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Help understanding LM43600 boost cap role

Could someone elaborate on what would happen if you loaded a boost cap (the cap connected to pin 3 and the high side switch) that was too low.  I accidentally loaded a 470pF instead of a 470nF and am wondering if this could damage the part.  I saw a lesser output voltage (~3.0 instead of 3.5) on the output.  After replacing the part it looks like it is running fine.  I'm just wondering how much risk is involved in not replacing this part.  Would this be a major cause for concern?

Also, why is there a charge pump necessary for the high side switch?


  • Hi,

    Yes, a sufficient value for bootstrap capacitor is required in order for the circuit to turn on properly. The role of bootstrap capacitor is as a power supply for the high side gate drive for buck configuration. When the Low Side FET is turned on, the bootstrap capacitor is charged since the Switch node is close to GND potential. 

    If you do not have a sufficient value for the CBOOT, therefore you might run into an issue where the High Side FET cannot be turned on for a desired amount of time or could also risk overheating the high side FET because you do not have sufficient gate drive on the FET. The worst case is when you have a high duty cycle condition (where the on time of the High Side FET is longer than the turn on duration of the low side FET). In a buck, this is when VIN is close to VOUT. 

    All in all, yes you do need at least a 0.47uF capacitor as recommended on the datasheet with at least 10V rating to compensate for temperature variation and stuff

    Here is somewhat an app note of the role of boot capacitor. It's not a synchronous circuit where you have FET on High Side and Low Side but the role of the capacitor is similar



  • Thanks for the reply that is helpful.
  • Just a further note; In configurations where N-type switches are used as the high side switch in a conventional positive buck, the gate drive signal needs to be higher than Vin to fully turn on the switch. The boost cap creates this voltage source by charging the cap during the switch off-time, and making use of the available higher voltage during the on-time.