0
\$\begingroup\$

I have an N-channel MOSFET called BUK9Y07-30B,115 and a bootstrap IC called PMD2001D,115. I'm going to connect the bootstrap IC with the gate of the MOSFET. But I don't know if I need a resistor before and after or only before or after.

I know that the gate of the MOSFET have a capacitance and it takes time for it to charge up. When the gate is fully loaded, then current can pass from drain to source. Current don't flow through gate to source. Instead, when the operator stop using the gate, e.g lower the voltage, then the current at the gate is going back where it comes from. Just as a capacitor.

So my question, if I want a very fast CMOS-shaped PWM at the gate at the MOSFET. Do I really need a resistor before or after or both the MOSFET/Bootstrap IC?

Because, I know if I place e.g 10k resistor before the bootstrap IC, then it will increase time to power up the NPN transistor and discharge the PNP transistor when PWM goes to 0V. The same thing happens to the other side of the bootstrap IC.

But is it required?

enter image description here

\$\endgroup\$
10
  • 1
    \$\begingroup\$ The labels NPN and PNP do not match the corresponding schematic symbols. The symbols show PNP on top and NPN on bottom. It would be nice if you could re-edit the image to remove this inconsistency. I suspect that the symbols are right and the labels are wrong. \$\endgroup\$
    – mkeith
    Apr 28 at 23:11
  • \$\begingroup\$ @mkeith Either way works though for a gate driver with various advantages and disadvantages to both. If I were doing it I would build it with the labels rather than the symbols. \$\endgroup\$
    – DKNguyen
    Apr 28 at 23:14
  • \$\begingroup\$ @DKNguyen I guess you are right. Although with only 3.3V, I think the inverting version is probably safer (will get closer to rails). So you opened my eyes a bit. But I do still think the OP should produce an unambiguous schematic/and labels. And since the arrangement of the BJT's will affect polarity of the output, it really should be straightened out. \$\endgroup\$
    – mkeith
    Apr 28 at 23:17
  • 1
    \$\begingroup\$ @mkeith Fixed! Thanks for notice it! \$\endgroup\$
    – MrYui
    Apr 28 at 23:22
  • \$\begingroup\$ Is your MOSFET actually triggerable by 3.3V between the gate-source? You don't care about the threshold voltage; that's when the MOSFET just barely starts to turn on. You want the gate-source voltage to achieve the rated RDson. \$\endgroup\$
    – DKNguyen
    Apr 28 at 23:27
1
\$\begingroup\$

The diagram you have shown on its own is not bootstrapping. It is just a low-side gate driver.

But for what you drew, you do not strictly need the MOSFET gate resistors. It has its uses to control the rise/fall times, ringing, and noise (which may prevent some circuits from working properly), but since no static current though it's not strictly needed.

You do need the base resistors for the BJTs if the PNP is on top and a NPN is on the bottom because there is static current and will short like a diode. If you have a NPN on top and a PNP on the bottom, then a base resistor is not needed because each transistors will prevent a base-emitter short in the other transistor.

\$\endgroup\$
7
  • \$\begingroup\$ Ok! So only one resistor before the low-side gate driver? 1k? 160 Ohm? 10k ohm? \$\endgroup\$
    – MrYui
    Apr 28 at 23:14
  • \$\begingroup\$ Assuming 0.7V base-emitter drop, that leaves 3.3V-0.7V = 2.6V of excess voltage from the PWN pin to appear across the resistor. Pick a resistor that limits the current to what you want when 2.6V is across it. Maybe 1mA? \$\endgroup\$
    – DKNguyen
    Apr 28 at 23:15
  • \$\begingroup\$ Hmm...Perhaps I could use 160 ohm resistor, only because I'm having a lot of those at my PCB e.g for LED-diode. \$\endgroup\$
    – MrYui
    Apr 28 at 23:19
  • \$\begingroup\$ 160 Ohms is too low. A typical 3.3V GPIO can only provide 5mA max, and that's pushing it so you don't actually want it to be supplying that. I'm sure many on here would consider a 1mA base current excessive in the sense the BJT probably saturates with a lower current. \$\endgroup\$
    – DKNguyen
    Apr 28 at 23:26
  • \$\begingroup\$ With the circuit as shown even the base resistor is not essential as they are arranged as emitter followers. One disadvantage of that is reduced swing to the gate of the FET. You need one that turns on fully @~2.8V. This is somewhat marginal. \$\endgroup\$ Apr 28 at 23:48

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.