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I am playing around with Mosfet specifically irf 540.According to the datasheet ,Vgs is about 10 volt.I added 2 resistors to 2 different circuits to see if there is a difference in the component's saturation. I applied then 3 volt to both gates .Surprinsingly the first circuit saturated and the other didn't.I have got then serveral questions:

  1. Why did the Mosfet saturate in the first configuration and not in both?
  2. Why did it saturate with 3 volt and not 10 ?
  3. Does this circuit really need a resistor following drain or source?
  4. What additional considerations to take in if i want to drive a load which draw at max 1 A at 24 V?

Here is a schematic of my simple circuit: enter image description here

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marked as duplicate by Daniel Grillo, JRE, Matt Young, Dave Tweed Jul 25 '15 at 12:27

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    \$\begingroup\$ The threshold is the voltage between gate and source. Measure those voltages in both cases. \$\endgroup\$ – PlasmaHH Jul 22 '15 at 14:29
  • \$\begingroup\$ Vgs is as seen by the FET. | Vgs cct 2 = Vin -(I_Q1 x R2) . As I rises Vgs decreases until a balance is reached. \$\endgroup\$ – Russell McMahon Jul 22 '15 at 15:48
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  1. In the second circuit the more current goes thropugh the MOSFET the higher will be the source potential \$V_S\$ because the same current goes through \$R2\$ which causes a voltage drop of \$V_S = R I\$.
    \$V_{GS} = V_G - V_S\$ thus the effect of \$V_G\$ is limmited by it self (negative feedback).
  2. The value for \$V_{GS}\$ you found in the datatsheet probably is the max. voltage allowed between gate and source
  3. It depends... if you want current limitation it is a good idea to have the resistor below the source (2nd circuit)
  4. If the load is inductive put a diode across the load
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A real IRF540 has a Vgs(th) of 2V to 4V, so its quite possible that it would start conducting at the lower end of that range.
Its also quite possible that whoever typed value into the sim model you're using chose the low end of the range. Vgs of 10V is only an example given to show an Rds(on) spec to try to sell the part.

So, with your relatively high value of R1, seeing the IRF540 saturate in your 1st circuit is not unexpected considering its possible Vgs(th) of 2V.
In circuit 2, as soon as the IRF begins to conduct the voltage across R2 will rise leading to Vgs no longer being the 3V you applied - consequently preventing the IRF from ever saturating.

The resistor represents your load, so no additional resistor is necessary.

Circuit 1 is the configuration normally used when you want the IRF to behave as a switch, because you can make it saturate with a voltage lower then your supply.

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