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I want to implement buck converter with conventional design (not synchronous etc) like below.
enter image description here

And I want to implement the converter for two cases. One is to use BJT as switching device and another is to use MOSEFET as swtiching device.

What I really want to know is that how much PWM voltage mangitude (Vgate-using MOSFET, Vbase-using BJT) should be when I use MOSFET or BJT as switch. Of course it must depend on the input voltage.

And to know the voltage magnitude, I thought I must determine the mode of switching device when it is assumed as on. To clearly deliver my question, I added a table.

enter image description here

What mode of MOSFET and BJT should be when it is on? For MOSFET, triode or saturation? For BJT, saturation or active?

I will really appreciate if someone answer this question and give me reason for that selection.

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  • \$\begingroup\$ Note that using a N-FET as a high side switch requires some mechanism to keep Vgs up as the source rises when the FET is conducting. \$\endgroup\$
    – Unimportant
    Commented Aug 23, 2022 at 8:54
  • \$\begingroup\$ What is 'high side switch' for N-FET? Why I need to keep Vgs up? Is that because of the source voltage being positive value when FET is conducting? @Unimportant \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:28
  • \$\begingroup\$ It's a high side switch since it's switching the positive side of the supply. When the FET is turned on it's source will rise toward the positive supply voltage as it starts conducting. So the gate must rise along, otherwise Vgs gets smaller. Vgs is what keeps the FET turned on. Without some mechanism such as a bootstrap capacitor all you get is a source follower. \$\endgroup\$
    – Unimportant
    Commented Aug 23, 2022 at 13:25

2 Answers 2

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What I really want to know is that how much PWM voltage magnitude (Vgate-using MOSFET, Vbase-using BJT) should be when I use MOSFET or BJT as switch. Of course it must depend on the input voltage.

It will not depend on the input voltage in a well designed SMPS but rather operate at a fixed voltage for MOSFET. For a BJT, the story is very similar, the base current will not depend on the input voltage but in an advanced design, it could be dependent on peak current. Let's ignore this for now and call it constant base current.

For a MOSFET, you need your Vgs to be at least Vgsth, the threshold voltage for the MOSFET you've chosen, but in practice more than that. 10 V is common for low voltage MOSFETs and 15 V for high voltage dito. You want to fully switch the MOSFET on and never stay in the linear region.

Here is a datasheet for a MOSFET picked at random: enter image description here

Below 2.5 V Vgs will be a disaster. More specifically, you want to look at the transfer characteristic:

enter image description here

You want Vds to be as small as possible, so you choose a Vgs to be high enough to satisfy this. At 3 V, the curve starts to straighten out. If your current isn't too high, this plus margin would do the job. The higher your current, the higher you need to go in Vgs to not have excessive losses.

As long as you stay below absolute max Vgs with margin, there is little penalty in going unnecessary high Vgs, so you will often find gate voltages twice what the transfer characteristic calls for in commercial power supplies.

For a BJT, you will be about 0.7 V above the emitter voltage, with a base current strong enough to stay away from the linear region, effectively Ipk < Ibase*hfe(min).

No one designs buck converters with BJTs these days, so you probably want to focus your efforts on MOSFETs and the gate drive of them.

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  • \$\begingroup\$ I got point of your explanation. And but Why should I avoid linear region for bjt? Is collector emitter current not enough for buck on operation? \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:18
  • \$\begingroup\$ You want to avoid linear region because of losses in the BJT. Losses are bad. Both voltage and current rating is enough to make a buck converter with BJT but base drive current requirement, forward voltage drop is too high for low output voltages and switching losses are higher than for MOSFETs. Crack open some commercial products and you will find MOSFETs almost everywhere and BJTs only on super high volume cost sensitive products with high input voltage, LED bulbs, fluorescent ballasts and perhaps a few low power SMPS adapters. \$\endgroup\$
    – winny
    Commented Aug 23, 2022 at 11:39
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For a switch-mode circuit you’ll want the transistor to be either on or off, with as little as possible in between. For a FET you’ll want a gate voltage that allows at least the maximum anticipated peak current to flow (this voltage depends on the FET you’re using). For a bipolar you’ll want a base current that allows at least the anticipated peak collector-emitter current to flow, this is given by dividing the peak current by the Hfe of the transistor. In both cases you’ll need to look at the datasheet for the particular device you want to use. As a guess, you’ll want 5 to 10V for a FET or around 2% of the peak current for a bipolar

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  • \$\begingroup\$ Is hfe is on the datasheet? And I have plan to implement synchronous buck converter with L289N(half birdge), and I'll design with Vin(input dc voltage of buck) = 5V, average current of inductor as 0.2A. with that given condition, how can I set pwm voltage to the gate? I wil use leftmost two complemtary bjt. \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:24
  • \$\begingroup\$ html.alldatasheet.com/html-pdf/22440/STMICROELECTRONICS/L298N/… \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:24
  • \$\begingroup\$ this is the datasheet of L298N. I \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:24
  • \$\begingroup\$ I ask this because I can't find the graph of the bjt characteristic! \$\endgroup\$
    – lsi
    Commented Aug 23, 2022 at 11:30

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