On applying a potential of -1 V at the grid | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,for $V_g = -1 \ V$,the relation is $I_p = 0.125 V_p - 7.5$.For $V_g = -3 \ V$ and $V_p = 300 \ V$,the plate current $I_p = 5 \ mA$.To find t

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Given,for $V_g = -1 \ V$,the relation is $I_p = 0.125 V_p - 7.5$.For $V_g = -3 \ V$ and $V_p = 300 \ V$,the plate current $I_p = 5 \ mA$.To find the amplification factor $\mu = \left( \frac{\Delta V_p}{\Delta V_g} ight)_{I_p = 	ext{const}}$,we first find the plate voltage $V_p'$ required to maintain $I_p = 5 \ mA$ at $V_g = -1 \ V$.Substituting $I_p = 5 \ mA$ in the given equation:$5 = 0.125 V_p' - 7.5$$12.5 = 0.125 V_p'$$V_p' = \frac{12.5}{0.125} = 100 \ V$.Now,the change in plate voltage $\Delta V_p = 300 \ V - 100 \ V = 200 \ V$.The change in grid voltage $\Delta V_g = -1 \ V - (-3 \ V) = 2 \ V$.Therefore,the amplification factor $\mu = \frac{\Delta V_p}{\Delta V_g} = \frac{200 \ V}{2 \ V} = 100$.

Similar Questions

The mutual characteristic curves of a triode are as shown in the figure. The cut-off voltage for the triode is.....$V$.

The plate voltage of a triode is increased from $200 \, V$ to $225 \, V$. To maintain the plate current,a change in grid voltage from $5 \, V$ to $5.75 \, V$ is required. The amplification factor is:

The amplification produced by a triode is due to the action of

How many electrodes are in the tetrode?

In the grid circuit of a triode, an input voltage $E = 2\sqrt{2} \cos \omega t$ is applied. If the amplification factor $\mu = 14$ and the plate resistance $r_p = 10 \text{ k}\Omega$, find the $r.m.s.$ current flowing through the load resistance $R_L = 12 \text{ k}\Omega$ in $\text{mA}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module