MOSFETs as switches

[rtarbell]

Concerning fundamentals of power MOSFETs used as switches:

I have seen several websites and datasheets on power MOSFETs that suggest to apply a gate-source voltage of >10V to turn a power MOSFET on as a switch. I have not seen any mention of what Vds (the drain-source voltage) is.

The equation for the two different ON states of a MOSFET are:

Saturation (amplifier mode): Vgs - Vth < Vds
Triode (switch mode): Vgs - Vth > Vds


==> When a MOSFET is used as a switch, and Vgs is initially 0 (so the switch is off), there is a certain Vds across the MOSFET.

HOW can we be sure that when we turn the MOSFET on by applying 12V as Vgs (usually by a gate driver), that the MOSFET will be in switch mode and not in amplifier mode? Either way the MOSFET will conduct, but if it gets "stuck" in amplifier mode, the MOSFET will dissipate more power than in switch mode.

[quasi]

Assumming only the one supply rail, the only way to guarantee saturation with a MOSFET is to have the load between the supply rail and the drain. This way you can guarantee that the gate voltage will be sufficient relative to the source.

Cheers

[kwantam]

It depends on the rest of the circuit. Obviously you can't expect to operate the MOSFET in linear if you connect it from your power supply to ground. On the other hand, imagine that you connect a resistor in series with the drain to V_dd, and connect the source to ground. Now when you turn on the MOSFET, as long as you turn it on hard enough that the voltage drop across the resistor spans the rail, you'll have 0 V_ds. In practice, there is a minimum channel resistance (Rds,on on the datasheet), which means you won't have exactly 0 V_ds.

An example: say you have a 1k resistor and a 10V supply. As you start increasing V_gs, the drain voltage will drop. At some point, V_ds will become less than V_gs-V_t, and now your MOSFET is in linear. To calculate the necessary Vgs for a particular operating point, start with the assumption that your MOSFET is already in linear (consider: why can you do this?):

(triode equation)
i_d = k' * (V_gs - V_t - V_ds/2)*V_ds

Vds = .2 (for example. We can choose this arbitrarily)
i_d = 9.8/1k = 9.8mA

So:
V_gs = 9.8e-3/(.2*k') + V_t + .1

Just remember, it's always monotonic, so if you make V_gs greater than this, V_ds will only get smaller. This is also the reason that you can assume that your MOSFET was in linear in the first place. A formal proof of this would probably make use of the fact that the i-v characteristics are continuously differentiable at the transition between linear and saturation.

Another circuit to consider: replace the resistor with an inductor, as in a boost converter. Now what happens?

-kwantam

[cliffforrest]

Originally posted by rtarbell


HOW can we be sure that when we turn the MOSFET on by applying 12V as Vgs (usually by a gate driver), that the MOSFET will be in switch mode and not in amplifier mode? Either way the MOSFET will conduct, but if it gets "stuck" in amplifier mode, the MOSFET will dissipate more power than in switch mode.

Not a helpful way of looking at it, IMO.

In the above case Vds will be whatever the external circuit imposes or lets it be, given that the FET is trying to be a resistance of Rds for the given Vgs:

If the external available voltage and available current (ie considering the RL:Rds) is such that the DS channel cannot saturate, the device will be in a linear mode and have higher dissipation. Otherwise Rds will win and the device saturates.

Look up high-side mosfet driving for a better understanding.

[rtarbell]

Thank you kwantam for your insiteful analysis! I understand now! I see that with most power FET characteristics, it does not take much Vgs to ensure that the device is in its ohmic region. A Vgs of ~10V will be sufficient for most power FETs to act as switches.