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Why put a step-down transformer after a vacuum tube in an a valve amplifier?
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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?)
amplifier vacuum-tube
asked 54 mins ago
Jacob Garby
262
add a comment |Â
up vote
5
down vote
favorite
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?)
amplifier vacuum-tube
asked 54 mins ago
Jacob Garby
262
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
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?)
amplifier vacuum-tube
asked 54 mins ago
Jacob Garby
262
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?)
amplifier vacuum-tube
amplifier vacuum-tube
asked 54 mins ago
Jacob Garby
262
asked 54 mins ago
Jacob Garby
262
asked 54 mins ago
Jacob Garby
262
asked 54 mins ago
Jacob Garby
262
asked 54 mins ago
Jacob Garby
262
262
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
8
down vote
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
$$Z_out = fracDelta VDelta I$$
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:
$$frac40 mAsqrt2 cdot frac120 Vsqrt2 = frac4.8 W2 = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_out = frac120 V40 mA = 3000 Omega$$
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.
answered 41 mins ago
Dave Tweed♦
108k9129232
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
1
See edit above.
– Dave Tweed♦
9 mins ago
add a comment |Â
up vote
5
down vote
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.
answered 26 mins ago
Olin Lathrop
276k28328774
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
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
$$Z_out = fracDelta VDelta I$$
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:
$$frac40 mAsqrt2 cdot frac120 Vsqrt2 = frac4.8 W2 = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_out = frac120 V40 mA = 3000 Omega$$
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.
answered 41 mins ago
Dave Tweed♦
108k9129232
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
1
See edit above.
– Dave Tweed♦
9 mins ago
add a comment |Â
up vote
8
down vote
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
$$Z_out = fracDelta VDelta I$$
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:
$$frac40 mAsqrt2 cdot frac120 Vsqrt2 = frac4.8 W2 = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_out = frac120 V40 mA = 3000 Omega$$
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.
answered 41 mins ago
Dave Tweed♦
108k9129232
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
1
See edit above.
– Dave Tweed♦
9 mins ago
add a comment |Â
up vote
8
down vote
up vote
8
down vote
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
$$Z_out = fracDelta VDelta I$$
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:
$$frac40 mAsqrt2 cdot frac120 Vsqrt2 = frac4.8 W2 = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_out = frac120 V40 mA = 3000 Omega$$
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.
answered 41 mins ago
Dave Tweed♦
108k9129232
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
$$Z_out = fracDelta VDelta I$$
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:
$$frac40 mAsqrt2 cdot frac120 Vsqrt2 = frac4.8 W2 = 2.4 W$$
If you use a lower plate voltage, the available power is reduced proportionally.
Note that this works out to an output impedance of:
$$Z_out = frac120 V40 mA = 3000 Omega$$
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.
answered 41 mins ago
Dave Tweed♦
108k9129232
edited 2 mins ago
answered 41 mins ago
Dave Tweed♦
108k9129232
answered 41 mins ago
Dave Tweed♦
108k9129232
answered 41 mins ago
Dave Tweed♦
108k9129232
108k9129232
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
1
See edit above.
– Dave Tweed♦
9 mins ago
add a comment |Â
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
1
See edit above.
– Dave Tweed♦
9 mins ago
1
1
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
So am I right in thinking the the transformer basically reduces the impedance, to one which is correct for the speaker?
– Jacob Garby
30 mins ago
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.
– Dave Tweed♦
28 mins ago
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.
– Dave Tweed♦
28 mins ago
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?
– Jacob Garby
25 mins ago
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?
– Jacob Garby
25 mins ago
1
1
See edit above.
– Dave Tweed♦
9 mins ago
See edit above.
– Dave Tweed♦
9 mins ago
add a comment |Â
up vote
5
down vote
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.
answered 26 mins ago
Olin Lathrop
276k28328774
add a comment |Â
up vote
5
down vote
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.
answered 26 mins ago
Olin Lathrop
276k28328774
add a comment |Â
up vote
5
down vote
up vote
5
down vote
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.
answered 26 mins ago
Olin Lathrop
276k28328774
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.
answered 26 mins ago
Olin Lathrop
276k28328774
answered 26 mins ago
Olin Lathrop
276k28328774
answered 26 mins ago
Olin Lathrop
276k28328774
answered 26 mins ago
Olin Lathrop
276k28328774
276k28328774
add a comment |Â
add a comment |Â
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