hi guys
i use a lm2577-adj for boost the voltage of a 1cell li-ion battery (3.8v-2600mah)
i use circuit that is in first page of lm-2577 datasheet and set voltage to 6.4 volt.
then i connet the circuit to four 1watt led's that i series each two of them and then parallel two strings.so for them i need 6.4 volt and 700ma current.(Pout=4.48)and P in should be about 3.8V*1.3A=4.94W
but when i connect the battery output voltage became to 5.5 volt and 200ma of current(Pout==1.1watt)
so what is problem?why voltage drop to 5.5 under load?

 

Post your schematic.

 

 

The most likely problem is an inductor with too high a resistance or too low a saturation current. It needs a resistance of way less than an ohm with a saturation current of at least few amps.

 

i use a 100uh 2A inductor.should i change it?
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can lm-2577 boost input=3.8V/1.3A to about output=6.4V/700MA ?
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please tell me maximum input current of lm2577 at 3.8 volt input.

 

Maximum output current of 1.2 amps. See page 16 of the TI datasheet.

 

Maximum output current of 1.2 amps. See page 16 of the TI datasheet.

i khnow.but at 1.2apms output what is maximum input current?
and why voltage drop? may because of amount of inductor(100uh) is false.may it?

 

i use a 100uh 2A inductor.should i change it?
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can lm-2577 boost input=3.8V/1.3A to about output=6.4V/700MA ?
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please tell me maximum input current of lm2577 at 3.8 volt input.

Let's start with your second question first. You say that you need an output power of 6.4 volts times 0.7 amps which is about 5 watts. 5 watts divided by 3.8 volts is about 1.3 amps. Since there are other losses in the circuit, I would design for more like 2 amps of input current.

That is the _average_ current needed from your battery. The inductor must have a peak current of something like -- very roughly -- 4 times the average current. This means the inductor should have a _saturation_ current rating of at least 8 amps.

In addition, the resistance of the inductor must be small enough to prevent large I*R losses at the average current. For example, to keep the inductor from dissipating more than a watt, the resistance would have to be less than 1/4 ohm. Note that the inductor must be physically large or it will get quite hot dissipating the 1 watt in my example.

The output switch of the LM2577 cannot withstand anywhere near the 8 amp peak inductor current given in my estimates.

A word of warning... the above are only guidelines and you should consult the data sheet of the LM2577 to see what they recommend.