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So I'm having fun with IR2101 MOSFET/IGBT half-bridge driver. I've checked typical schematic and tried to implement “happy path” scenario. enter image description here

I have created a very same circuit as in application notes, and seems like bootstrap circuit does not work. I get the following values:

Vcc = 12 V, VB=11 V, Vs = 11 V when HIN=LIN=0, and the same but Vb = 11 V and Vs = 9 V when HIN=LIN=1. Vgs = 11 V in this case.

I've tried circuit simulator (yes, IR2101 even has PSPICE model) and result is even more funnier...enter image description here

Can anyone suggest what can go wrong here? Capacitor is 1uF which should be more than enough for gate charge of 30nC. I also thought maybe diode is not fast enough, but it's schotky diode with very fast recovery.

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    \$\begingroup\$ There's no bridge supply voltage, and VS and Q2 are wired wrong. You're also driving both gates high at the same time, a violation. \$\endgroup\$
    – Tim Williams
    Commented Aug 5 at 13:32
  • \$\begingroup\$ Note also that the input "Hin" needs PWM ... for the "bootstrapping" circuit to work. \$\endgroup\$
    – Antonio51
    Commented Aug 5 at 14:01
  • \$\begingroup\$ Specifically, the source of your high-side device (Q1) must be pulled down to 0V so that the bootstrap capacitor can charge. Normally, the drain of the low-side device would be tied to the source of the high-side device to implement a half-bridge. If you're putting the load between the two MOSFETs, then you probably don't really need two MOSFETs anyways. \$\endgroup\$
    – W5VO
    Commented Aug 5 at 21:44

1 Answer 1

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This is just a misunderstanding on how the VB boost works only when the low side has PWM and the high side is then used for enable or direction/polarity with dual-half or full-bridge.

Then the datasheet specs will apply.

VB = VS +10 min to VS +20 max High-side floating supply absolute voltage

You were just getting VB < VS from this diode-cap "voltage-doubler circuit" at static or DC levels.

Be careful not to deviate from any design requirements in the datasheet.

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  • \$\begingroup\$ Are you suggesting that this circuit will work only in case we have some periodic signal? So that capacitor is chaged when it's pulled to ground by low side? \$\endgroup\$
    – seeker
    Commented Aug 5 at 13:32
  • \$\begingroup\$ yes A voltage doubler requires a repetitive pulse to charge the capacitor and PWM is the proper way to regulate average %output current. \$\endgroup\$
    – D.A.S.
    Commented Aug 5 at 13:39
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    \$\begingroup\$ Thanks, as it turned out that's not the only mistake I've made. Anyways, thanks for help! \$\endgroup\$
    – seeker
    Commented Aug 5 at 14:38

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