I know this question might sound obvious, but still I cannot find an answer.

What is the purpose and best-fit applications for field effect transistors?

I mean, we have bipolar junction NPN transistors. It can be used to control a high-current circuit with a low-current signal, for example enabling relay through a microcontroller pin output. The most important characteristics (please, no holywars on this statement) is working voltage, hfe and power dissipation. We may assume that a NPN transistor with hfe = 50 for fixed voltage and base-emitter current 10mA passes up to 500mA from collector to emitter. In general, we can say that collector-emitter current is determined by base-emitter current.

[Disclaimer: I am not sure about the following statements and making it clear is the purpose of posting this question]: Ok, now let's take a look at FET transistors. Source-drain current is determined by gate-drain voltage:

enter image description here

The second chart (above letter б) is the dependency of source-drain current of gate-drain voltage. So,

  • While bipolar junction transistors' "useful" current amplification is determined by current between two pins, FET transistors' current amplification is determined by voltage between two pins;
  • FET transistor consumes much fewer energy because gate-drain resistance is very high;

Assuming this two statements are correct, I do not quite understand how do I want to use this transistors, and when should I prefer them to bipolar.

Thanks in advance.

  • 4
    \$\begingroup\$ Your last 2 points cover 99% of the reasons you'd use a MOSFET over a BJT. \$\endgroup\$ Commented May 5, 2014 at 17:13
  • \$\begingroup\$ @IgnacioVazquez-Abrams Can you please give some more details? I can conclude of these two points only one thing, or guideline - I should use MOSFET when I do not want/cannot afford wasting current on BJT, and basically that means that MOSFETs are always better (forgetting about possible price difference). \$\endgroup\$ Commented May 5, 2014 at 17:15
  • 2
    \$\begingroup\$ Use a Mosfet when it's cheaper, faster, can carry more current, or requires less components for your circuit. It really depends on your application and what the current prices are for related components. For example, a MOS inverter doesn't require two extra resistors on the input side unlike a bjt inverter. \$\endgroup\$
    – horta
    Commented May 5, 2014 at 17:18

4 Answers 4


If you wish to avoid holy wars you'll need to avoid making simplistic and incomplete statements :-).

Bipolar transistors are current driven.

MOSFETs are voltage driven.

In both cases the spread of parameters during manufacturing is such that a circuit will almost always rely on feedback to produce a given voltage or current gain.

MOSFETs tend to be slightly more costly at the very bottom end for "jelly bean" applications. But, for switching more than a few ~100mA, MOSFETs are usually as cheap or cheaper than functionally equivalent transistors, are easier to drive from a uC (microcontroller) as a digital switch than bipolar transistors and tend to have very significantly superior on characteristics.

An "on" bipolar transistor exhibits a saturation voltage. This can be several tenths of a volt and to get it much under 0.1V usually requires a high base to collector current ratio that is undesirably high. At 1 A a 0.1 \$V_{sat}\$ (saturation voltage) dissipates 0.1 W and is the equivalent of a R = V/I = 0.1/1 = 100 \$m\Omega\$ transistor. But at 10A the figures are 1 Watt dissipation and 10 \$m\Omega\$. The 0.1V is very difficult to achieve at higher current levels.

The \$R_{DSon}\$ (Drain-Source on resistance) of MOSFETs is typically under 0.1 \$\Omega\$ and you can get devices with 10 \$m\Omega\$ or even sub 1 \$m\Omega\$.

As switching speeds rise MOSFETs need a gate driver to charge and discharge the gate capacitance. These can be relatively cheap.

More soon ....

