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  1. #1
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    Using 18650 li-on battery to boost 5v

    Hi i am trying to boost 18650 battery voltage to 5v in order to charging mobile phone.I used mc34063.When i connected to cell phone it shows charging but there is no current flow.I connected circuit to dc load but output voltage drops to 2.5 volt.I just tried to get 200ma but it drops to 2.5v.When i increase the current voltage decreases more.I installed the circuit following the steps in this site.

    Click image for larger version. 

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    here is the graph of current voltage at output.

    Here is the circuit that i built:

    Click image for larger version. 

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    Here is the detailed circuit information:
    https://circuitdigest.com/electronic...ircuit-diagram

    What must i do to correct output voltage.Do you have any suggestions?If i can boost battery to 5v it would be better but i couldnt find a way.Help me.
    Thanks

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  2. #2
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    Re: Using 18650 li-on battery to boost 5v

    That's a badly designed circuit. 34063 needs a current limiting resistor at pin 8, refer to data sheets. Suggested dimensioning rule is Ic/Ib=20, about 62 ohms for 3.6V input and 1A peak current.

    Apart from this point, don't expect mind blowing performance from ultra cheap 34063 switcher.


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    Re: Using 18650 li-on battery to boost 5v

    I have noticed that your R1 and R2 resistors are not properly selected. Try and get R2 to be exactly 3 times R1.
    -------------
    --Akanimo.



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    Re: Using 18650 li-on battery to boost 5v

    Sir how did you calculated 62ohms.I will connect 180 ohm to pin8 to try it.

    I used 6.8k and 2.2k as formula says Vout=((6.8k/2.2k)+1)*1.25) so resistor chosen properly.vout=5v when there is no load.When i connect the dc load vout decreases so much.



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    Re: Using 18650 li-on battery to boost 5v

    how did you calculated 62ohms.I will connect 180 ohm to pin8 to try it.
    180 ohm is the suggested value for 12V input and 1A Ipk. Read https://www.onsemi.com/pub/Collateral/AN920-D.PDF



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    Re: Using 18650 li-on battery to boost 5v

    Quote Originally Posted by elessar95 View Post
    ...
    I used 6.8k and 2.2k as formula says Vout=((6.8k/2.2k)+1)*1.25)...
    I'd prefer 2.7k and 8.2k; or 3.3k and 10k

    vout=5v when there is no load.When i connect the dc load vout decreases so much.
    What would most likely cause this is current sense just like FvM has mentioned. For your application, the current sense resistor that you are using is too high and so the controller switches too early.
    Last edited by Akanimo; 21st April 2019 at 11:06.
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  7. #7
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    Re: Using 18650 li-on battery to boost 5v

    Current sense may be reduced down to 0.2 if you want maximal output current, efficiency will probably drop.



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    Re: Using 18650 li-on battery to boost 5v

    actually i used 0.22 ohm and 0.33 ohm both.Even i used jumper as a very small resistor.Result is the same around vout=2.5v at 0.2a dc load.Am i follow right? Rsc=0.22ohm?



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    Re: Using 18650 li-on battery to boost 5v

    Hi,

    I assume there is a PCB layout problem...as so often.


    Klaus
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    Re: Using 18650 li-on battery to boost 5v

    I am using breadboard to change components easily



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    Re: Using 18650 li-on battery to boost 5v

    Quote Originally Posted by elessar95 View Post
    I am using breadboard to change components easily
    Breadboard is not suitable for SMPS. A veroboard is preferable. You can use pogo pins with it for easy replacement if you want.
    -------------
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    Re: Using 18650 li-on battery to boost 5v

    Hi,

    I agree with Akanimo. Breadboard is useless.
    Just read the design considerations in the datasheet.

    Klaus
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    Re: Using 18650 li-on battery to boost 5v

    Quote Originally Posted by FvM View Post
    That's a badly designed circuit. 34063 needs a current limiting resistor at pin 8, refer to data sheets. Suggested dimensioning rule is Ic/Ib=20, about 62 ohms for 3.6V input and 1A peak current.
    Sorry i don't get it.Is that Rsc for current limit resistor?I calculated it and its less than 1ohm.The other thing i didn't understand that is What is Ic/Ib?Collector current/base current for what?
    Thanks.



