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I found the definition of peak efficiency:

Peak efficiency=\$ \frac{P_{out,max}}{P_{in,max}} \$,at first,i thought if i have higher peak efficiency,and i have the output voltage,which is much higher than input.For example,if the peak efficiency of A and B are 30% and 85%,and both inputs are 1V,so the output A should be smaller than output B

But i compare the value with two paper:

A:peak efficiency:73% ; input:50mV ;output:1.2V

B:peak efficiency:85% ; input:0.6V ;output:0.9V

So now i am confused that what is the meaning of peak efficiency,for a boost converter,if its efficiency is higher than the other,what does it mean? this boost conversion is better? this boost has less power consumption?

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  • \$\begingroup\$ You have to read specs carefully : "peak" efficiency is measured at the most favourable working point for the convertor. What matters is the efficiency under the conditions where you are using it. So if A is 73% efficient under your input and output voltages at your load current, and B is 85% efficient at some other condition, don't blindly choose B; check what it does under your actual conditions : it may be less efficient than A in your application. \$\endgroup\$ – Brian Drummond Feb 10 '18 at 12:48
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Power efficiency is the ratio of output power to input power and may not at all coincide with a higher output voltage being produced. You need to factor in current as well as voltage to calculate power efficiency. Peak power efficiency occurs at different points for differing boost converters. Some may produce a peak efficiency at a higher output voltage than others but this is not obvious from the topology or a glance at the circuitry.

For example,if the peak efficiency of A and B are 30% and 85%,and both inputs are 1V,so the output A should be smaller than output B

No, they could both produce the same output voltage but "A" does so less efficiently and therefore consumes more current from the incoming supply to feed the same voltage and load as "B".

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  • \$\begingroup\$ so according to your meaning,' "A" does so less efficiently and therefore consumes more current from the incoming supply to feed the same voltage and load as "B".',B has less power consumption? \$\endgroup\$ – Shine Sun Feb 10 '18 at 11:59
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    \$\begingroup\$ Yes, "B" has less power consumption. \$\endgroup\$ – Andy aka Feb 10 '18 at 12:01

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