Amp basics: loading

[xtian]

@V2, I appreciate your post. It's not obvious to me why the plate load resistor is in parallel with the subsequent vol pot. The plate resistor is tied to B+, and the vol pot is tied to ground.

[JazzGuitarGimp]

@V2, I appreciate your post. It's not obvious to me why the plate load resistor is in parallel with the subsequent vol pot. The plate resistor is tied to B+, and the vol pot is tied to ground.

The power supply output (B+ to GND) is very low-impedance, so when considering stage loading, B+ and GND are considered one and the same,

Edit: Actually, I should have said: when doing any sort of circuit analysis, B+ and GND are considered one and the same.

[vibratoking]

@V2, I appreciate your post. It's not obvious to me why the plate load resistor is in parallel with the subsequent vol pot. The plate resistor is tied to B+, and the vol pot is tied to ground.

This is a good question. There is a basic circuit concept that seems to get missed quite often. For AC signals, power and ground look the same from an analysis point of view. You can mentally replace every power supply with ground when doing an AC analysis. Now do you see the parallel connection?


Lou beat me to it.

[matt h]

(deleted)

[xtian]

Excellent insight. Thanks, gents.

[V2]

yes, what they said. Sorry for being late to reply.

[tubeswell]

@V2, I appreciate your post. It's not obvious to me why the plate load resistor is in parallel with the subsequent vol pot. The plate resistor is tied to B+, and the vol pot is tied to ground.

This is a good question. There is a basic circuit concept that seems to get missed quite often. For AC signals, power and ground look the same from an analysis point of view. You can mentally replace every power supply with ground when doing an AC analysis. Now do you see the parallel connection?


Lou beat me to it.

In terms of what the signal 'sees', both the B+ supply and the ground look one and the same. (This is because the filter caps shunt any AC, that otherwise would get into the power rail, to the ground return)

The more resistors you connect to (either) the B+ and the ground, the more paths there are for current to drain away from the AC signal, and the more powerful the signal has to have been to begin with, in order to preserve the signal's integrity. In this regard load resistors are like springs on a trampoline pad, the more springs you have connecting the pad to the frame, the harder you need to bounce on the pad to get it to move.

Impedance-bridging between gain stages affects signal attenuation (and Matt referred to this earlier). Apart from the plate resistor (Ra) and the grid load (Rl) for any following stage (including tone stacks), the internal plate-resistance (ra) of the tube is also a factor in the tube's output impedance (Zo), in that internal plate-resistance affects transconductance (gm). This includes load resistances on both sides of any AC-coupling.

So the signal gets injected into the tube at the control grid in the form of an AC voltage swing, the plate-resistance determines the resulting plate current (i.e.; transconductance happens), and the resulting plate current is drained away from the plate by the load (plate resistor and other load resistors in parallel), and the energy that's leftover in the signal makes it to the next stage's grid.

[jazbo8]

I have Merlin's preamp book.

The subject is well covered in the book... perhaps you can point to specific chapters or sections that you do not understand, so others can explain it (or at least try to).

[xtian]

I'm actually reading it again, after starting this thread. Section 1.17 "The AC Load Line" explains exactly what @V2 said about anode resistor and the resistance to ground following the coupling cap.

Merlin's book should be subtitled, "Don't panic; you will have to read this book five hundred times, but you will gain new understanding each time."

[jazbo8]

I'm actually reading it again, after starting this thread. Section 1.17 "The AC Load Line" explains exactly what @V2 said about anode resistor and the resistance to ground following the coupling cap.

Merlin's book should be subtitled, "Don't panic; you will have to read this book five hundred times, but you will gain new understanding each time."

Glad you got a handle on it, it's definitely well worth re-reading from time to time.

[sepulchre]

Merlin's book should be subtitled, "Don't panic; you will have to read this book five hundred times, but you will gain new understanding each time."

Got That right!

[roberto]

There is nothing magic about a 10x ratio. That's just some rule of thumb developed for those that don't want to understand.

I would like to add "and neither want to experiment". I can say very few of my amps' loads are following this rule. I'm often around half of it.

I've answred to the eq load here: https://tubeamparchive.com/viewtopic.ph ... 725#314725

And here you can find an intesting reading: http://www.aikenamps.com/index.php/desi ... amplifiers

[pdf64]

Just to mention that the 'load impedance ~10x > source impedance' thing is known as 'impedance bridging' http://en.wikipedia.org/wiki/Impedance_bridging

Agree that much interstage coupling in guitar amps is somewhat below that ratio.
The most extreme example in regular guitar amps being the 50k load presented by the trem intensity control on the end of the BF Fender vibrato channel.

The effect of paralleling up plate resistance, plate resistor and load is an example of where I find the Norton equivalent circuit to be more helpful than the Thevenin.

The Trainwreck 3rd stage is another example where a heavy load is used, possibly in order to limit gain / avoid additional series resistance required by a potential divider, signal voltage swing, bias excursion recovery etc.
Pete