Question about LTM8050

Hello Kazu,

You should read the DN1021 - How to Produce Negative Output Voltages from Positive Inputs Using a uModule Step-Down Regulator. ( http://www.analog.com/media/en/reference-design-documentation/design-notes/dn1021fa.pdf )

Iout(neg) =< (1-DCmax) * Iout(pos)
Where: 
Iout(pos) - the maximum output current of IC
DCmax = (|Vout|) / (Vin(min)+|Vout|) = (15V)/(24V + 15V) = 0.38462
Iout(pos) = 2A (You can see in the datasheet)

=> In this case maximum of negative current = Iout(neg) = (1 - 0.38462) * 2A =  1.23076A

Best regards
Adam

Hi Kazu,

Adam is correct, you need a higher current rated part when using a buck for negative output.

When using a buck for negative output, you are using it as an inverting buck-boost.  This means that the load current is a fraction of the total current leaving the Vout pin.  You can check the Vout pin current in your simulation to verify this is true.