In some high current buck converters, there are multiple SW pins for output current. How the current gets divided between them? Example, lets consider a buck convertor has 5 SW pins and sourcing 30 A current, whether the current 30 A gets divided to 6 A each a pin. Is it correct or the current may vary?
\$\begingroup\$
\$\endgroup\$
4
1 Answer
\$\begingroup\$
\$\endgroup\$
2
Assuming you mean multiple pins on the IC package for a single switch node, the IC designers typically try to manage the layout and bonding (if applicable) to try to get the current to share equally between all the pins. This is to avoid violating maximum current/pin design rules. Otherwise there could be reliability issues with (E.g.) electromigration, temperature rise or excessive voltage drops.
Of course it will not be perfect, so there will be small differences in current sharing.
-
\$\begingroup\$ Of course, it also depends on the PCB layout. If, for instance, one pin isn't connected on the PCB (not that you should ever do that!), clearly that pin can't have any current through it and the current sharing will be significantly less perfect. \$\endgroup\$– HearthCommented Mar 28, 2023 at 4:01
-
\$\begingroup\$ Also, you could imagine a case where all pins do not share the main current. For example, one pin could be a Kelvin connection to the switch node for purposes of inductor current sensing. \$\endgroup\$– John DCommented Mar 28, 2023 at 18:41
divided to 6A each a pin... only if each pin has identical load and each load draws 6 A \$\endgroup\$