determining wattage

[blueguitar]

Hi everyone.

This may be a simple question for most of you, or maybe I'm missing a piece of the picture here, but I was hoping for some insight on low wattage designs. I've been studying a few schematics of early amps (champs, princtons, early marshalls, etc), and from what I can tell, the wattage produced is determined by the tubes used in the preamp and power amp. The champ and princton for example are incredibly simple designs with a single preamp and output tube and come out around 5 watts. I understand that at a given wattage, the actual dB output varies a bit due to other design factors such as speaker sensitivity, etc. I've seen a few very low wattage amps out there (1 watt, 1/2 watt, etc), and was wondering how an amp with 1 preamp tube and 1 poweramp tube (like the champ and princton) can be designed to only produce 1/2 a watt. Is it simply a matter of adding the appropriate resistors to the chain, the way a basic attenuator would?

Like I said, I'm sure I'm missing a piece of the puzzle and would appreciate a learning opportunity here

I love the idea of the very low wattage amps like the gilmore jr for example. Just want to understand a bit more before I build one.

thanks,
blueguitar.

[R.G.]

Good question.

It helps to remember that a power amp is best considered to be a wart on the power supply, and to let some of the power out to the speaker under special circumstances. You can't put out more power than the DC power supply can make, and with losses you can't even put out as much as it makes.

It is possible to only use a small fraction of the DC the power supply makes - just turn down the volume knob. The problem with that is that guitarists are looking for the odd and unusual things that happen when you use nearly all the available power from the DC power supply.

So the answer to getting tube power sound at lower wattages is to use a smaller, less capable power supply. Maybe also a smaller, less capable tube so the tube is working nearer it's upper end when the power supply is too.

That's the concept.

[R.G.]

Forgot to add: the reverb output to drive a tank is sometimes a small power amp. Some tanks have 8 to 16 ohm input coils. So the reverb driver section of a tube amp can be and was in the past a small fractional-watt power amp on its own.

I have built micro amps from Fender reverb output transformers driven by 12A?7 tubes, and liked the sound. Hughes and Kettner did something similar with some of their products a decade or two ago.

[TransLucid]

To add to RG's comments:

The max. output power is determined by the OT primary impedance and the supply voltage. The tube just controls the current. You can theoretically use any tube (as long as you don't exceed it's max. plate dissipation rating) for any power amp. You just choose the supply voltage and primary impedance that limits the output power to the desired level.

If you look at the Fender reverb driver mentioned by RG, it is operating well below the maximum power that can be handled by the 2 AT7 sections, because of the supply voltage and OT primary impedance chosen.

[Enzo]

remember that a tube, as they say in other parts of the world, is a valve. It doesn't MAKE power, it controls it.

To get a 1 watt amp instead of a 5 watt amp, we give the tube less to work with. The exact same circuit will put out very different amounts of power when you run it on 475 volts versus 275 volts.

It isn;t about the parts, it is about the circuit. A tube doesn't act alone, it is part of a system.

[blueguitar]

Thanks for all the replies. I love this forum. So as far as my understanding goes, decreacing the voltage supplied to the plate will determine how much is available for the tube to work with. Decreace the plate voltage, and you decreace the amount of signal gain the tube can produce. If this is correct, (and please tell me if I'm wrong) can you still drive a tube to distortion while using lower plate voltages? Can an underdriven preamp tube supply enough current to produce output tube distortion?

[Austin]

Thanks for all the replies. I love this forum. So as far as my understanding goes, decreacing the voltage supplied to the plate will determine how much is available for the tube to work with. Decreace the plate voltage, and you decreace the amount of signal gain the tube can produce. If this is correct, (and please tell me if I'm wrong) can you still drive a tube to distortion while using lower plate voltages? Can an underdriven preamp tube supply enough current to produce output tube distortion?

A preamp tube that is coupled with a capacitor to the output tube does not really supply any current at all to the output tube because when the bias is negative relative to the cathode of the power tube the grid does not draw current. All the preamp tube does is swing voltage, it makes an ac sine wave that varies in amplitude(voltage) compared to ground, and the only real load it has is working into the resistor from ground to the signal coming from the cap. Reducing the voltage to an amplifier that was designed with a higher voltage reduces the power for several reasons, one reason is that the preamp may swing less voltage and this is similar to lowering the volume knob, another reason is that as the power tube voltage is lowered it's internal resistance decreases and this makes the ideal load for it also decrease- but the primary stayed the same. So if you could adjust the output transformer primary impedance to match the now lower internal resistance of the tube you might be able to obtain a similar amount of power up to a point, however by keeping it the same you have made the transfer of energy into the output transformers primary less efficient as it is no longer matched. Having a higher load impedance than optimum also changes the way the tube controls the current flowing through it and can change the harmonic content/distortion of it.

I heard Eddie Van Halen used to use a variac auto transformer to lower the mains to help get his legandary tone, "brown sound".

