In modern electronics, this circuit is a differential stage; tubes lovers call it a Long-Tailed Pair. It is used to drive push-pull output stage with two signals in opposite phase and same amplitude.
This version is special because of the current source: practically, this is realised at best cost with either a MOSFET or a BJT. Of course, some may say that putting a semiconductor in a tube amplifier is nonsense, but it has many advantages:
- you define accurately the current value
- the gain stage is higher than if you use the traditional cathode resistor (the long-tail)
- it does not impact the sound or the global noise
To use the solver, make sure to understand how to bias it:
- set power supply Vb once for all
- define the common DC bias for grids: Vbias; put for instance 0 or 10;
- define the offset voltage between grid 1 and 2: dVg (0 for quiescent state); if you put for instance 1, it will mean that Vg1=50+1=51V whilst Vg2 remains at +50V; if you put dVg=-1, then grid of T1 will be at 50-1=49V, the grid of T2 remaining stuck at +50V. This the way to unbalance the stage and see how it impacts anode voltages. You may see the gain by putting increasing values, e.g. 10m, 100m, 1, etc.
(zero for equilibrium)