Miller Effect (Slew rate current)

What is a driver anyway? Isn't it just another tube in the chain of amplifying stages? Well, yes, sort of. A driver is indeed just another stage when you refer to it's total gain in the system. A driver is ussualy only called a driver when the next stage is the output stage; the tube that delivers the output power.

Besides amplifying the signal in regards to the voltage swing, the driver can also have to perform some other tasks. At this position in the chain, the driver will usually swing quite a large voltage, as the output tube requires a big swing on it's grid to reach it's maximum output. For example, the 300B requires around 120Vpp or more, while the 71A requires only around 60Vpp for full output power.

 

Here an extra factor comes into the game, namely the input capacity of the next stage (tube). Even if this capacity is in the pico range (a lot of zeroes behind the dot!), it starts to have it's affects when voltage swing and frequency increases. The formula for this effect is along the lines of:

© VT52.com
 

Where:
Upeak (V) is the voltage swing on the grid of the power tube (0.5xVpp)
f (Hz) is the frequency at which the current is calculated; our maximum frequency of concern, i.e. 30kHz
Cgc (pF) is the capacity of the output tube between grid and cathode
Cga (pF) is the capacity of the output tube between grid and anode
A is the gain of the stage; when using inductive loads this is usually equal to the µ of the tube

 

In this formula you can enter the pF values, you don't have to calculate back to F (14pF = 0 . 000 000 000 014 F).

To give an example, below I have used the 71A and 300B both at 30kHz and 40Vpeak (80Vpp) grid swing.

 

71A:
A = µ = 3
Cga = 7.4pF
Cgc = 3.7pF
U = 40V
f = 30kHz

Ig = 2 x π x 40V x 30kHz x (3.7pF + 7.4pF(3.0 + 1)) / 1.000.000.000 = 0.25mA

 

300B:
A = µ = 3.8
Cga = 15pF
Cgc = 9pF
U = 40V
f = 30kHz

Ig = 2 x π x 40V x 30kHz x (9pF + 15pF(3.8 + 1)) / 1.000.000.000 = 0.61mA

yet at 70V swing the 300B requires 1.07mA
 


NOTE: To be sure the driver can handle these currents, multiply by 5 to see what bias current the tube should have. In case of the 300B, the driver should be biased at a minimum of 5mA in order to operate without any problems.

 

These examples go to show that the choice for a driver tube has to be considered not only for the gain it will give, but also if it will stay stable at high frequencies while driving the output tube. Using an ECC83, 30, 1H4 or similar tubes as a driver for the 300B is just asking for trouble, while they may have no problems with the smaller types of tubes like the 71 or 31 etc.

I have to point out that this is a "worst case" scenario, as the maximum grid swing will not often be the normal level you are playing (I hope for your neighbor's sake!). Despite this being at it's worst/maximum, it does not mean the amp should not be able to handle this situation. We may not hear 30kHz, but the amp has no problem in reproducing these signals (taking for granted that the iron that is used is of a good quality to pass these frequencies). That little bit of reserve the driver had might be swallowed up by something just outside the audible range, yet has enough influence on the tube to mess up the audible as well.

 

Conclusion:
Although these calculations are not all-revealing, they do tell a bit about what is required of the driver and that not just any tube can be used as a driver for any other given tube without possible consequences. "Better to prevent something than to fix it later" (in dutch it sounds much nicer and it is shorter too :).