, 98, but that's because I rounded.)

Now, for the sake of this calculation, keep the voltage constant and recalculate for 4 ohms and you get this: P=(28 * 28)/4, or 200 watts (196 really.)

This works because most solid state amps are considered constant voltage sources and change the output by varying the current across the load.

Does this make sense?

[Guy 13]

Ohm's Law is why.....  Let's do some math (and this is simplified, so don't piss and moan that the numbers aren't exact)

According to Ohm's Law, power can be calcualted as the square of the voltage divided by the resistance.

P=(E*E)/R

Now let's say you have an 8 ohm load and 100 watts.  Determining the voltage gives a rounded value of 38 volts.

And plugging the numbers into the calculation as such P = (28 * 28)/8 gives you 100 watts (okay, 98, but that's because I rounded.)

Now, for the sake of this calculation, keep the voltage constant and recalculate for 4 ohms and you get this: P=(28 * 28)/4, or 200 watts (196 really.)

This works because most solid state amps are considered constant voltage sources and change the output by varying the current across the load.

Does this make sense?

Hi Bob and all Audio Circle members.
Yes Bob, it make lots of sense!
Thanks, your explanation is simple and clear.
Now I understand.
If the load (Speakers Ohms) on a solid state amplifier vary, that mean the power output will vary and therefore for some frequencies, the output in decibels will vary, which is not good, because it won't be linear.
Now, does that make sense?
Guy 13