# Mosfet as a switch.

#### electr

Joined May 23, 2009
49
Hello.
When using mosfet as a switch, we want its voltage drop to be minimal when its close.
In that case, should the mosfet be in triode state or saturation state, when its close?

Thank you very much.

#### hobbyist

Joined Aug 10, 2008
887
When any transistor is used as a switch, you want it to be at cutoff, or saturation.

Because a switch is either all on or all off, so with a mosfet, mainly because there is a lot of current flowing and you want the least amount of voltage across it to keep it from overheating. P=E x I so the smaller the volt drop. the less power consumed.

#### electr

Joined May 23, 2009
49
As you can see in the below graph, Vds is lower when mosfet is in linear region than in saturation region.
So why would you want it to be in saturation region? #### hobbyist

Joined Aug 10, 2008
887
I'm probably not understanding your graph, but with a load applied then the load drops much of the voltage and the mosfet drops the remainder, wich would be small if it is saturated. Isn't the bottom numbers talking about the voltage applied to the mosfet, not talking about the voltage drop across it???? I'm confused so let's wait till someone with knowledge about this can chime in..

#### beenthere

Joined Apr 20, 2004
15,819
If you look at the data sheet for any FET, you will see that the resistance from D to S is lowest when turned fully on. Insure the G - S voltage is 10 - 15 and there you are. When used as a switch, the device is either fully on or fully off. Behavior in the linear region does not matter.

For instance, an IFR1404 has a resistance of .004 ohm when turned on. That lets it pass 202 amps as long as the gets switched on fast enough.

#### Audioguru

Joined Dec 20, 2007
11,251
The same graph appears on a few different websites so it is probably correct.
It sounds like the terms Linear and Saturated are mixed up and I don't know why.

#### electr

Joined May 23, 2009
49
If you look at the data sheet for any FET, you will see that the resistance from D to S is lowest when turned fully on. Insure the G - S voltage is 10 - 15 and there you are. When used as a switch, the device is either fully on or fully off. Behavior in the linear region does not matter.

For instance, an IFR1404 has a resistance of .004 ohm when turned on. That lets it pass 202 amps as long as the gets switched on fast enough.
According to the graph, VDS in stauration is larger then VDS in linear.
In saturation, VDS > (VGS-VT),
in linear, VDS < (VGS-VT).

Moreover, If I'lll set VGS to 10V-15V, and assuming VT=1V, then in order for the mosfet to be in saturation region, VDS will need to be above 10V.

#### eblc1388

Joined Nov 28, 2008
1,542
According to the graph, VDS in stauration is larger then VDS in linear.
In saturation, VDS > (VGS-VT),
in linear, VDS < (VGS-VT).

Moreover, If I'lll set VGS to 10V-15V, and assuming VT=1V, then in order for the mosfet to be in saturation region, VDS will need to be above 10V.
You are not reading the graph correctly.

It is obtained first by applying a fixed gate voltage, then changing voltage across the Drain-Source pins of a MOSFET, probably using a power supply, and measuring the current Ids. As Vds is obtained from a power supply, you can set its output voltage to whatever voltage you want, within limits. The Vds is an input parameter, not a result.

Now in the graph, different Vds give the same 25A with Vgs-Vth=5V, so what is Vds for 25A? 5V or 8V, or you just picks the minimum as answer?

The graph shows that the Ids(drain-Source) current will no longer increase with a certain gate voltage even if one increase the voltage between Drain-Source to higher level. i.e. you can't make the MOSFET pass more current even if you use a higher Vds voltage.

So at most one can get 25A with a 5V (Vgs-Vth) gate voltage, even with Vds=10V or more applied. The available current increases to 36A if (Vgs-Vth) is 6V.

Therefore the MOSFET is operating in a "saturate region" giving a behavior like saturation but it does not mean the MOSFET itself is actually saturated. For that one would need to apply about 10V to the gate to get the MOSFET really saturates.

Joined May 28, 2009
508
You can use a JFET as a switch. When there's a bias voltage on the Gate, the JFET will not conduct between the Source and Drain. Take the voltage away and it will conduct. If you don't have a bleeder resistor between the Gate and ground, the JFET may continue to conduct when bias voltage returns due to internal capacitance that needs to be bled off.

#### electr

Joined May 23, 2009
49
You are not reading the graph correctly.

It is obtained first by applying a fixed gate voltage, then changing voltage across the Drain-Source pins of a MOSFET, probably using a power supply, and measuring the current Ids. As Vds is obtained from a power supply, you can set its output voltage to whatever voltage you want, within limits. The Vds is an input parameter, not a result.

Now in the graph, different Vds give the same 25A with Vgs-Vth=5V, so what is Vds for 25A? 5V or 8V, or you just picks the minimum as answer?

The graph shows that the Ids(drain-Source) current will no longer increase with a certain gate voltage even if one increase the voltage between Drain-Source to higher level. i.e. you can't make the MOSFET pass more current even if you use a higher Vds voltage.

So at most one can get 25A with a 5V (Vgs-Vth) gate voltage, even with Vds=10V or more applied. The available current increases to 36A if (Vgs-Vth) is 6V.

Therefore the MOSFET is operating in a "saturate region" giving a behavior like saturation but it does not mean the MOSFET itself is actually saturated. For that one would need to apply about 10V to the gate to get the MOSFET really saturates.
Thank you very much!
So what is the advantage in getting the mosfet into saturation region when the switch is close?

In a boost converter, they recommend to use a mosfet in saturation region as a switch when close.
But what would happen if the inductor current tends to rise above the maximal value that the mosfet allows in its saturation region? (for a given VGS voltage).

#### eblc1388

Joined Nov 28, 2008
1,542
So what is the advantage in getting the mosfet into saturation region when the switch is close?
Operating a MOSFET in saturation means it has the lowest resistance between source and drain and the power dissipated as heat is minimized.

But what would happen if the inductor current tends to rise above the maximal value that the mosfet allows in its saturation region? (for a given VGS voltage).
No idea what will happen.

Why is it possible that the inductor current can rise above that limited by the MOSFET in the first place?