Question

The plate resistance of a triode is 8 kΩ and the transconductance is 2.5 millimho. (a) If the plate voltage is increased by 48 V and the grid voltage is kept constant, what will be the increase in the plate current? (b) With plate voltage kept constant at this increased value, by how much should the grid voltage be decreased in order to bring the plate current back to its initial value?

Answer 1

If the plate voltage is increased by 48 V and the grid voltage is kept constant,then the increase in the plate current will be 6mA. This can be calculated as follows:

• We know that,

        Dynamic resistance of triode = r(p)= ΔV(p)/ △I(p)

• ∴ Subsituting the values  r(p)=8 kΩ , △V(p)=48 V

         8000 = 48 /△I(p)

         △I(p) = 48/8000

                  = 6mA

b) With plate voltage kept constant at this increased value, the grid voltage to be decreased in order to bring the plate current back to its initial value is 2.4V . This can be calculated as follows:

• We know that,

        Mutual inductance of a triode valve = g(m) = ΔI(p)/ △V(g)

• ∴Subsituting the values g(m)=2.5 millimho=.0025 mho, △I(p) =6 mA =.006 A

        we get,

         0.0025 = 0.006 / △V(g)                      

        △V(g) = 0.006/0.0025

                   = 2.4 V

Answer 2

Explanation:

In the question, it is given:

Plate resistance,

$\mathrm{rp}=8000 \mathrm{k} \Omega=8 \mathrm{k} \Omega$

(a) Plate voltage change, $\delta V p=48 V$

Plate resistance’s formula:

$r P=\delta V P \delta I P V G=\text { constant }$

$\Rightarrow \delta I p=\delta V p r p V G=\text { constant }$  

$\Rightarrow \delta I p=0.006 A=6 \mathrm{mA}$

(b) Now, the kept constant is Vp.

The plate current’s change,

$\delta ! \mathrm{p}=0.006 \mathrm{A}=6 \mathrm{mA}$

Transconductance,$g m=0.0025 \mathrm{mhosVG}$

= δIpgm

At constant plate voltage, 2.4 V.