My electronics professor is currently teaching DC-to-DC converters. He decided to drop us some theoretical homework and just postpone the experiment. He gave us some questions. This is one of those that I need more insight on.

enter image description here

Reference: https://www.electrical4u.com/buck-boost-converter/

I don't have quite a good grasp yet on the topic but I do know that buck converter causes a step down in the output voltage while boost converter causes a step up in the output voltage.

I am aware that a MOSFET and Schottky diode is used for the buck or boost converters. I am not sure if they are the only specific transistor and diode used. But if they are or if at least they are the common type, why is it so? Honestly I've already seen some answers in other websites but I would love to ask it here as well since the answers constructed here can be really insightful.

What I've read is that MOSFETs are used because they are switching transistors and Schottky diode is used because it allows higher switching speeds. I just mentioned this because I don't want to sound like I'm literally asking for my homework to be done without any effort. I just put this here so that I have at least some input.

Edit: Here is a more accurate basic buck topology. enter image description here

  • \$\begingroup\$ For high voltage stuff other types of transistors might be used. \$\endgroup\$
    – Pete W
    Dec 19, 2020 at 22:06
  • \$\begingroup\$ Just a quick comment on the schematic you have shown. This does not represent the most basic BUCK topology. The polarity of the output voltages looks wrong to me, but the main point is that it is not the most simple buck converter. \$\endgroup\$
    – mkeith
    Dec 19, 2020 at 22:19
  • \$\begingroup\$ Okay, I did not notice that immediately. Thank you for the clarification. \$\endgroup\$
    – AndroidV11
    Dec 19, 2020 at 22:21
  • \$\begingroup\$ Here is the basic buck topology. electronics.stackexchange.com/questions/276989/… \$\endgroup\$
    – mkeith
    Dec 19, 2020 at 22:23
  • 1
    \$\begingroup\$ That picture is wrong. That is an inverting buck converter so the polarisation markings are wrong \$\endgroup\$
    – user16222
    Dec 20, 2020 at 0:04

2 Answers 2


MOSFET's are used because the voltage drop across the MOSFET, when on, is lower than the voltage drop across a BJT. This makes the MOSFET more efficient as a switching device.

Schottky diodes are used because the voltage drop across the Schottky diode under forward bias is lower than the voltage drop across an ordinary silicon diode, once again, improving efficiency.

In some cases, instead of a Schottky diode, a second MOSFET may be used. If that is the case, then the second MOSFET would normally be switched on and off in complimentary fashion to the first MOSFET. This is referred to as "synchronous" operation. Synchronous operation is generally more efficient than using a Schottky diode.

  • 1
    \$\begingroup\$ @Hearth I altered the wording slightly. If you can't be pedantic on an engineering forum, then where CAN you be pedantic. \$\endgroup\$
    – mkeith
    Dec 19, 2020 at 22:06
  • 2
    \$\begingroup\$ @mkeith On a math forum. \$\endgroup\$
    – DKNguyen
    Dec 19, 2020 at 22:08
  • 1
    \$\begingroup\$ Besides their lower forward voltage drop compared to a bipolar diode, Schottky's are also significantly faster, particularly when turning off. While not a big deal at low switching frequencies, when an SMPS's switching frequency is hundreds of kilohertz or higher, a Schottky diodes help to reduce the losses in the entire power train. \$\endgroup\$
    – SteveSh
    Dec 19, 2020 at 22:11
  • 1
    \$\begingroup\$ @mkeith: I was about to up-vote until I noticed your using a period at the end of an interrogative sentence. Grammar Nazis can be as pedantic as engineers or mathematicians. :-) \$\endgroup\$ Dec 19, 2020 at 22:17
  • 2
    \$\begingroup\$ @mkeith: Fortunately, most engineers (and mathematicians) do seem to have senses of humor, an area in which many grammarians seem to be seriously deficient. \$\endgroup\$ Dec 19, 2020 at 22:24

It's all about picking the right tool for the job.

For the top switch, you could use a bipolar transistor (BJT), PMOSFET, NMOSFET, GeNFET, IGBT, etc. Parameters to consider are:

  • Conduction losses, ie current multiplied by voltage across the device while it is conducting. Since current is fixed by the design, it's all about voltage. For a BJT it will depend on VceSSat ; for a MOSFET it's about RdsON.
  • Switching losses, which depend on switching speed, ie gate to drain/source capacitance and stored charge for BJTs
  • Driving losses, ie how much base/gate current it needs
  • Drive voltage in some cases: for example if you want to boost the voltage from one single AA battery, you can't use a FET which needs several volts Vgs to turn on ; you need a very low threshold voltage FET or a low VceSAT BJT.
  • Simplicity and requirements of the driving circuit, especially regarding PMOS vs NMOS. For example a high side NMOS switch requires a bootstrap cap, or a charge pump if you want it to stay on continuously ; a PMOS does not but it will have worse efficiency.
  • Cost, of course. Note that cost is not just the cost of the component. A more expensive component that offers better efficiency may end up being cheaper by not requiring a heat sink, for example. Thermal management is expensive.

For a MOSFET of a specific voltage, the figure of merit is RdsONQg. Qg is gate charge, how much charge you have to put into the gate to make it switch, and RdsON is the FET resistance when it's conducting. You want low RdsON for low conduction losses, but that means a larger silicon chip (more area), which means more capacitance, which means higher Qg, slower switching and higher gate drive current. This depends on FET construction and each generation of FET tends to be better designed than the previous one, so you get a better RdsONQg product. There are other compromises. Higher voltage FETs will have worse RdsON. Ultra low gate threshold voltage means thinner oxide so the maximum allowed gate voltage is lower, making the FET more fragile.

A few decades ago MOSFETs weren't as good as they are now, so the switching device would often be a BJT.

Then low-medium voltage FETs (say up to 40V) evolved, Qg and RdsON went down, so BJTs began to be replaced because FETs offered a better compromise.

It took longer for FETs to get better at higher voltage, so BJTs were still the choice for mains-powered switching supplies, but now they are being replaced by FETs too. At even higher voltages, there's a competition between IGBTs, FETs and SiC FETs.

Likewise low input voltage converters require low gate threshold FETs, which took longer to come to market, so BJTs stayed in use there for a while.

Modern FETs just switch much faster than BJTs, require less gate drive, and have lower conduction losses, so for BJTs it's pretty much over in the switching converter game.

So the answer is: use the part that provides the best compromise for relevant parameters. As technology evolves and new components are developed, the optimum choice may change.

As far as the diode is concerned, in a switching converter its recovery time is important because that determines how long it needs to turn off. So you need a diode with low recovery time and low recovery charge. Schottky diodes are the fastest, plus they have lower voltage drop, but they tend to have higher leakage current and their performance gets worse at higher reverse voltage. Replacing the diode with a FET gives much lower voltage drop, so lower conduction losses, but it adds complexity as the FET needs to be driven, so you have another compromise.

The best choice depends on how much losses you actually have in the diode. If you make a buck converter that converts 12V to 1V to power a CPU, its duty cycle will be very low so the diode would conduct 92% of the time, plus the diode's forward voltage is 60% of the output voltage, which means the diode losses will be about 50% of output power. So replacing the diode with a FET really increases efficiency. However if you make a buck converter from 24V to drive a 20V LED string, then duty cycle will be low and the diode's drop is only 3% of the output voltage, so replacing the diode with a FET would only gain a few % efficiency. Probably not worth the extra cost.


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