Power Tube Grid Drive Voltage

[lowell]

How much grid drive voltage should 6L6 get? Or should I say what's the max? In thinking about it I've concluded that 50% of the AC voltage on the grids must not be greater than the voltage from G1 to the cathode, cause then we'd get grid current limiting right? So max grid drive is dependent on how the tube is biased. Let me know if I'm on the right track here.

Also, in the RCA receiving tube manual there is a figure under typical operation called Peak AF Grid No.1 Voltage. Is this the peak AC grid drive voltage when the tube is operating per the "typical operation?" If so then this figure must be 50% of the peak to peak AC voltage on the grids, right?

[Merlinb]

You are correct. For maximum unclipped output power (i.e., clean output), the peak AC grid voltage is equal to the bias voltage (maybe a little less, if you REALLY wanted to avoid any grid current). That's what the datasheet will tell you.
For a guitar amp however, the sky is the limit (within reason of course). The more signal to hit it with, the more overdrive you get!

[mooreamps]

So max grid drive is dependent on how the tube is biased.

Yes, well that's only half of it. The other half is the output impedance of the stage driving it. Let's say ; for example ; it's a 12ax7 and lets say it's load resistor is 100K ; and lets say the grid resistor on the 6L6 is 100K.

So, here is what we have. Let's say the max output voltage from the 12ax7 is 25 volts ; unloaded [very important]. Then when driving it 100K load, the output voltage is dropped between it's internal plate resistance; [66K], and it's plate load resistor ; [100K] in parallel with the grid load resistor of the power tube [100k}. So , you have a 66K driving impedance dropping 25 volts across a 50K effective load impedance. It will drop about 13.2 volts across the internal plate impedance, leaving 11.8 volts to the grid of the power tube.


A lower driving impedance ; like from a 12AU7 ; will drop less voltage across itself, and deliver more voltage to the power tube.


-g

[lowell]

Ok right on. Thanks guys. G Moore thanks for the enlightening impedance info. How do you calculate voltage drop when you have a nominal voltage and in/out impedance figures? Finally, is this somewhat easily transferred to phase inverters? Specifically a concertina.