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I'm using the high side of an IR2110 driver. If I set the duty cycle to 100% the circuit burns out, I haven't how can I calculate the max duty cycle for the driver in a given circuit?

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    \$\begingroup\$ A circuit diagram and what burns out are necessary for a reasonable answer. \$\endgroup\$
    – Peter Smith
    Commented Jun 9, 2016 at 13:07
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    \$\begingroup\$ @PeterSmith Actually, it is not really necessary. In his case, the boot capacitor discharges and the upper fet stops conducting, which leads to a burn, be we can answer without having to know what is burning exactly. \$\endgroup\$
    – dim
    Commented Jun 9, 2016 at 13:17

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The duty cycle may not be 100% because the driver, like most of its type, uses a charge pump (bootstrap capacitor) to supply the high-side gate voltage (which has to be at least \$V_\mathrm{GS(th)}\$ above the voltage being controlled for an enhancement mode N-channel MOSFET). If left on indefinitely, the capacitor discharges through parasitic current paths, the gate voltage drops, and the high-side MOSFET enters the linear region and burns out due to excessive power dissipation. If you're unlucky, wild oscillation may occur in the process, potentially destroying the rest of the circuit. (Remember that MOSFET capacitances change strongly with \$V_\mathrm{DS}\$, so you can still get very severe oscillations long before you would expect the degree of discharge to be problematic based on a simple calculation.)

So, the duty cycle in itself is not the important factor, but rather the on-time. You can calculate the maximum safe on-time from the gate leakage current, the required gate charge for proper operation, and the capacitance of the bootstrap capacitor. In practice, you may need to measure \$V_\mathrm{GS}\$ using an oscilloscope (obviously, exercise due care and use high-voltage probes if you are working at 500V) and experiment to determine the parameters, since I don't expect that e.g. gate leakage vs. temperature is going to be specified in the MOSFET datasheet. Some of this leakage will occur through the driver itself, which is similarly unlikely to be specified in great detail.

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    \$\begingroup\$ There is also a minimum off-time, because the boot capacitor needs some time to charge again through the diode. \$\endgroup\$
    – dim
    Commented Jun 9, 2016 at 13:15
  • \$\begingroup\$ @dim good point, although the minimum off-time will be rather short (at least in relation to the maximum on-time) given that this is a high-current path. It is worth checking, of course, but will not pose a real problem in many application. \$\endgroup\$
    – Oleksandr R.
    Commented Jun 9, 2016 at 13:22
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Short answer : you can set the duty cycle "very close" to 100%, but not equal to 100%. You just have to ensure the high side transistor has time to go off and the output voltage drop to zero sufficiently long for the bootstrap capacitor (the one between Vb and Vs) to be recharged. So, approximately, the time when HIN must stay low at each period could be approximated by: $$t_{HIN-off} = t_{swtich-off} + 2 * R C$$ where:

  • \$t_{swtich-off}\$: transistor + driver switch-off duration (depends on tf, toff, gate resistance, transistor characteristics...)
  • R:estimated resistance of the Vcc circuit
  • C: bootstrap capacitor
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This subject is explained very well by IR at: www.irf.com/technical-info/designtp/dt04-4.pdf

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