# Finding Vs from the following DC, common source amplifier

#### u-will-neva-no

Joined Mar 22, 2011
230
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!

#### jegues

Joined Sep 13, 2010
733
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!
Since we are dealing with an amplifier, I would begin by assuming saturation and after the analysis is complete recheck that we are indeed operating in saturation.

Write an equation for Vgs in terms of Id, and solve for them both using the saturation equation for Id and check that Vds > Vgs-Vt.

The reason he uses 3 as opposed to 3k is because he is assuming Id is the order of mA.

Last edited:
• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
Thanks jegues. Was the analysis for:

$$V_S -0 = I_D.3K\Omega$$
correct? It seems strange just to assume a current in the mA region. I must be wrong with the analysis or missing a step out..

#### Vahe

Joined Mar 3, 2011
75
The current being in the $$\tex{mA}$$ region is not strange. One piece of information that you have not provided (and is not shown in the figure that you have posted) is the value of $$K$$. If, the given value of $$K$$ is given with the units of $$\text{mA/V}^2$$, then it is very reasonable for the current to be in the $$\tex{mA}$$ range.

Best regards,
Vahe

• u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
K is in the mA range. i did the calculation and realises that the working out is wrong. You must not ignore the 3kohm value!

#### jegues

Joined Sep 13, 2010
733
K is in the mA range. i did the calculation and realises that the working out is wrong. You must not ignore the 3kohm value!
If are consistent with all your calculations, keep all the resistances and currents in mA range will work fine.

• u-will-neva-no