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I'm going to be using two IR2113 ICs to control each side of an H-Bridge (intended for an Inverter application.) From the application note, the expression to find the bootstrap capacitor is as follows

$$ C > \frac{2[2Q_{g} + \frac{I_{qbs(max)}}{f} + Q_{ls} + \frac{I_{cbs(leak)}}{f}]}{V_{cc} - V_f - V_{LS} - V_{Min}} $$

Following are the the values of the parameters that I was able to find:

  • Qg, Gate charge of High Side FET = 63nC.
  • I(qbs), Quiescent current for high side driver circuitry = 230uA.
  • Q(ls), Level shift charge required per cycle = 5nC
  • Frequency of Operation = 50Hz for one side, 20kHz for the other
  • Vcc, Supply Voltage, 12V
  • Forward voltage drop across bootstrap diode = 1.3V
  • Voltage drop across low side FET, 1.5V

I was not able to find the following values:

  • I(cbs - leak), Bootstrap cap. leakage current. Am I correct that if I use a ceramic capacitor, this value would be value and therefore can be ignored in the above expression?

  • V(Min), the application note states this is the minimum voltage between the Vb and Vs. Unfortunately, by looking at the suggested schematic (see below) I'm unable to understand what value should I be using for this.

enter image description here

For what its worth, by assuming that I can ignore capacitor leakage current and Vmin, I used the above expression and the values I found for a frequency of 50Hz. The capacitor size came out to be approximately 1uF. Does the capacitor need to be bigger than this in practice? If so, is there a rule of thumb saying how big? Is there a drawback to having a large bootstrap capacitor?

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    \$\begingroup\$ I THINK Vmin is the lower gate minimum drive voltage between LO and pin 2 in fig 2. (Or mayb Vs and HO - which should be the d=same. This should be in the data sheet but I'd ot expect it to be > 1V and maybe much less. Too high and some FETs will start to turn on (only ones wity very low Vth). // At a glance 1 uF sounds very high but you may be correct. That's for the 20 kHz side presumably. 50 Hz side will be much smaller I think. // As they note - only wity Al caps do you ned to worry about leakage. \$\endgroup\$ – Russell McMahon Aug 19 '11 at 10:48
  • \$\begingroup\$ 1uF was for 50Hz. I just redid my calculation and the answer turns out to be the same. 20kHz ought to be smaller I think, as frequency is dividing the rest of the terms. The value for 20kHz is 0.03uF. \$\endgroup\$ – Saad Aug 19 '11 at 11:00
  • \$\begingroup\$ Additional application notes: Fairchild Fairchild AN-6076, Bootstrap Circuit Design Manual. [Just piling up references.] \$\endgroup\$ – Nick Alexeev Apr 15 at 2:22
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The application note is clear on ceramic vs. electrolytic capacitors:

"Factor 5" (bootstrap capacitor leakage current) "... is only relevant if the bootstrap capacitor is an electrolytic capacitor, and can be ignored if other types of capacitor are used."

The application note also refers to DT98-2a (which itself refers to DT04-04) which despite focusing on IGBTs has this to say:

enter image description here

"To size the bootstrap capacitor, the first step is to establish the minimum voltage drop (\$ \Delta V_{BS}\$) that we have to guarantee when the high side IGBT is on.

If \$V_{GE(min)}\$ is the minimum gate emitter voltage to maintain, the voltage drop must be:

\$ \Delta V_{BS} ≤ V_{CC} −V_F −V_{GE(min)} −V_{CE(on)} \$

under the condition:

\$V_{GE(min)} > (V_{BS(UV)}- \$)

where \$V_{CC}\$ is the IC voltage supply, \$V_F\$ is bootstrap diode forward voltage, \$V_{CE(on)}\$ is emitter-collector voltage of low side IGBT and (\$V_{BS(UV)}-\$) is the high-side supply undervoltage negative going threshold."

With some interpretation, we can see that \$V_{GE(min)} => V_{GS(min)}\$ and \$V_{CE(min)} => V_{DS(min)}\$.

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  • \$\begingroup\$ Thank you very much. One question though - suppose I require a capacitor of 1uF. Should I use a 1uF cap. or around 5uF? Is there a disadvantage to using a larger cap? \$\endgroup\$ – Saad Aug 20 '11 at 13:37
  • \$\begingroup\$ @saad Not that immediately comes to mind, other than the fact that the \$ 5 \mu F\$ cap will be larger. \$\endgroup\$ – Adam Lawrence Aug 22 '11 at 15:27

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