# Using 18650 li-on battery to boost 5v

1. ## 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. here is the graph of current voltage at output.

Here is the circuit that i built: 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  Reply With Quote

2. ## 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.

1 members found this post helpful.  Reply With Quote

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3. ## 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.  Reply With Quote

4. ## 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.  Reply With Quote

5. ## 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  Reply With Quote

6. ## Re: Using 18650 li-on battery to boost 5v Originally Posted by elessar95 ...
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.  Reply With Quote

7. ## 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.  Reply With Quote

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8. ## 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?  Reply With Quote

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

Hi,

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

Klaus  Reply With Quote

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

I am using breadboard to change components easily  Reply With Quote

11. ## Re: Using 18650 li-on battery to boost 5v Originally Posted by elessar95 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.  Reply With Quote

12. ## 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  Reply With Quote

13. ## Re: Using 18650 li-on battery to boost 5v Originally Posted by FvM 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.  Reply With Quote

14. ## 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.  Reply With Quote

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

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.

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Also tell us the output current that you want for the charger and your switching frequency.  Reply With Quote

16. ## 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?  Reply With Quote

17. ## 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.  Reply With Quote

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18. ## Re: Using 18650 li-on battery to boost 5v Originally Posted by elessar95 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. Use calculated value of inductance.

- - - Updated - - - Originally Posted by FvM 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.  Reply With Quote

19. ## Re: Using 18650 li-on battery to boost 5v Is that location for current limiting resistor? 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.  Reply With Quote

20. ## 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.

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

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