I'm researching valve amplifiers. I found this schematic for one:
So the input is amplified by the first valve, and then the amplified signal is amplified again by the second valve, right?
My question is, why is the voltage being stepped down before going to the speaker? It seems pointless to me, increasing the voltage with the valves and then decreasing it again. All the schematics I can find online do this. Why?
(Is the 300V rail at the top related to the transformer? If not, what's it for?)
Best Answer
It's a question of impedance.
The anode (plate) voltage of the tube varies over a wide range, while the current varies over a much smaller range. If you define output impedance as
$$Z_{out} = \frac{\Delta V}{\Delta I}$$
This usually works out to a fairly high number for a typical vacuum tube, on the order of thousands of ohms.
On the other hand, most speakers have a low impedance — on the order of 4 to 16 Ω — which means they want a relatively higher current change coupled with a relatively smaller voltage change.
Note that in both cases, you're talking about the same amount of power (voltage × current), which is what the amplifer is really achieving — an increase in signal power from input to output.
The transformer provides this impedance change. It trades off a high voltage swing for a high current swing. Without it, you'd get only a tiny fraction of the available signal power actually delivered to the speaker, limited by the relatively low current in the tube.
From a comment:
The 300V power supply is required for much the same reason: The output of the impedance of the tube is inherently high.
The 6V6 tube is rated for 50 mA plate current (average), which means that the signal current swing must be less than about ±40 mA (peak). Similarly, the tube is rated for a plate voltage of 250 V (nominally, but it is frequently overdriven in this respect), so the signal voltage needs to be less than about ±120 V (peak).
The signal power available at the output is therefore the RMS current multiplied by the RMS voltage, or:
$$\frac{40 mA}{\sqrt{2}} \cdot \frac{120 V}{\sqrt{2}} = \frac{4.8 W}{2} = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_{out} = \frac{120 V}{40 mA} = 3000 \Omega$$
To drive an 8Ω speaker, you'd use a 3000Ω:8Ω transformer (19.4:1 turns ratio), which would give you 4.38 VRMS and 548 mARMS at the speaker.