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Here are the given values:

  • \$U_B = 12V\$

  • \$R_1 = 100k\Omega\$

  • The transistor is not "loaded" meaning, the Source-Drain current and the current through \$R_D\$ are the same.


Here is what is needed:

Find \$R_2\$ so that \$U_{GS}\$ is \$6V\$. I have already calculated this and \$R_2\$ should equal \$100k\Omega\$.

Then the assignment asks if the current through \$R_D\$ is \$50mA\$ the voltage across \$U_{DS}\$ should be \$6V\$ aswell.

This is where I don't know any further. If the current should be \$50mA\$ the \$R_D\$ should equal \$240\Omega\$, but then theres no way \$U_{GS}\$ is going to be \$6V\$.

enter image description here

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    \$\begingroup\$ Why would not it be 6V? 6V on the gate so the MOSFET is in saturation region. UB is 12V and 50mA is flowing through RD. Therefore, 240KOhms should be the value of RD. You are measuring the potential across G and S where S is GND. 6-0= 6V... \$\endgroup\$
    – 12Lappie
    Jan 12, 2017 at 20:10
  • \$\begingroup\$ I dont exactly understand, what you mean by "measuring the potential across G and S", can you elaborate ? \$\endgroup\$
    – zython
    Jan 12, 2017 at 20:33
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    \$\begingroup\$ The source of the MOSFET is connected relative to GND. Therefore it's potential is zero. The potential on the Gate is 6V. The difference between the two will give you the Vgs. Or Ugs in this case. \$\endgroup\$
    – 12Lappie
    Jan 12, 2017 at 20:58

1 Answer 1

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The schematic shows an Enhancement mode Nch. The circuit exceeds the typical threshold for Vgs(th) = 0.5~4V putting the drain into the "linear" mode where linear and saturated modes are opposite to BJT's. So BJT's Vce drops <1V when saturated while FET's drop in voltage to a low RdsOn when in "linear mode".

Since you calculated correctly Rload to be 240, much higher than typical RdsOn <10, your conclusion is also correct.

Thus 240 Ohms is the sum of Rd and RdsOn

There is no way the drain can be 5V.

It will be reduced by the R ratio of RdsOn/(RdsON+Rd) * V+ (near 0V)

Here a Nch MOSFET is shown with some arbitrary Drain current for Vgs above Vth on the x axis. and thus is a linear voltage controlled resistor until that resistance curve saturates or flattens out.

Rds will pull down the 240 Ohms by the some unknown amount based on the rated RdsOn @Vgs=6V If RdsOn was 10 Ohms Vd= 0.5V

Again this x axis is the Vgs-Vth difference. enter image description here

  • A saturated BJT has Vce<1V
  • A saturated MOSFET has Vds > saturation threshold of a parabolic curve where Id will not increase with Vds. Thus in Saturation mode with large Vds, Id=constant current controlled by Vgs.

    • In the "linear" region a MOSFET will act as a Voltage control switch of some RdsOn with low Vds to the left of the parabola.
  • The X axis scale could be in xx mV for millohm type MOSFETs.

  • another graph here shows in log-log Vds vs Id for a Pch enter image description here
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