amplification in vacuum tubes

[vikram789]

hi, we say amplification µ=plate resis.(rp) x gm = δVp/δVg=plate voltage/grid voltage
but im confused where is the amplification taking place it is just that to maintain the same plate current we have to EXTERNALLY increase the plate voltage,δVp, in response to an decrease in grid voltage,δVg. Also i have one more doubt that, since we do not make the grid voltage positive w.r.t. cathode so amplification cannot take place as the presence of -ve voltage on grid will on restrict the no. of electrons reaching the plate.

 

[Audioguru]

Modulating the negative grid voltage causes the vacuum tube to conduct more and less plate current.
Ohm's Law causes the voltage in the plate resistor to change as the plate current in it changes. Then there is amplification.

 

[FvM]

gm in the said formula is a quantity valid for a specific bias point, thus it already considers the present grid and anode voltage. By the way, the formula is valid for any electronic component that can be decribed as a controlled current source, also BJT or FET. You can consult a tube datasheet to learn about gm values achievable under specific bias conditions.

Also, the formula describes small signal behaviour. Large signal amplification can be analyzed using datasheet characteristic curves or non-linear models.

 

[vikram789]

@ Audioguru

how can the current be increased as we r only applying -ve voltage to the grid and that wiill only decrease the flow of current

 

[Audioguru]

A voltage signal capacitor-coupled to the grid causes the grid voltage to become less negative and more negative. When the grid is less negative then the vacuum tube conducts more.
When it conducts more then its plate current increases.