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I am trying to decide on what H-bridge driver (for MOSFET) I should pick. I am completely new to this field so I might say something wrong.

So I decided to make a full H-bridge using 4 N-Channel MOSFETs, IRF1405. In order to do this I need an H-bridge driver to turn on the high side.

I found one full H bridge drive IRS2453(1)D(S) that can drive both sides and has internal bootstrap FET.

But I also saw IRS2001, but this is a half bridge driver (so I assume I need to get 2 of these for a full H-bridge driver) and it also requires bootstrap which I'm not sure how to pick which capacitor and diode to use.

So can someone help me choose, or if there is any other alternative I'll gladly hear your suggestion.

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  • \$\begingroup\$ Selection of a driver is usually driven by the MOSFETs being driven. If you don't say what MOSFETs you are using, this question isn't answerable. \$\endgroup\$ – Phil Frost Aug 17 '13 at 2:54
  • \$\begingroup\$ oh I didnt know that , I updated my thread I am using IRF1405 \$\endgroup\$ – subz Aug 17 '13 at 2:59
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Your first concern in selecting a gate driver is to find one that can drive enough current to switch your selected MOSFETs fast enough for your application. As a rough estimation, you can divide the total gate charge of your MOSFET by the current the driver can sink/supply.

$$ t_{on}=\frac{Q_g}{I_g}$$

Using the worst-case values for IRF1405 and the slower of your two gate drivers, IRS4253:

$$ \require{cancel} \begin{align} t_{on} &= \frac{260 \cdot 10^{-9} C}{180 \cdot 10^{-3}A} \\ &= \frac{260 \cdot 10^{-6} \cancel{C}}{180 \cdot \cancel{C}/s} \\ &= 1.44 \mu s \end{align} $$

Off is faster, because this driver (which is typical) can sink more current than it can source:

$$ \require{cancel} \begin{align} t_{on} &= \frac{260 \cdot 10^{-9} C}{260 \cdot 10^{-3}A} \\ &= 1 \mu s \end{align} $$

If your switching frequency is 10kHz, each switch period is \$1/10000 = 100\mu s\$ and you will spend \$ (1.44\mu s + 1\mu s) / 100\mu s = 2.44\%\$ of that time switching. Probably acceptable, but you should calculate your switching losses and check.

Also, keep in mind this calculation is an approximation. The current specified in the gate driver datasheet is current into a short circuit, but a MOSFET gate isn't that. Unlike a short circuit, the gate voltage rises as it is charged, which will reduce the current the driver can provide. Also, your layout may introduce more inductance and resistance than there was in the test circuit the manufacturer used, further reducing current. Consequently, your actual switching losses may be higher than this calculation suggests.

In selecting the bootstrap capacitor, you want to make sure it's significantly bigger than the gate capacitance it will be charging, so that the bootstrap voltage doesn't sag appreciably when you switch. It also needs to supply whatever leakage current there is as long as you keep the high-side switched on. You can calculate these leakage currents, or just make the bootstrap capacitor way bigger to be safe. 100 times bigger than the gate capacitance should be good, so at least \$26\mu F\$. Bigger doesn't hurt much, so round up to a standard value or whatever you already have in the BOM or stock.

Since this capacitor is the power supply for the high-side gate current, you also want it to be very low impedance. It wouldn't hurt to parallel your big capacitor with some smaller \$100nF\$ decoupling capacitors very near the gate driver(s).

Selecting a bootstrap diode isn't terribly difficult. It needs to be able to withstand the reverse voltage when the H-bridge is switched high. Also keep in mind that you will lose the diode's voltage drop from the gate voltage. A Schottky diode might be nice for this reason, but depending on your circuit, you may not find one that can take the reverse voltage. A simple 1N4148 can take reverse voltage up to \$100V\$.

The reverse recovery time of the diode can also be relevant if your are switching very fast; 1N4148 has a reverse recovery time of \$4ns\$, so you will have to have the H-bridge switched low for significantly longer than that for the bootstrap capacitor to have time to recharge between cycles.

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  • \$\begingroup\$ umm when i calculate Ton on my calculator i get 1.44 microseconds not milli, is there a step im missing ? \$\endgroup\$ – subz Aug 17 '13 at 5:59
  • \$\begingroup\$ @subz no, I just can't subtract 3 from 9. I'll edit the question in a bit, but you are right. \$\endgroup\$ – Phil Frost Aug 17 '13 at 10:16
  • \$\begingroup\$ @subz also, I had the bootstrap capacitor calculation 10 times too big. Apparently, I was up too late when I wrote this answer. I didn't feel stupid at the time... \$\endgroup\$ – Phil Frost Aug 17 '13 at 11:26
  • \$\begingroup\$ just one more question how did u calculate the bootstrap capacitance i , because the gate capacitance is 260nC , so C = Q/V and if my gate is charged to say 24V isnt C = 10nF, and 100x this is 1uF or am I wrong here ? Thanks \$\endgroup\$ – subz Oct 5 '13 at 3:06
  • \$\begingroup\$ @subz \$nC\$ is a unit of charge, not capacitance, but otherwise your reasoning seems sound. Looks like I might have read the wrong value out of the datasheet -- input capacitance (\$C_{iss}\$) is listed as \$5.4nF\$. You can also calculate a "capacitance" as you did and get a similar number -- the difference is due to miller effect and other non-linear behaviors. Either way doesn't make much difference: the point is simply to make the bootstrap capacitor much bigger so the gate charge doesn't significantly lower the bootstrap voltage. \$\endgroup\$ – Phil Frost Oct 8 '13 at 18:48

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