What is grid voltage measured with respect to?

[Fischer777]

When a vacuum tube data sheet states that it will conduct a given amount of milliamps with a designated grid voltage (the anode and cathode voltages being fixed in this example), is this grid voltage given with respect to the anode plate or cathode? I find it confusing because the grid voltages are generally written as being negative, and most circuit diagrams show the most negative part of the circuit being designated ground (thus all voltages and voltage drops are given in terms of positive voltages).

 

[Jony130]

Typically when we measure the voltage we do it with respect to GND. In this case we typical connect grid via resistor (R2) to GND and put resistor (R3) between cathode and GND to create negative grid voltage with respect to cathode.
http://www.wiredwithtubes.com/tech_info/voltage_current_calculate.gif
The current flowing through the tube will now create a voltage drop over the R3 resistor and with the grid referenced to zero, the tube will be properly biased by negative voltage Vg = - Ia*R3.
We have the same situation in JFET transistors.

 

[Fischer777]

Okay, thanks, that makes a lot of sense. I do have one other question though, and that is wouldn't the plate resistance also have to be taken into effect when calculating the voltage drop? So if one had R3, the plate resistance (Rp), and a voltage of V, would it be more correct to say Vg=-(V/(R3+Rp))*R3?

 

[Jony130]

No, we don't have to include Rp resistor. The controls voltage is grid voltage with respect to cathode. So for example if we have a 100V power supply voltage and we read from tube data sheet that for Vg = -1V we have 5mA anode current for Va = 50V.
So from the ohm's law R3 = 1V/5mA = 200Ω and Rp = (100V - 50V)/5mA = 10KΩ.
And we select R2 (grid resistor) in order to achieve desired input impedance.

 

[Fischer777]

Would something like what I have in the picture attachment work? The thinking is that because the plate resistance is 5kΩ (in the given situation) and the cathode biasing resistor is 5kΩ, the voltage drop across the grid will be 15V. The combination of the 15500Ω resistor and the 16000Ω resistor create a current of .952mA, which when multiplied by 15500=14.76V, roughly -.24V. If the tube data sheet says that -.25Vg when Va=15 permits 3mA to go through, then 15V/3mA=5kΩ roughly.

This situation is hypothetical, but I'm wondering if I'm understanding the concept at all

 

[Jony130]

Yes, It should work as you described.