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?

These are the tasks:

  • 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:

enter image description here

This is the characteristic (Figure 6): enter image description here

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. enter image description here

Marked Solution:

If translation is nessecary please comment below enter image description here enter image description here

  • \$\begingroup\$ 100 view and no answers. did i make a poor effort in explaining my problem ? \$\endgroup\$
    – zython
    Commented Jan 16, 2017 at 5:19
  • 2
    \$\begingroup\$ Question looks well-formed, but we're a little lazy and don't like doing maths. \$\endgroup\$
    – pjc50
    Commented Jan 16, 2017 at 15:55
  • \$\begingroup\$ 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 \$\endgroup\$
    – zython
    Commented Jan 16, 2017 at 18:20

1 Answer 1


$$\text {You already obtained the correct value for }R_2.$$ $$\text {I don't know how you calculated } R_D, \text {but it should be calculated as follows:} $$ $$ \text {At } U_B = 12V,$$ $$ R_D = (U_B - U_{DS})/50ma = (12V - 6V)/50ma = 120 ohms.$$ $$U_{Dmax} \text { is obtained when }U_{DS} = 0, (12V - 0) = 12V.$$ $$I_Dmax = U_{Dmax}/R_D = 12V/120 = 100ma$$ $$\text {At }U_{DS} \text{= 6V, } U_{GS} \text{= 6V, } \text {and }I_D=50ma, \text {it is operating in the "saturation" region.}$$. $$\text {The saturation region ($U_{GS}$ = 6V), is any point to the right (higher voltage) of $U_{DS}$=5V.}$$

$$\text {To shift the operating point into the linear region, you have to decrease the drain }$$

$$\text {current to around 32ma, which gives }U_{DS} = 2V.$$ $$\text{I calculated the gain }\beta\ \text {to be about 8ma/V.$$ Using this, and the given formula,}$$ $$\text {one obtains: $I_D$ = 8 [7-1 -(2/2)] x 2 = 80ma.}$$

I hope that by understanding how I obtained the various answers, your errors and confusion will be gone.


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