# Choosing resistor values to operate a Half Bridge Driver

I am designing a 10W (18-12V) synchronized buck converter with a Half Bridge Driver and 2 MOSFETs as shown below. The Driver get a PWM signal from a programmed micro-controller (a 15KHz, 3.5V pulse at HIN and LIN). The Driver IC is the IRS2003(S)PbF. After the inductor there is a capacitor (not shown here). I feel confident about these values. However, for the MOSFET driver, I can't see from the datasheet any instructions on choosing the resistors' values. I know I must protect the IC which has a maximum Output current of 290mA for the high side driver (HO I think). The MOSFETS used are the IRL60B216. The MOSFET datasheet shows that the gate resistance is 2 ohms. How do I determine the resistors' values? where do I look at within the datasheets?

Update:I tried this out. As in the answer below: "When designing a mosfet driver the important thing is ensuring those capacitors are charged and discharged quickly while not overly stressing whatever is driving them. The IRS2003 has a minimum dead time of 400ns and so you need then to switch well within that number." I got this waveform for the Vgs of the MOSFETs, the upper one is for the higher side.

I got this waveform for the dead time

which I'm assuming is wrong because the charging and discharging of the equivalent MOSFET's "capacitors" are taking more than 400nS. I'm staring the 18V input a 0V and increment it by .1V, which is causing a short circuit as I exceed the current rating of the power supply after 2V. What can I do to prevent this? I tried 10 ohm resistors and it improved the dead time a little. but I can't go below that I think.

The gate resistance is really the equivalent gate input resistance before the "capacitors" that make up the gate itself.

As such Rg really only comes into effect while switching and has no DC component.

When designing a mosfet driver the important thing is ensuring those capacitors are charged and discharged quickly while not overly stressing whatever is driving them. The IRS2003 has a minimum dead time of 400ns and so you need then to switch well within that number.

The size of the resistor chosen can therefore not be too large.

It is difficult to tell from your schematic if PV+DWN is 12 or 18V. So I will assume 18V here. Derating the max output of the IRS2003 to say 250mA, with a voltage of ~18V when you pull down Q3 or Q4, that means you need a resistor of at least $72\Omega$

Obviously, if that rail is actually 12V you can scale that back appropriately.

Remember this current will only be required for charging and discharging the gate capacitances so its effect is short-lived provided you do not go overboard with the switching frequency.

However, it's prudent to actually do some tests with live components to verify the switching times are within the required window.

• you may be curious as to why I'm changing it to be right. However I'm even more curious that you're changing it back to be wrong. I won't change it again, this is not an edit war. Both final and penultimate paragraphs now use it s correctly. – Neil_UK Apr 8 '17 at 4:32
• Hi @Neil_UK. I have no idea, I did a few little edits but was not aware you changed them in the mean time. I didn't think SO allowed that... maybe if it's the original author it over-rides. Sorry for any frustration that might have caused you. Thanks for noticing and fixing my errors though. I appreciate it. – Trevor_G Apr 8 '17 at 14:26
• @Trevor Thank you This makes sense. The MOSFETs planed to be used are the IRL60B216, which can sink 300A at 18V Vds. Is this selection appropriate considering the need for a maximum of .6A thourgh the switch? could it sink more current and damage the half bridge driver? – Nadim Apr 8 '17 at 17:39
• @Nadim, ya I looked at the mosfet specs, I think you should be ok. The IRS current are totally separate from the load currents. However, the IRLs are not exactly the fastest mosfets in the world. Their inherent switching times eat up a good proportion of that 400ns. So keeping that R value on the low side is prudent. – Trevor_G Apr 8 '17 at 17:39
• @Trevor I added an updated question above, can you please answer? – Nadim Apr 12 '17 at 17:39

The resistor value is determined by Vcc. The driver spec sheet says each driver can source at least 130mA and sink at least 270mA, although it gives typical values of 290mA and 600mA respectively*. Using the most conservative value, 130mA, the derived resistor value is 12V/130mA = 92 ohms. In practice, you can pick a smaller resistor value as the driver will tend to perform close to its typical specs ($I_{MAX}$=290mA --> R=41 ohm) and also because the current will rapidly fall off as the MOSFET gate is charged/discharged.

*The sink/source currents are for a Vcc of 15V, but a Vcc of 12V should be reasonably close.

• Why do you use Vcc to calculate the resistor value? The outputs at the driver pins are rising from 0 to VB+.3V and 0 to Vcc+.3 for the high side and low side respectively, but I don't see the relevance. I think @Trevor got it right above by using the 18V from the MOSFET's drain. – Nadim Apr 8 '17 at 16:34
• @Nadim I had read PV+DWN as 12V, and calculated from there. I had used Vcc for expediency's sake because it was explicitly labeled as 12V. Trevor's answer is correct when that rail is 18V – Alex Apr 8 '17 at 18:09