1
\$\begingroup\$

Is the below formula, present on page 26 of this buck converter correct to use?

enter image description here

Or should I use the formula mentioned in the 3 paragraph of this accepted answer?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ The question is incomplete. You do not actually say what you are wanting. If it is the power dissipation in a buck regulator given power in and power out (So dissipation = Pin-Pout) then Olin's answer is correct. [Olins answers are correct 99.9876% of the time]. || The formula above has the units of power but is not obviously meaningful. || Iload x Vsat is the dissipation in the output switch when on. || Iq is undefined - if it is I quiescent then its the standby power. This may or may not be relevant when loaded. || Giving a source reference would help. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 15, 2023 at 11:50
  • 1
    \$\begingroup\$ @RussellMcMahon, thank you for the comment. So, the formula in the TI datasheet is incorrect for calculating the power dissipation of the buck converter? \$\endgroup\$
    – Freshman
    Commented Sep 15, 2023 at 12:13
  • 1
    \$\begingroup\$ I've now seen where the formula came from in the datasheet / app note and what it means in context. || Both answers SHOULD give approximately the same answer. || Olin's is easier by measurement. Measure Power in. Measure power out. Difference is dissipation . || TI formulat assumes most dissipation is due to losses in switch in IC. They assume that Iin = Iload x Vout/Vin. OK if transformer is lossless and switching losses small. THEN Dissipation in switch = Iin x V_switch_saturated. Substitute Iin as above and you get their expression. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 15, 2023 at 12:32

1 Answer 1

Reset to default
1
\$\begingroup\$

The power dissipation in a buck regulator (or any regulator) given Pin (power in) and Pout (power out) is
Dissipation = Pin-Pout
So Olin's answer is correct.
[Olins answers are correct 99.9876% of the time].

The TI formula is also "correct enough" but is more theoretical.
Both answers SHOULD give approximately the same answer.
Olin's is easier by measurement.
Measure Power in.
Measure power out.
Difference is dissipation.

TI formulat assumes most dissipation is due to losses in switch in IC.
They assume that Iin = Iload x Vout/Vin. OK if transformer is lossless and switching losses small.
THEN Dissipation in switch = Iin x V_switch_saturated.
Substitute Iin as above and you get their expression.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Sure, thank you very much for the answer. In my case, Input Voltage is 24V. Input current is 1A. Output Voltage is 5V. Output current is 3A. So, Input Power - Output power = 24W - 15W = 8W. This is a very high power dissipation right, even for a buck converter? Any suggestions to avoid this high power dissipation? \$\endgroup\$
    – Freshman
    Commented Sep 15, 2023 at 14:55
  • 1
    \$\begingroup\$ @Freshman Fi 6.9, page 9 in the datasheet indicates an efficiency of about 75-80% for those figures. Your 8w losses for 24w in is 67% efficient - somewhat lower than their figures. || Fig 6.8 suggests switch losses (Vsat x Iload x Vo.Vin) is substantially less than 6w or 8w. || Possible improvements may be found in inductor losses (wire thickness, core material, ...), cooling, Schottky diode losses, ... . \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 16, 2023 at 1:24
  • 1
    \$\begingroup\$ 1N5822 - fig 10.2 page 4 is very marginal here. It's a 3A diode and your Iout is 4 or 5A. It's Vf is around 0.6V - OK in many uses but maybve several watts dissipation. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 16, 2023 at 1:26
  • 1
    \$\begingroup\$ Exmple only - This diode is about as good as you reasonably get. Cost is $1.27 in 1 quantity. Still not perfect. || What inductor are you using? \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 16, 2023 at 1:31

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