2
\$\begingroup\$

I am just starting to use MOSFETs more frequently. Rather than use suggested parts in circuits that others design, I want to understand them better so I can design my own circuits. I typically use them to drive inductive loads from microcontroller pins.

I'd like to see if my understanding is correct with the following:

  1. The \$V_{GSS}\$ (Gate-Source Voltage) listed under Absolute Maximum Ratings is the maximum voltage that can be applied to the gate with respect to the source. (And thus should be avoided.)

  2. \$V_{GS}\$ (Gate Threshold Voltage) is the voltage at which the gate starts to turn on, but does not mean that the MOSFET (as a switch) is fully conducting. It is given with min and max values, which means that the starting point may vary from part to part.

  3. To determine the gate voltage at which the MOSFET is fully "on" the \$R_{DS(on)}\$ (Drain-Source On-Resistance) value can be examined: the "test conditions" show voltage values at which the FET is fully on.

I don't think my understanding of #3 is quite right. Olin provided a great answer which is where I derived that point from.

Consider a Fairchild FQP30N06L; the datasheet shows two test voltages (10 and 5V) for the \$R_{DS(on)}\$ at different resistances (varying only by <10 mΩ).

Based on this, I'm confused about how to determine what voltage I should supply at the gate for the FET to be considered fully on. Can you explain what I need to look at on the datasheet to calculate this properly?

\$\endgroup\$
4
\$\begingroup\$

Point 1 and point 2 are quite accurate I would say. For point 3, I always search for the following graph when I'm looking for the likely conduction capabilities of FETs: -

enter image description here

It tells me that if I put 3.5V on the gate and I want to pass 10A drain current there is likely to be 0.5V dropped across the device leading to a power dissipation of 5W. This is an on-resistance of 50 m\$\Omega\$. However, if i used a gate drive of 10V, I can expect the voltage dropped by the device to be about 0.25V i.e. a power dissipation of 2.5W or an on-resistance of 25 m\$\Omega\$.

Regarding the state of "fully-on" you can see that there is no magic state that is reached at some arbitrary gate voltage but I'd say that at about 4.5V gate drive, the benefits of driving with a higher gate voltage are diminishing.

Figure 3 in the data sheet (which shows the graph which you deduced the on resistance variation of 10 m\$\Omega\$) is a derivation of the graph (figure 1) in my answer. Obviously, Figure 1 carries more information because it covers a wider range of gate voltages.

\$\endgroup\$
  • \$\begingroup\$ This is a very well described and practical answer. There can be a benefit of driving with a higher gate voltage - if the gate driver is up to the job of swinging the gate capacitance up and down in voltage with fast rise times. A higher voltage swing on the gate can result in faster turn on time for high speed switching applications. \$\endgroup\$ – Michael Karas Oct 28 '13 at 0:42
1
\$\begingroup\$

The rule of thumb equation for MOSFET behavior should be the Level 2 MOSFET spice model equation, which states:

$$ I_{DS} = \beta (V_{GS}-V_t)V_{DS}-V_{DS}^2 $$

for

$$ V_{gs} > V_{t} $$

MOSFET's operate in three relevant operating modes (the irrelevant one is the subtreshold region), cut-off, saturation and linear region. When you want to operate the MOSFET as a switch, you want to be in the linear region. The condition for being in the linear region is:

$$ V_{GS} - V_{t} > V_{DS} $$

If you take the derivative of the above function with respect to \$ V_{DS} \$ to find the equivalent on resistance:

$$ g_{out} = \frac{\delta I_{DS}}{\delta V_{DS}} = \beta (V_{GS}-V_t) - 2 V_{DS} $$

$$ r_{out} = \frac{1}{g_{out}} = \frac{1}{\beta (V_{GS}-V_t) - 2V_{DS}} $$

Finally, in the equation for the output resistance, you can safely assume that \$ V_{DS} \$ will be really small compared to the \$ \beta(V_{GS}-V_t) \$ term. So: $$ r_{out} \cong \frac{1}{\beta (V_{GS}-V_t)} $$

This equation tells me that, if I bias the MOSFET in the linear region, by having a greater overdrive voltage, I will have a constant resistance that is inversely proportional to my overdrive voltage. That is until, my current through that resistor increases to such a degree that my drain source voltage exceeds to break my assumption where I said the drain source voltage was really small.

That is why, you usually overdrive the MOSFET for switch operation, to the highest voltage available in your system, or the highest voltage the device gate will allow without breaking down.

\$\endgroup\$
  • \$\begingroup\$ When you want to operate the MOSFET as a switch, you want to be in the linear region Shouldn't it be saturation? \$\endgroup\$ – m.Alin Oct 28 '13 at 15:12
  • \$\begingroup\$ No, it is the linear region of operation you'd like the device to be in for a switch. The saturation region approximation for the MOSFET is a voltage controlled current source, and it's used mainly for current source and amplifier configurations. \$\endgroup\$ – deadude Nov 4 '13 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.