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Like diode and BJT having around 0.6V drop, is there any voltage drop across the MOSFET drain and source when the MOSFET is turned on? In the datasheet, they mention diode forward voltage drop, but I assume that it for the body diode only.

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    \$\begingroup\$ The 0.6V drop on a BJT is from base to emitter (analogous to gate to source for a MOSFET). The MOSFET's drain to source voltage is analogous to the BJT's collector to emitter voltage. Vce and Vds are not necessarily 0.6V. \$\endgroup\$
    – Null
    Commented Dec 19, 2014 at 15:44
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    \$\begingroup\$ Yes they do, but it's not a saturation voltage (about 0.6v) like in the case of a BJT. Rather, it behaves like a resistor (when turned on hard), so the voltage drop can be much smaller than the "analogous" BJT on voltage drop. Very useful for low voltage circuits. \$\endgroup\$ Commented May 17, 2016 at 10:14

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The MOSFET behaves like a resistor when switched ON (i.e. when Vgs is large enough; check the data sheet). Look in the data sheet for the value of this resistor. It's called Rds(on). It may be a very small resistance, much less than an Ohm. Once you know the resistance, you can calculate the voltage drop, based on the current flowing.

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    \$\begingroup\$ Can you add for clarity's sake, that the Vgs (Gate->Source) voltage must be of high enough magnitude for that Rds(on) value in the datasheet to be valid \$\endgroup\$
    – KyranF
    Commented Dec 19, 2014 at 15:48
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MOSFET: When the gate voltage is large with respect the the threshold voltage Vth, the voltage drop from drain to source is linearly dependent on current (for small voltages << Vth of the MOSFET), so it behaves like a resistor. The resistance is less when the MOSFET is more enhanced, so a more positive voltage on an n-channel MOSFET gate relative to the source. The equivalent resistor might be tens of ohms for a small MOSFET down to milliohms for a large power MOSFET. From the 2N7000 datasheet you can see that for a gate voltage of 4V and a Vds < 0.5V the resistance is a couple of ohms (typical, worst case would be much more than that). So typically at 50mA, it would drop maybe 100mV. (The resistance Rds(on) is the slope of the curves near the origin). Rds(on) increases greatly with high temperature, so be careful with using 25°C specifications. If you don't give it enough gate voltage (many MOSFETs are specified at 10V, some at 4.5, and fewer at 1.8 or 2.5) you may get a much higher Rds(on).

enter image description here

BJT: Voltage drop from collector to emitter is dependent on current but not linearly. At low current and with high base current, the BJT might have a voltage drop of tens of millivolts. From the 2N3904 datasheet you can see the characteristics when Ib = Ic/10. You can see that at, say, a current of 50mA it has a voltage drop of about 90mV so quite similar to the 2N7000. Vce(sat) is the relevant specification. It's pretty stable with temperature, but you must give it plenty of base current for the expected collector current. If you don't give it enough base current, the voltage from collector to emitter can increase greatly. At more than the base voltage it's no longer considered to be saturated.

enter image description here

One interesting difference between the two is that the MOSFET drops almost exactly zero voltage at zero current, whereas the BJT drops perhaps 10 mV at zero collector current (assuming you put some reasonable current in the base- that's not reflected in the above curve). That makes the MOSFET generally a superior switch for precision instrumentation applications where 10mV is a big deal.

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  • \$\begingroup\$ You write "the voltage drop from drain to source is linearly dependent on current". But I think it's the opposite. It's the drain current that is linearly dependent on the drain-source voltage. \$\endgroup\$ Commented Jan 8, 2022 at 22:46
  • \$\begingroup\$ @AbdelAleem you can think of it that way, it’s not wrong, both statements are true, however normally the load resistance is much higher than Rds(on) and the supply voltage is more-or-less fixed, so it is much closer to a constant current than a constant voltage being applied to the MOSFET, if that makes sense to you. \$\endgroup\$ Commented Jan 8, 2022 at 22:50
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I think you are comparing two different things.

The 0.6V drop you typically see in a BJT is the B-E (base to emitter) junction.

For a mosfet, a similiar analogy doesnt exist. G-S (gate to source ) will always be whatever the gate voltage is with respect to the source.

For a BJT collector to emitter, that will vary depending on your collector current and collect or emitter resistor.

For a mosfet, there is a paramater called Rds(on) which is the resistance between source and drain. So the D-S (drain to source) voltage, like the C-E voltage will vary depending on current.

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  • \$\begingroup\$ When they say "drain-source voltage", do they mean the voltage drop across the drain-source terminals of the MOSFET? \$\endgroup\$ Commented Jan 7, 2022 at 14:50
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    \$\begingroup\$ Yes, just that. \$\endgroup\$ Commented Jan 7, 2022 at 14:58

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