I will like to understand the term efficiency in a SMPS design. Is there a way to determine the best efficiency for a particular design?
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\$\begingroup\$ The "minimum efficiency that can be tolerable" part of your question is entirely subjective and (on a bad day) could lead to your question being closed. I would edit that part of the question out. \$\endgroup\$– Adam LawrenceJun 3, 2013 at 12:27
1 Answer
What don't you understand?
Efficiency is simply a function of the ratio of power out to power in.
A Switch-mode power supply that draws 5W in and supplies 4W of output power has an efficiency of 80% (\$\frac{4}{5} = 0.8\ \$).
A linear regulator generally has very poor efficency. Take, for example, a linear regulator that is producing 5V from a 12V source, with 1A of draw:
\$P_{in} = 12V * 1A = 12W\$
\$P_{out} = 5V * 1A = 5W\$
\$\text{Efficiency} = \frac{5W}{12W} = 0.4166... = ~41.6\%\$
I'm ignoring the power consumption of the regulator itself here, since it will be small in proportion to the overall power.
Yikes! 40% efficiency! That's terrible. It also means the other ~60% of the power (7 watts!) will be dissipated as heat, so you'll need a big heatsink to get rid of the waste-heat.
Now, most switchers have efficiencies in the 70-90% range. Lets look at the above 12V-5V converter with a 80% efficient switcher.
\$P_{out} = 5V * 1A = 5W\$
\$P_{in} = P_{out} * \frac{1}{0.8} = 6.25W\$
\$I_{in} = \frac{6.25W}{12V} = ~0.52A\$
Well, since the efficiency has gone up, the current drawn on our input is now less then the current out the output! This is possible because, remember, energy is conserved (minus the efficiency losses), not current or voltage.
You can also see that the dissipation in the switch-mode converter is only 1.25W (\$P_{in} - P_{out} = \text{dissipation}\$).
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\$\begingroup\$ Linear efficiency can be poor at heavy loads, but compare much better to switchers at light loads and are often much less complex, cheaper, and use less parts. \$\endgroup\$ Jun 3, 2013 at 12:30
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\$\begingroup\$ @Madmanguruman - Absolutely! There are still plenty of applications where linear regs are a much better choice. \$\endgroup\$ Jun 3, 2013 at 13:37
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\$\begingroup\$ Maybe it would be more accurate to say that energy is conserved. \$\endgroup\$ Jun 3, 2013 at 14:50
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3\$\begingroup\$ @ConnorWolf No. Power is the derivative of energy, or energy is the integral of power. Power is to energy as acceleration is to velocity. Power is measured in watts or joules per second, while energy is measured in joules or watt*seconds. Related, sure, but synonymous, not at all. \$\endgroup\$ Jun 3, 2013 at 17:23
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1\$\begingroup\$ @PhilFrost - I think I was using the more colloquial definition, where they are (perhaps incorrectly) used interchangeably. In a precise physics context (which we should be striving for), you are correct (and I'm editing my answer as such). \$\endgroup\$ Jun 4, 2013 at 14:29