transformers/output impedance

Transformer is basically an impedance converter. To set proper conversion ratio we need to know speaker's impedance and recommended tube "plate load impedance". If we don't know speaker's impedance we can measure DC resistance and multiply it by about 1.25 (impedance is mostly resistive). Plate loading we can find in tube's datasheet (or plate-to-plate loading impedance in Push-Pull configuration).
We need our transformer to reflect speaker impedance (secondary side) to plate loading impedance (primary side). Transformer impedance ratio is a square of turns ratio.

Example: Plate load impedance is 10,000 ohms and speaker impedance is 4 ohms. Necessary transformer impedance ratio will be 2,500. Turns ratio will be square root of 2500 = 50. Voltage and current ratios will follow turns ratio. Secondary voltage change will be 50 times smaller than primary but secondary (speaker's) load current will be 50 times smaller on primary side.

As you can see it would require very big voltage changes on primary to get decent output voltage, hence power delivered to speaker. We can lower this ratio by finding tubes operating at lower plate loading impedance or to put tubes in parallel.

That's right, except for the losses in transformer. This impedance cannot be simply measured with ohmmeter because transformer does not operate at DC but behaves according to ohm laws for the CHANGES in voltage. It is called reflected impedance. Transformer has to be large enough to carry desired power. Cross section of the center of the core is proportional to square root of power and is roughly 1 square inch for 50W (for 200W it will be 2 sq inch etc.). We want as few turns as possible to avoid losses in copper but we have to provide minimum number of turns per volt. This number is inversely proportional to square area of the core and is smaller for large transformers. Knowing maximum voltage and turns per volt we can establish absolute number of turns. We select core material, size and number of turns to prevent core saturation, when magnetic flux is higher than max flux density rated for given core.

All this is also related to frequency. It is easier to saturate transformer at lower frequencies and it has to be taken into account for size of core and number of turns per volt.

Absolutely right Bifwynne. That's why audio is more of the art than science. We can calculate but have to match and listen.

Bifwynne, some people believe that transformers and capacitors have no place in signal path. Eliminating output transformer is nice but requires some descent output impedance to drive speakers and it is usually achieved by placing a lot of large tubes in parallel. Unfortunately heaters take a lot of power. 6AS7G that Atmasphere was using takes 2.5A at 6.3V =15.75W per tube. 20 tubes will dissipate 315W. If you prefer stereo listening it will be 630W. Adding main circuit will result in more than 1kW - a space heater (nice in the winter). On the other hand we audiophiles are eager to sacrifice for great sound, and I've heard it is great.

Output impedance and ability to drive low impedance are two different things. My amplifier has DF=4000 at low frequencies but cannot drive speakers below 3 ohms.

I cannot speak of tube amps but in SS amp feedback always reduces output impedance. I made this small example a while ago as a proof:

Let’s take amplifier that has gain of 30 (31.6dB). When input voltage is 1V output voltage is 30V. Output voltage drops (for whatever reason) 1V under 1A load to 29V. That's 1ohm output impedance (DF=8).

Now, let's build this amp with gain of 300 but feed 3% of the output voltage back to the input in opposite phase. As a result amplifier’s output is the same 30V as before but input is the difference between 1V and 3% of 30V = 0.1V Let’s verify (1V-0.03*30V)*300=30V

Let’s load this amplifier with 1A. Our voltage drop inside is still 1V under 1A load, but output voltage will be higher than 29V because we subtract less from the input. Output voltage will be 29.9V and output impedance will be 0.1V/1A=0.1ohm (DF=80). Let’s verify. (1V-0.03*29.9V)*300-1Vdrop=29.9V.

Output impedance dropped 10 times. Expression 1+B*Aol is known as “Improvement Factor”. In our case B (“Feedback Factor”) = 0.03 (3%), Aol (“Open Loop Gain”) = 300 thus Improvement Factor = 1+0.03*300=10

Adding NFB WILL add more power to lower load impedance because output impedance is lower. I'm sure Ralph has maximum power in mind.

It is important to realize that it doesn't matter in above example why voltage initially dropped by 1V at 1A load. It could be output impedance or bandwidth limit (or anything else). NFB will reduce output impedance, increase bandwidth, improve linearity - hence THD and IMD.