Why do they use this?

[Pedjoh]

Excuse me for a perhaps weird question.
But, I have noticed that for instance Ampeg use another version for connecting the grid/ cathode resistors.
They connect the cathode with 2 resistors in serie and in that point between these resistors the grid resistor is connected (instead of to ground).

So, can anyone understand what I mean and explain it to me?

Great!

/pedjoh

[Wakculloch]

It'll either be for a cathode follower, or an application needing a large cathode resistor. In either case the lower resistor is the load, and the upper resistor provides the bias voltage to keep the grid at the correct operating point with respect to the cathode.

[bbartilson]

Excuse me for a perhaps weird question.

So, can anyone understand what I mean and explain it to me?

No problem. We can do weird.

I don't really understand the question though. Are you talking about the way they sometimes wire phase inverters? If not that, you'd have to post a schematic or part of a schematic for me to look at.

There are several reasons for referencing the control grid to a point other than ground.

There's also the possibility that the schematic is drawn poorly. Two circuits which are electrically identical can appear different to the untrained eye. Not saying you're reading it wrong, but that's a possibility as well.

The implementation I'm thinking of uses a single triode for the phase inverter, and uses the plate for one signal and the cathode for the other phase to drive the other output tube. In this case the tube sits 'in the middle' of the supply voltage in order to allow sufficient voltage swing for both phases to work.

Not sure if that's what you're talking about or not.

-Bill

[Enzo]

I think that is what he means, Bill, like your phase inverter example.

All tubes need to have the proper voltage relationship between grid and cathode - we call that the bias. Note that the relationship is between those two elements, not necessarily to ground. In the input stage of a typical Fender or something, the cathode will have a 1500 ohm resistor between the cathode and ground for example. The current through the tube causes a volt or so to appear at the cathode. the grid will have a large value resistor to ground, but without any current flowing there, it will be at zero volts.

the net result is that the cathode is a volt more positive than the grid. Or looking the other way, the grid is a volt more negative than the cathode. Either way it is the same 1 volt relationshop. We say there is a volt of bias.

In some circuits like the split load phase inverter, or a cathode follower, the cathode is not going to be close to ground, bet we still need that relationship to the grid. SO we take the same thing our old Fender had - a 1.5k cathode resistor and a 1 meg grid resistor, and where they come together, instead of ground, we find the half load of the split, or the whole load of the cathode follower.


Try this: imagine a table with one leg an inch too short. You would have to "bias" the short leg somehow to keep the table level. A hunk of wood maybe. Now further imagine we put this table in an elevator. On the ground floor, the table might as well be anywhere - the tabletop is three feet off the floor (the ground) - and sitting level. Now run the elevator up to the tenth floor. Now the tabletop is 103 feet above the ground. But we still need that 1" bias for the short leg if the table is to sit level.

So this is roughly analogous to those resistors. If you want the cathode to sit at 100 volts, you still need a couple resistors to set the grid bias. So between the cathode and ground, we now have the 1500 ohm resistor, and we have the 100k load resistor. Cathode is at 100v, and at the bottom end of the 1.5k resistor we are down to about 99 volts. (Yes, I know my numbers are not precise). SInce no current flows through the grid, a resistor from that 99v point to the grid will not drop the voltage any, so there will be 99v on the grid - one volt less than the cathode. There is our bias, even though the whole mess sits at 100v above ground.

[Pedjoh]

Thanks for your reply's guys!
I'll try to present a schematic to clarify what I mean.

It's for example in the schematic for Ampeg Svt preamp I found at another
thread here in this forum.
http://users.aol.com/portaflex/schems/svtpream.gif
The stage where midrange boost/cut is connected have these resistors. 560r/7k5 , 47k/6k8 and 1k/47k.

/pedjoh

[Enzo]

OK, look at V5 - a classic cathode follower. Note the 1 meg resistor to the grid and the 1k cathode resistor that meet at their bottom ends. The current through the 1k establishes the voltage across that resistor, and so sets the bias.
Then the 47k is the load for that tube.

You could turn that stage into a plain old vanilla gain stage by replacing the 47k with a wire to ground and inserting the 47k between the plate at pin 1 and the power supply. Not that we would want to do that here, but just looking at it like a random circuit.

The righthand side of V4 is another cathode follower. The 47k and 6.8k resistors are its load. It is broken into two resistors so it can make a voltage divider and reduce the signal level leaving the stage.

[Pedjoh]

Great, now I'm beginning to getting it!

Found some more info at tubecad.com and together with your's
I think it's clear for me now.

Thanx!

/pedjoh