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Understanding N-Channel MOSFETs

Cirkit

Oct 28, 2015
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I'm confused by the VGS(TH) and current rating of N Channel MOSFETs.

For instance, the attached datasheet for a Fairchild FQPF20N06L quotes an ID current of 15.7A and a VGS(TH) of between (1.0V to 2.5V). However the graph in Figure 2 of the datasheet seems to indicate that a VGS voltage in excess of 10V is required for an ID of 15.7A?

The graph also seems to indicate that below a VGS of approximately 2.1V, no drain current flows?
 

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Harald Kapp

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Vgs(th) is the threshold voltage above which the MOSFET starts conducting. That is by no means the point where the transistor is fully on. The drain current varies in triode mode with Vgs and Vds according to
upload_2017-6-8_19-4-57.png (source Wikipedia, other equations are valid in other modes of operation)

To fully turn the MOSFET on VGS >> Vgs(th) is required, therefore the datasheet states Vgs = 10 V for this mode of operation.

For Vgs < Vgs(th) ID is ~0 (apart from some small leakage). The characteristics (graphs) shown in a datasheet can only show typical values, or in some cases extremes, to. They can never cover each single parameter variation. Vgs(th) is a rather variable parameter (1.0 V .. 2.5 V), as you have noticed. Drawing a curve for every value of Vgs(th) would make the graph a meaningless black blob. Therefore a typical value is used to chart the characteristic.

A good design compensates these variations e.g.by using feedback or (not so good) trimming mechanism. THe worst case scenario uses selected transistors where from a batch of components only those are used that match some given paameters with small tolerances. This is very costly and therefore avoided.

The specific MOSFET you refer to is explicitly designed for switching applications. It is therefore easy to ensure Vgs >> Vgs(th) regardless of t he exact value of Vs(th) as there are only two states: on and off. This kind of MOSFET shall be used in linaer applicaion only with caution and under careful observation of the SOA (safe operating area, figure 9).
 

OBW0549

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Vgs(th) is the gate-to-source threshold voltage at which drain-to-source conduction begins. Below that voltage, little drain current flows (normally just a few microamps); as the gate voltage is raised above Vgs(th), more and more drain current flows. Needing a Vgs of 10V (or more) to get an Id of 15.7A sounds about typical.
 

(*steve*)

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In the datasheet you should find a graph of Id vs Vds for various values of Vgs.

You normally (in switching applications) want to use the device in the triode mode (where the graph is steeply rising) rather than in the saturation region (where the graph is close to horizontal).

The easiest way is to draw a horizontal line on the graph representing your current plus some safety margin (at, say, 200% of your planned Id). Then, the lines which cross it are candidates. The point at which each crosses (Vds) multiplied by your actual current gives an approximation of the power lost in the device. This must be within the capabilities of the device (and also feeds into heat sinking calculations).
 

Cirkit

Oct 28, 2015
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Thanks for all the responses, I have a better understanding of what VGS to use in switching applications. My intention is to use a TO220 transistor for PWM dimming of LEDs. So switching, but at high frequency. The VGS would be from a microcontroller so I'm guessing no where near 10V. Can anyone recommend a suitable part that would switch around 5A to 6A with a VGS of <3V to guarantee the transistor is fully on?

Steve, why is it better to use a MOSFET in triode mode for switching applications? I would have thought this would cause heating due to higher RDS(ON)?

You normally (in switching applications) want to use the device in the triode mode (where the graph is steeply rising) rather than in the saturation region (where the graph is close to horizontal).
 

Harald Kapp

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Any 'logic level MOSFET' with suitable paramters for Vds and Ids can be considered. Other parameters (e.g. gate capacitance etc.) may be of relevance dependinmg on the details of the application (e.g. witching frequency).
why is it better to use a MOSFET in triode mode for switching applications?
Assuming the same drain-source current Ids:
  • In triode mode Vds < Vgs-Vgs(th). A low drain-source voltage causes low power dissipation.
  • In saturation mode Vds > Vgs-Vgs(th) therefore power dissipation is high.
The nomenclature is a bit misleading as in bipolar transistors saturation is the operating area where Vce is minimal, therefore power dissipation is very low. With MOSFETs saturation means that the channel has been (more or less) fully created by Vgs >> Vgs(th) and the current is modulated by the drain-source voltage.
This Wikipedia articel describes the basics. For more detail you will have to study the relevant literature. Here is an entry point with references at the end.
 

Audioguru

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All new people are confused with the threshold voltage of a Mosfet. It is when the current in the Mosfet is very low, usually only 0.25mA. I tell the new people that the threshold voltage is when the Mosfet is almost turned off, not almost turned on.

The triode or saturation modes of Mosfets are the opposite of ordinary transistors so they are confusing. I say the Mosfet is a switch that is fully turned on or it is an amplifier.
 
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