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I'm struggling to understand couple of basic points regarding MOSFETs.

Below is an output characteristics of an N-channel MOSFET: Signal diagram

And below are definitions of \$ R_{DS(On)} \$ from two different sources:

Equation on the on-resistance

Explanation of the on-resistance

Is the power equation \$ P = I_{D}^2 * R_{DS(On)} \$ only valid for the linear region? I can see that the resistance is almost constant in linear region so it makes sense to use \$ P = I_{D}^2 * R_{DS(On)} \$ formula to calculate the power.

  1. But what if the MOSFET is in the saturation region? \$ V_{DS} \over I_{DS} \$ is not constant with increasing \$ V_{DS} \$ since Id almost saturates.

    It seems \$ R_{DS(On)} \$ is meaningless in saturation region. Which formula should be used for calculating the power in saturation region? \$ P = I_{D}^2 * R_{DS(On)} \$ or \$ P = V_{DS} * I_{DS} \$ ? If power for both regions are calculated different ways, should we each time check the saturation condition \$ V_{GS} > V_{Th} \$ and \$ V_{DS} > V_{GS} - V_{Th} \$ and depending on that decide which formula to be used?

  2. Secondly how does the current flow after pinch off point since the depletion layer blocks/closes the inverison layer as in the below illustration:

Physical construction of a MOSFET

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The concept of Ron only makes sense where the MOSFET behaves like a resistor, which is in the linear region. In saturation the transistor is modelled by a current source (with a resistor in parallel to account for channel length modulation). So in saturation you wouldn't use something like Rds(ON).

1) The power is always given by $$P = V_{ds} \cdot I_{ds}$$ so it is sufficient to use the correct value for the current. Rds,on is not required but it could be used to determine the current for small values of \$V_{ds}\$.

2) The depletion region does not block the flow of current. It results in a field pointing from the drain to the channel. Electrons that reach the end of the channel get swept to the drain by this field.

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  • \$\begingroup\$ so P = Id^2 * Rds(ON) is never used when calculating the power? \$\endgroup\$ – HelpMee Nov 7 '16 at 20:54
  • \$\begingroup\$ It could be used, of course. But for a general expression, that could be used in a model for example it's better to use only the current. \$\endgroup\$ – Mario Nov 7 '16 at 20:58
  • \$\begingroup\$ my question was is it only used for linear region P = Id^2 * Rds(ON) . does this equation make no sense for saturation region? \$\endgroup\$ – HelpMee Nov 7 '16 at 21:01
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    \$\begingroup\$ The concept of Ron only makes sense where the MOSFET behaves like a resistor, which is in the linear region. In saturation the transistor is modelled by a current source (with a resistor in parallel to account for channel length modulation). So in saturation you wouldn't use something like Rds(ON). \$\endgroup\$ – Mario Nov 7 '16 at 21:04
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    \$\begingroup\$ If you are using a MOSFET as a switch, you will probably know how much current you need to switch, but you don't really care about Vds. So in that case, it is easier to just calculate I^2 Rds. You would do this, for example, when checking the thermals to figure out whether you need a heatsink, and if you do, what its thermal resistance must be. \$\endgroup\$ – mkeith Nov 8 '16 at 7:00
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Linear RDS equation
Yes, the equation for RDS is derived in the linear mode of operation and is only valid for this case. For example, when used as a switch.

Power in Saturation Region
Yes, you will need to first determine what mode of operation the MOSFET is under. During saturation, power through it will be equal to VDS * IDS (which equals IDSAT for a long-channel approximation, where IDS has slope of 0 after VDSAT point). Because IDSAT is roughly constant, power will increase linearly with VDS from that point out. Remember this is an approximation that is not good for short-channel devices.

Pinchoff
The depiction of the channel cutting off can be misleading. While the depletion region means there are no mobile carriers, there is an electric field established that, when a carrier enters the field, will be swept up and create a current.

Some bedtime reading for you if you want to know more.

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  • \$\begingroup\$ It's called depletion region for a reason. There is no thin layer of charges. \$\endgroup\$ – Mario Nov 7 '16 at 21:00
  • \$\begingroup\$ Can a MOSFET be used as switch in saturation mode as well? \$\endgroup\$ – HelpMee Nov 7 '16 at 21:03
  • \$\begingroup\$ You're right there are no mobile carriers in a depletion region, but there are indeed carriers there once they are swept into the field (definition of a current). I'll update my answer to be clearer. \$\endgroup\$ – Joel Wigton Nov 7 '16 at 21:15

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