Lm2577t-adj

Discussion in 'General Electronics Chat' started by yoosefheidari, Oct 31, 2013.

  1. yoosefheidari

    Thread Starter New Member

    Oct 26, 2013
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    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?
     
  2. #12

    Expert

    Nov 30, 2010
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    Post your schematic.
     
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  3. yoosefheidari

    Thread Starter New Member

    Oct 26, 2013
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  4. RichardO

    Well-Known Member

    May 4, 2013
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    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.
     
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  5. yoosefheidari

    Thread Starter New Member

    Oct 26, 2013
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    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.
     
    Last edited: Oct 31, 2013
  6. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Maximum output current of 1.2 amps. See page 16 of the TI datasheet.
     
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  7. yoosefheidari

    Thread Starter New Member

    Oct 26, 2013
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    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?
     
  8. RichardO

    Well-Known Member

    May 4, 2013
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    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.
     
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