The plate current in a triode is given be `I_p=0.004 (V_p+10V_g)^(3//2)mA`
where `I_p,V_p and V_g` are the values of plate current , plate voltage and grid voltage , respectively. What are the triode parameters `mu,r_p and g_m` for the operating point at `V_p = 120 ` volt and `V_g =-2` volt

Options

A. `10, 16.7 kOmega , 0.6 ` m mho
B. `15, 16.7 kOmega , 0.06 ` m mho
C. `20, 6 kOmega , 16.7 ` m mho
D. None of these

Answer

A

Explanation

`I_p=0.004 (V_p+10V_g)^(3//2)mA`
`implies(DeltaI_p)/(DeltaV_g)=0.004[3/2(V_p+10V_g)^(1//2)xx10]xx10^(-3)`
`implies g_m = 0.004 xx3/2(120+10xx-2)^(1//2)xx10xx10^(-3)`
`implies g_m=6xx10^(-4)` mho = 0.6 mho
Comparing the given equation of `I_p` with
`I_p=K(V_p+muV_g)^(3//2)` we get `mu=10`
Also from `mu=r_pxxg_mimplies r_p=(mu)/(8_m)=10/(0.6xx10^(-3))`
`implies r_p = 16.67 xx10^(3) Omega =16.67kOmega`