0
\$\begingroup\$

In DC/DC buck converters datasheet dropout voltage vs output current is given like the graph seen below. To find the worst case minimum input voltage, lets say at 3.3V output and 1A output current, to this converter i would look at the 105C line and see there will be a 1.5V dropout. Therefore if the minimum input voltage for proper operation is 5V than i have to provide a minimun of 6.5V for proper operation at the highest temperature condition. Am interpreting this correctly?

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ What you write is true assuming that the dropout voltage stays the same between Vout = 3.3 V (this graph) and Vout = 5 V (for which we have no data). The question is if that is a fair assumption. At higher input voltages buck converters charge the inductor faster so with a higher current. This might cause more voltage drop. \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 10, 2017 at 14:52
  • 1
    \$\begingroup\$ But, with higher voltages you get better turn-on characteristics from MOSFETs (if used). \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2017 at 15:19

1 Answer 1

Reset to default
1
\$\begingroup\$

Actually, the ~1.5V dropout voltage refers to the minimum difference between actual input and output voltage, not nominal input voltage and output voltage. Thus you'll need at least \$ 1.5V + 3.3V = 4.8V \$ input voltage at 1A output current.

Similarly, if you want to run your circuit at 105C and draw 3A, you'll need at least \$2.2V + 3.3V = 5.5V \$.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.