Hey everyone!
I have attached the following image that will help with my problem. I have the solution but would like to ask a few questions.
I first have to find the DC current gain. I have two equations to choose from which are:
\[I_D=\frac{1}{2}K(V_G_S  V_t)^2\] (saturation region)
or:
\[I_D=K[(V_G_S  V_t)V_D_S  \frac{1}{2}(V_D_S)^2]\] (Triode region)
The solution uses the saturation formula. Would it be possible to explain the reason why this is so?
In addition, \[V_G_S \] is unknown but this can be written as:
\[V_G_S = V_G  V_S\]
I know how to find V_G (potential divider) but am confused how \[V_S = 3I_D\].
I thought that \[V_D = V_S\] so from the figure shown:
\[V_S 0 = I_D.3K\Omega\]
so:
\[V_S = I_D.3K\]
Please let me know where I am going wrong and thank you for reading!
I have attached the following image that will help with my problem. I have the solution but would like to ask a few questions.
I first have to find the DC current gain. I have two equations to choose from which are:
\[I_D=\frac{1}{2}K(V_G_S  V_t)^2\] (saturation region)
or:
\[I_D=K[(V_G_S  V_t)V_D_S  \frac{1}{2}(V_D_S)^2]\] (Triode region)
The solution uses the saturation formula. Would it be possible to explain the reason why this is so?
In addition, \[V_G_S \] is unknown but this can be written as:
\[V_G_S = V_G  V_S\]
I know how to find V_G (potential divider) but am confused how \[V_S = 3I_D\].
I thought that \[V_D = V_S\] so from the figure shown:
\[V_S 0 = I_D.3K\Omega\]
so:
\[V_S = I_D.3K\]
Please let me know where I am going wrong and thank you for reading!
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