[blueguitar]

So let me try again here.

decreacing the supply voltage will decreace the amplification factor (gain) in the preamp tube, however the tube will distort easier, or at lower levels of AC voltage swing from the guitar, especailly prominant in the second stage preamp. Is this correct?

[Austin]

So let me try again here.

decreacing the supply voltage will decreace the amplification factor (gain) in the preamp tube, however the tube will distort easier, or at lower levels of AC voltage swing from the guitar, especailly prominant in the second stage preamp. Is this correct?

I think these are all good questions somebody else could maybe answer better but I'll try:

Well if you decrease the supply voltage you also decrease the heater voltage too and that has an unknown effect because nobody experiments with that much but basically the cathode would become less capable of emitting electrons. The other things that happen really depend on the rest of the circuit like: if you had cathode bias like most amps, the voltage going down at the plate would cause the current to go down through the tube and then at the same time the voltage drop across the cathode bias resistor would get smaller making the amount of bias less which would cause more current to flow again. If you had fixed bias via battery, the current through the tube would go down with the plate voltage then.

Distortion however can occur for many reasons.

[Enzo]

I think we have to differentiate between dropping the mains voltage versus dropping various power supplies on the secondary side. Leave the heater voltage where it needs to be, so to speak.

And welcome to the forum, Austin, if we haven;t said so already.

[Enzo]

Distortion however can occur for many reasons

Especially in an election year.

[tubeswell]

... the only real load it has is working into the resistor from ground to the signal coming from the cap. ...

The AC load on a gain stage is not just from the following stage's grid resistor (Rg), but also from the stage's own plate resistor (Ra), which is in parallel with Rg (in terms of AC load). But not only that, the stage's own plate resistance (ra) adds to the AC load, because its in series with the Ra||Rg. And if you have an unbypassed cathode resistor (Rk), then the whole shebang is in series with that as well (in terms of AC load) (or if the Rk is partly bypassed, the whole shebang is in series with that part of the frequency spectrum that doesn't get the benefit of the gain boost).

So for a fully bypassed triode loaded:

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra]

And for an unbypassed triode loaded (it must therefore be*);

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra+Rk(u+1)]

* I couldn't find this formula in Merlin's book; I deduced it taking into account the formula for gain for stage with an unbypassed Rk.

[blueguitar]

The AC load on a gain stage is not just from the following stage's grid resistor (Rg), but also from the stage's own plate resistor (Ra), which is in parallel with Rg (in terms of AC load). But not only that, the stage's own plate resistance (ra) adds to the AC load, because its in series with the Ra||Rg. And if you have an unbypassed cathode resistor (Rk), then the whole shebang is in series with that as well (in terms of AC load) (or if the Rk is partly bypassed, the whole shebang is in series with that part of the frequency spectrum that doesn't get the benefit of the gain boost).

So for a fully bypassed triode loaded:

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra]

And for an unbypassed triode loaded (it must therefore be*);

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra+Rk(u+1)]

* I couldn't find this formula in Merlin's book; I deduced it taking into account the formula for gain for stage with an unbypassed Rk.

Thanks for the information tubeswell. I've always appreciated your comments. I hate to be this guy, but could you please explain this in english? I'm totally new to this, and I'm just trying to get an understanding. Still trying to get all the pieces to the puzzle, much less put them together. Sorry to ask you to dumb it down for a new guy, but . . .

[J M Fahey]

I suggest you do what I did over 43 years ago and found very useful:
get Jack Darr's "How to service your guitar amplifier" and read it at your own speed.
There's a couple reprints on sale , maybe you can get a used one at Amazon or something similar (or EBay, Craigslist) or you can download a condensed version of it, without the schematics.
Very practical, feet on the ground approach.
It will open your mind.
Get it and tell us what you think.
Good luck.

[Austin]

The AC load on a gain stage is not just from the following stage's grid resistor (Rg), but also from the stage's own plate resistor (Ra), which is in parallel with Rg (in terms of AC load). But not only that, the stage's own plate resistance (ra) adds to the AC load, because its in series with the Ra||Rg. And if you have an unbypassed cathode resistor (Rk), then the whole shebang is in series with that as well (in terms of AC load) (or if the Rk is partly bypassed, the whole shebang is in series with that part of the frequency spectrum that doesn't get the benefit of the gain boost).

So for a fully bypassed triode loaded:

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra]

And for an unbypassed triode loaded (it must therefore be*);

Gain = [u(Ra||Rg)]/[(Ra||Rg)+ra+Rk(u+1)]

* I couldn't find this formula in Merlin's book; I deduced it taking into account the formula for gain for stage with an unbypassed Rk.

The load on a gain stage as I think about it is how I should have stated it, the load it works into, it usually is just a resistor though it could be a transformer or a tonestack. The other stuff like plate resistor and bias details are just how it gets its voltage gain and determine how much gain and how much current it could source. A lower plate resistor means less gain but more current could be sourced. Less bias voltage means more current through the tube (up to a point limited by the plate resistor) and more voltage drop across the plate resistor.