  • 1
    \$\begingroup\$ Well, you're right about simplistic statements but my knowledge doesn't allow me to write more details :( I have a few questions about your answer: "more costly" - you mean money or ...? What is "jelly bean applications"? What's "chepor"? Word is completely unknown ( \$\endgroup\$ Commented May 5, 2014 at 17:28
  • \$\begingroup\$ @AlexeyMalev: "as chepor cheaper" should be "as cheap or cheaper" (typo ...). \$\endgroup\$
    – Matt L.
    Commented May 5, 2014 at 17:44
  • 4
    \$\begingroup\$ Whether BJTs are current or voltage driven is a holy war. kevinaylward.co.uk/ee/voltagecontrolledbipolar/… \$\endgroup\$ Commented May 5, 2014 at 17:54
  • \$\begingroup\$ @SpehroPefhany- He rather spoils his point with his last few sentences, I thought. "All modelsare wrong. Some models are useful." - his last model is not very useful :-) \$\endgroup\$
    – Russell McMahon
    Commented May 5, 2014 at 21:43

The most obvious answer that springs to mind is when trying to switch a medium load. A BJT collector-emitter will saturate at possibly as low as 200mV whilst switching (say) 10 amps - power dissipation is 2 watt.

A decent MOSFET might have an on resistance of 5 milliohms and the volt drop will be 50mV at 10 amps - power dissipation is 0.5 watts. I'd choose the MOSFET!

If you go to a much more "powerful" application, the IGBT wins out because MOSFETs cannot readily achieve a low-enough on-resistance at (say) 500 amps but MOSFETs are creeping into this area year by year.

  • \$\begingroup\$ No dude I mean 5 milli ohms as in 5 one thousandths of an ohm or 0.005 ohms. \$\endgroup\$
    – Andy aka
    Commented May 5, 2014 at 17:32
  • 1
    \$\begingroup\$ Ah, you meant source-drain resistance? Sorry than. \$\endgroup\$ Commented May 5, 2014 at 17:33
  • \$\begingroup\$ "on resistance" is the source-drain resistance \$\endgroup\$
    – Andy aka
    Commented May 5, 2014 at 17:39

One great thing about MOSFETs that I've noticed is that they are much better for switching things. I can apply a voltage to the gate without worrying about limiting the current. However, when using a BJT, I need to put a current limiting resistor in series with the base. Otherwise, the switching signal can dump loads of current through the BJT to ground.

For example, if I need to switch a load that needs more current than my microcontroller can source, I'm going to reach for my small signal N Channel MOSFETs. I use them because I know the load-switching circuit that I make will not require a resistor in series with the gate.

Does this answer your question?

  • \$\begingroup\$ Yes, I think I got the answer I wanted. Thanks a lot. \$\endgroup\$ Commented May 5, 2014 at 17:24
  • 1
    \$\begingroup\$ You often should still throw a resistor on your gate line even with MOSFETS, for without it, for some parts, the gate capacitance can result in a very short, but very high current draw, blowing up the IO pin on your microcontroller. \$\endgroup\$ Commented May 5, 2014 at 17:43
  • \$\begingroup\$ @whatsisname Good point, thanks. I'm also considering way to limit gate-drain voltage to possibly limit source-drain maximum current? Or this is actually useless because the load will not use more power than it needs? \$\endgroup\$ Commented May 5, 2014 at 17:48
  • 1
    \$\begingroup\$ @whatisname that is an interesting idea. I don't think any of my microcontrollers have been ruined by sourcing too much current to the gate of a MOSFET. I think for the circuits that I've worked on (DIY stuff with through hole components) this isn't a real problem. But I have not done any testing to confirm or deny this. \$\endgroup\$ Commented May 6, 2014 at 2:40

MOSFETs are majority-carrier devices and can switch (especially off) much faster than BJTs.

Also, MOSFETs are voltage controlled and therefore don't require steady-state control current to keep them on. They do require large peak currents to charge and discharge the gate capacitance. (And the Miller capacitance as the FET transitions through the linear region.

When on, a FET is resistive. A bipolar transistor has a flatter saturation voltage. For similar size devices the FET will have lower conduction loss at low currents, and the bipolar will have lower conduction losses when its saturation voltage is less than the current*RDSon of the FET.

Bipolars are generally cheaper. (Not always though.)

IGBTs have some of the characteristics of both devices and can be a good choice too.

Which device you should use depends on the application requirements.


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