The following readings were obtained experimentally from a FET:

$V_{GS}$ 0.1V 0.1V 0.4V

$V_{DS}$ 5V 14V 14V

$I_D$ 8mA 3.3mA 7.1mA

Obtain (i) ac drain resistance, (ii) transconductance and (iii) amplification factor.

Question

The following readings were obtained experimentally from a FET:

[latex]V_{GS}[/latex] 0.1V 0.1V 0.4V

[latex]V_{DS}[/latex] 5V 14V 14V

[latex]I_D[/latex] 8mA 3.3mA 7.1mA

Obtain (i) ac drain resistance, (ii) transconductance and (iii) amplification factor.

Accepted Answer

(i) At constant [latex]V_{GS}[/latex] (i.e.at [latex]V_{GS}=0.1V)[/latex]
when [latex]V_{DS}[/latex] is increased from 5V to 14V,
the drain current [latex]I_D[/latex] increases from 8 to 3.3 mA.
Therefore, change in drain–source voltage, [latex]\Delta V_{DS}=14-5=9V[/latex]
Change in drain current, [latex]\Delta I_{DS}=3.3- 8=0.3mA[/latex]
Hence, ac drain resistance will be at constant[latex]V_{GS}[/latex]:
[latex]r_d=\frac{\Delta V_{DS}}{\Delta I_D}[/latex]
[latex] herefore[/latex] At constant[latex]V_{GS}[/latex],[latex]r_d=\frac{9V}{0.3mA}=30K\Omega[/latex]
(ii) At constant [latex]V_{DS}[/latex](i.e.at [latex]V_{DS}=14 V)[/latex],when [latex]V_{GS}[/latex] is changed from
-0.1 to-0.4 V the drain current [latex]I_D[/latex] changes from 8.3 to 7.1 mA.
The change in gate–source voltage will be
[latex]\Delta V_{GS}=0.4-0.1=0.3V[/latex]
The change in drain current will be
[latex]\Delta I_D=3.3-7.1=1.2mA[/latex]
Therefore, transconductance will be
[latex]g_m=\frac{\Delta I_D}{\Delta V_{GS}}[/latex]at constant[latex]V_{DS}[/latex]
[latex]g_m=\frac{1.2mA}{0.3V} =4mA/V=4000\mu mhos[/latex]
(iii) Amplification factor is given by
[latex]\mu =r_d imes g_m=(30 imes 10^3) imes (4000 imes 10^{-6})=120[/latex]

Question 3.11: The following readings were obtained experimentally from a F...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions