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I'm researching valve amplifiers. I found this schematic for one:

image

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?)

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    \$\begingroup\$ The loadline of the vacuum tube requires that stepdown impedance transformation. Running a (small) tube at 8 ohms will produce very low power. Running that same tube at 5,000 ohms (200 volts and 40 milliAmps) is a big success. \$\endgroup\$ – analogsystemsrf Sep 17 '18 at 16:39
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    \$\begingroup\$ It seems pointless to me .... so, avoiding a 300V potential at the speaker terminal seems pointless to you? \$\endgroup\$ – jsotola Sep 17 '18 at 19:08
  • \$\begingroup\$ @jsotola I see what you're getting at, but in that case, why provide a 300V potential in the first place? \$\endgroup\$ – Jacob Garby Sep 18 '18 at 17:11
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    \$\begingroup\$ the voltage is required for vacuum tube operation \$\endgroup\$ – jsotola Sep 18 '18 at 17:18
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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:

Any idea what the 300V rail is for? Is it simply a power supply for the valves? Why is it so high-voltage?

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.

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    \$\begingroup\$ So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker? \$\endgroup\$ – Jacob Garby Sep 17 '18 at 15:54
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    \$\begingroup\$ Yes, that's the idea. The impedance ratio is the square of the turns ratio. For example if you need a 1000:1 impedance ratio, you'd want roughly a 32:1 turns ratio. \$\endgroup\$ – Dave Tweed Sep 17 '18 at 15:56
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    \$\begingroup\$ Thanks, got it! Any idea what the 300V rail is for? Is it simply a power supply for the valves? Why is it so high-voltage? \$\endgroup\$ – Jacob Garby Sep 17 '18 at 15:59
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    \$\begingroup\$ See edit above. \$\endgroup\$ – Dave Tweed Sep 17 '18 at 16:15
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    \$\begingroup\$ A useful by-product is the reduction in the voltage across the speaker wires, in case an animal or person touches an exposed wire. \$\endgroup\$ – Andrew Morton Sep 17 '18 at 17:20
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In addition to what Dave Tweed said (+1), the transformer in this case also eliminates the DC bias current from going to the speaker, and decouples the common mode input and output voltages.

The plate current of V1 sits at a center value when idle. The input signal causes the plate current to go both up and down from the center value according to the peaks and troughs of the input signal.

Even if there was a speaker that was impedance-matched to the plate of the 6V6, the DC bias current thru it would not be desirable. The transformer also blocks DC, while passing the relevant AC parts of the signal.

Note that impedance matching is still the primary reason. Since a transformer is required for that anyway, the designer of the circuit made use of the fact that it also blocks DC, and that the common mode input and output voltages are decoupled. This latter fact allows one side of the speaker to be grounded, even though the transformer primary is tied to 300 V.

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Short Answer: Reduce output impedance to prevent significant voltage loading

For good bass response the speaker is a linear motor/generator with back EMF on kick drum pulses. Thus the output impedance must be much lower than the speaker. THis is also called the Dampening Factor= Zspeaker/Zout and is only 20 on cheap low power amps , 100 on good amps and 1000 on great power amps.

So what is it on a Vacuum Tube Amp?

  1. THat depends on the Tube Zout divided by turns ratio of transformer squared.

  2. So the impedance reduction of turns ratio n² reduces the high output impedance to somewhat lower than the speaker impedance.

  3. Without specs, its hard to guess but never as good as soldid state but infact the harmonic distortion from back EMF, not just the tube's soft limiting but of the poor damping factor may be "pleasant" to some guitar players but "muddy" to audio experts playing broad spectrum.

  4. Since turns ratio also reduces voltage by n, the tube voltage swing must be n times bigger than what the speaker sees

  5. e.g. thus maybe 9 times bigger swing and Vdc and /81 reduction of the high output impedance.. .perhaps more turn ratio... 20;1 Voltage ratio is 400:1 impedance ratio possibly giving a dampening factor of <10 ie poor D.F. so they often used 16 Ohm speakers.

  6. BTW Many Tube amp designs are much better than this one.

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    \$\begingroup\$ Dampening Factor is news to me. Thank you for the education \$\endgroup\$ – Reversed Engineer Sep 18 '18 at 9:55
  • \$\begingroup\$ DF is the inverse of Load Regulation error for any power supply except applied to audio so 1% load reg error = DF of 100 and DF<10 means load error>10% often from back EMF but also just efficiency loss for steady CW \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 18 '18 at 13:06
  • \$\begingroup\$ Aha - I see why that's a useful metric! \$\endgroup\$ – Reversed Engineer Sep 18 '18 at 14:53
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I need to correct your misleading terminology. It is an impedance matching power transformer, not a step-down transformer!

In order for you to understand the answer, you need to know:
1) The purpose of an amplifier, is to amplify power (not current or voltage).
2) Vacuum tube devices could only provide "small" currents, but could handle high voltages.
3) Vacuum tubes had impedances of K ohms, while speaker impedances were in the order of ohms.

Since P = VI, to provide the max power amplification with small current devices, one has to use the max voltage that the device can handle (this is the answer to your "why high voltages" question).
Since max power transfer between two devices occurs when their impedances match, the impedance matching power transformer was the ideal solution to this problem (and the other problems mentioned in the other answers).

The voltage rails of any circuit, are required because of the "conservation of energy law." Although signal power is being amplified, it comes at the cost of the power supplied by the voltage rails.

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  • \$\begingroup\$ Right, I think that makes sense. So, the total output impedance of the amplifier before the impedance transformer is just the impedance of the tube itself? \$\endgroup\$ – Jacob Garby Sep 21 '18 at 22:51

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