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    Re: Using 18650 li-on battery to boost 5v

    I'm talking about the pin 8 current limiting resistor, you have confirmed its necessity in post #4. Ic/Ib ratio is a matter of appropriate dimensioning of this resistor. If you had read AN920, you would know about.



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    Re: Using 18650 li-on battery to boost 5v

    Please also show how you calculated for your inductance.

    Rsc = 0.3/(2*Ioutmax*(1+ton/toff)) according to the datasheet.

    If your inductor is not capable of supplying enough charges during the main switch 'ON' interval, then the is output voltage will drop.

    - - - Updated - - -

    Also tell us the output current that you want for the charger and your switching frequency.
    Last edited by Akanimo; 24th April 2019 at 07:49.
    -------------
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    Re: Using 18650 li-on battery to boost 5v

    Ipk(switch)=2Iout((ton/toff)+1)=2*0.3*(1.31+1)=1.386
    Lmin= ((Vin(min)-Vsat)/Ipk(switch))*ton (3.2-1/1.386)*11.35*10^-6=18uH
    I was calculating Iout 500ma but Ipk current was equal to 2.31a and its too much for it.It can only handle 1.5A so i seleceted output 300ma and ipk(switch) is smaller than 1.5A refer to datasheet.
    Frequency is 50kHz
    Rsc=0.33/Ipk(switch)=0.33/1.386=0.23ohm.

    I use 100uH for inductor maybe its too high?



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    Re: Using 18650 li-on battery to boost 5v

    Sorry i forgot writing Ipk'switch
    Ipk'switch=(Vin-Vsat/Lmin)*ton(max)=(3.7-1/18.01*10^-6)*11.35*10^-6=1.7A
    Rsc=0.33/I'pk(switch)=0.33/1.7= 0.2 ohm again.



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    Re: Using 18650 li-on battery to boost 5v

    Quote Originally Posted by elessar95 View Post
    Ipk(switch)=2Iout((ton/toff)+1)=2*0.3*(1.31+1)=1.386
    Lmin= ((Vin(min)-Vsat)/Ipk(switch))*ton (3.2-1/1.386)*11.35*10^-6=18uH
    I was calculating Iout 500ma but Ipk current was equal to 2.31a and its too much for it.It can only handle 1.5A so i seleceted output 300ma and ipk(switch) is smaller than 1.5A refer to datasheet.
    Frequency is 50kHz
    Rsc=0.33/Ipk(switch)=0.33/1.386=0.23ohm.

    I use 100uH for inductor maybe its too high?
    See my calculations as attached based on datasheet.
    Click image for larger version. 

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    Use calculated value of inductance.

    - - - Updated - - -

    Quote Originally Posted by FvM View Post
    I'm talking about the pin 8 current limiting resistor, you have confirmed its necessity in post #4. Ic/Ib ratio is a matter of appropriate dimensioning of this resistor. If you had read AN920, you would know about.
    Did you get what FvM meant here? He calculated for the current limiting resistor that is connected across pin7 and pin8. It is not Rsc, do not confuse the two. So you need to have this resistor added to your circuit if you have not done so already.
    -------------
    --Akanimo.



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  19. #19
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    Re: Using 18650 li-on battery to boost 5v

    Click image for larger version. 

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    Is that location for current limiting resistor?

    Click image for larger version. 

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    And is this how should i calculate?

    I'm sorry i'm quite slow learner.I am using capacitor type and resistor type inductor.Should i use toroidal inductor for SMPS?

    Thank you.



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    Re: Using 18650 li-on battery to boost 5v

    Yes, that's the position of the current-limiting resistor.

    You can use anything as long as it is going to be able to store sufficient energy and then deliver the energy when you need it.

    - - - Updated - - -

    I think you need to select the inductor properly. The Kg method is a very useful tool when selecting one.
    -------------
    --Akanimo.



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