# Did I calculate this MOSFET circuit correctly?

Edit: I have updated the question with the marked answer

Follow up question to my question from two days ago:

How to calculate this resistor MOSFET circuit?

• Choose $$\R_D\$$ and $$\R_2\$$ that if a current of $$\50mA\$$ flows through $$\R_D\$$, and $$\U_{GS}\$$ and $$\U_{DS}\$$ are $$\6V\$$

• Calculate $$\I_{D,max}\$$ and $$\U_{D,max}\$$ and chart the characteristic into Figure 6

• Mark the operating point in Figure 6 in which region is it ? What has to be done to shift the operating point into the linear region?

• Calculate the factor $$\\beta\$$ if $$\U_{th}\$$ equals $$\1V\$$

• Calculate the draincurrent with the factor $$\\beta\$$ and the values: $$\U_{GS}= 7V\$$ and $$\U_{DS}=2V\$$ and mark the operating point in the characteristic in Figure 6

This is the circuit:

This is the characteristic (Figure 6):

There are the given values:

$$\U_B\$$ is $$\12V\$$

$$\R_1\$$ is $$\100k\Omega\$$

Here are my solutions/thoughts

I have calculated that $$\R_2\$$ has to be $$\100k\Omega\$$. But this is already where am not sure what is correct. I have been told (here, full disclosure) that $$\U_{GS}\$$ and $$\U_{DS}\$$ will always be equal in this circuit and in the case of $$\6V\$$ $$\R_D\$$ has to be $$\120\Omega\$$. I have mixed feelings about this, since I myself calculated $$\240\Omega\$$ and in the question earlier have been told that even $$\240k\Omega\$$ is correct.

Then to mark the characteristic, I would have marked either $$\12V\$$ or $$\6V\$$ and $$\100mA\$$ or $$\50mA\$$, why $$\100mA\$$ the task later asks, how to get the operating point into the ohmic region again, almost implying that I is in the saturation region initially. In this instance only $$\100mA\$$ and $$\12V\$$ would make sense, because thats the only instance the operating point is in the saturation area. I am confused.

Then in the next task, calculating $$\\beta\$$ should be no trouble, since I know all the variables, and only have to solve for $$\\beta\$$, but I need the correct operating point to be able to solve it.

Then in the last task, which I haven't done aswell yet, my solving way should be to simply solve for $$\I_D\$$ with the factor $$\\beta\$$ from the task earlier since I have all the variables.

### Marked Solution:

If translation is nessecary please comment below

• 100 view and no answers. did i make a poor effort in explaining my problem ? Jan 16, 2017 at 5:19
• Question looks well-formed, but we're a little lazy and don't like doing maths. Jan 16, 2017 at 15:55
• well i have solved it now to the best of my knowledge, i will update this poste with the correct marked solution as i have done in the past in about a week Jan 16, 2017 at 18:20

$$\text {You already obtained the correct value for }R_2.$$ $$\text {I don't know how you calculated } R_D, \text {but it should be calculated as follows:}$$ $$\text {At } U_B = 12V,$$ $$R_D = (U_B - U_{DS})/50ma = (12V - 6V)/50ma = 120 ohms.$$ $$U_{Dmax} \text { is obtained when }U_{DS} = 0, (12V - 0) = 12V.$$ $$I_Dmax = U_{Dmax}/R_D = 12V/120 = 100ma$$ $$\text {At }U_{DS} \text{= 6V, } U_{GS} \text{= 6V, } \text {and }I_D=50ma, \text {it is operating in the "saturation" region.}$$. $$\text {The saturation region (U_{GS} = 6V), is any point to the right (higher voltage) of U_{DS}=5V.}$$
$$\text {To shift the operating point into the linear region, you have to decrease the drain }$$
$$\text {current to around 32ma, which gives }U_{DS} = 2V.$$ $$\text{I calculated the gain }\beta\ \text {to be about 8ma/V.$$ Using this, and the given formula,}\text {one obtains: $I_D$ = 8 [7-1 -(2/2)] x 2 = 80ma.}