0
\$\begingroup\$

I am trying to build a power bank that has 3 output USB ports that can provide 2.1 amp each; it has 8 LiPo cells. So total of 2.1×3=6.3amps so basically 7 amps at 5.1v out.

I have built the setup on breadboard but cannot get more than 2 amps out of even one USB port. I have tried making the inductor bigger, in size, triple wire for lower resistance and larger diode and N channel MOSFET.

PWM is with STM8S microcontroller at switching speed of 64kHz using 3mH inductor. I thought that paralleling the inductor, diode and switch would be enough to give more amperage but I'm not seeing that as a result. Instead the result is just 2.1 amps total. Then Schottky diodes starts to become hot or the 3 logic level MOSFETs.

Can the boost converter be useful for low voltage, high amperage or do I have to use transformer based circuit? Or do I need 10 diodes and 10 MOSFETs and a huge inductor for greater current?

enter image description here

|improve this question
\$\endgroup\$
  • 3
    \$\begingroup\$ Please post the exact scheme. Theoretically, leave alone load regulation, there should be not much problems. In practice you have to measure few points with oscilloscope and get an idea what fails. It can be input voltage dipping, can be inductor saturation, who knows? I once had similar behavior with lipo overcurrent protection going active. \$\endgroup\$ – Gregory Kornblum Oct 1 '18 at 9:18
  • 4
    \$\begingroup\$ Over 500mA on solderless breadboard is asking for trouble, 8A is doomed. \$\endgroup\$ – Jasen Oct 1 '18 at 10:13
  • \$\begingroup\$ 3mH at 64kHz has an impedance of ??ohms and a DC resistance with losses of ?? watts and an SRF of ?? kHz? and max DC,AC current of ??? ( hint ... too much) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 1 '18 at 14:19
  • 2
    \$\begingroup\$ @TonyEErocketscientist, I don't think it's appropriate to fill the OP's question with empty placeholders. We should rather ask him for this information in the comments, and have him fill in the details himself. As it is, the question content containing placeholders is not useful. \$\endgroup\$ – Reinderien Oct 1 '18 at 17:41
  • \$\begingroup\$ @Reinderien I disagree, however I understand the rules. How are you going to teach them how to write a better question? Especially when few have experience of a Test Engineer or R&D Engineer for 40yrs yet everyone understands they need good info. But how to present concisely is how I showed it. It was a mistake to delete it IMHO for this newb. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 1 '18 at 21:10
4
\$\begingroup\$

Yes you can use a boost converter. This would be difficult design that required experience. 3.3V requires carefully choosing the MOSFET for threshold levels, most logic level MOSFETS really want 4.5V. 8A of current will produce a lot of heat in a diode so you will need a heatsink and probably a fan. The loop formed by the MOSFET, diode, and output capacitor needs to be small, any excess inductance will increase loss and noise. And finally the output capacitor current is discontinuous so you need capacitors that can handle the RMS current. For this application I would probably not use a diode but instead another MOSFET and find a way to bootstrap the 5V to run the controller and gate drive.

Yes, it can be done. But this is a very advanced project (to do well) without knowing a lot about power electronics.

|improve this answer
\$\endgroup\$
  • 1
    \$\begingroup\$ You can also replace the diode with another MOSFET, the mosfet will have less conduction losses and produce less heat. However you have to make sure that the mosfets are never on at the same time, so some more consideration needs to be taken into the design. Think these designs are calles "synchronous boost converters". So advantage: Less losses, less heat. Disadvantage: probably slightly higher cost and more complexity. \$\endgroup\$ – Remco Vink Oct 1 '18 at 11:36
  • \$\begingroup\$ I think TI has now a smart component that drives such mosfet automatically. Saw it on a seminar. \$\endgroup\$ – Gregory Kornblum Oct 1 '18 at 15:55
  • \$\begingroup\$ LM74670-Q1 i think \$\endgroup\$ – Gregory Kornblum Oct 1 '18 at 15:57
4
\$\begingroup\$

Convert 3v to 5v very high amperage using boost converter

If the input voltage is 3 volts and the inductor is 3 mH and the switching cycle is 15.625 us (inverse of 64 kHz) then the maximum switched current into the inductor at near enough 100% duty cycle is 3 volts x 15.625 us / 3 mH = 15.625 mA.

This clearly and obviously is the DCM (discontinuous conduction mode) case but, it is so far off meeting the requirements that it's obvious that the CCM (continuous conduction mode) case is impractical.

You have a value of inductance that is way too high. You need an inductance that is single digit micro henries.

Edited section to show the futility of a 3 mH inductor

If you did a simulation of the only possible technique to achieve 5 volts from 3 volts with a 3 mH inductor at 64 kHz switching (continuous conduction mode), you would find that the inductor current rating needs to be at least 6 amps: -

enter image description here

The top picture is the simulation circuit. The middle picture is the transient response showing output voltage (blue) and inductor current (red).

The bottom picture is focused in on the inductor current showing that it changes between 5.652 amps and 5.662 amps. This delivers a net power to the diode, capacitor and load of about 10.82 watts.

Now, if you tried to find a 3 mH, 6 amp inductor, you would be surprised how big and costly it would be. The largest current 3 mH inductor in Farnell was rated at 500 mA and have a DC resistance of about 3 ohms. The inductor you would need for this design requires at least 12 times the current and would cost several tens of dollars.

Of course you could wind your own BUT, the point is that you have chosen the wrong value inductor for this application. Choose something around 10 uH.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed♦ Oct 1 '18 at 11:43
  • 1
    \$\begingroup\$ Supposing the inductor is very big, but wouldn't work anyway in CCM mode? \$\endgroup\$ – Marko Buršič Oct 1 '18 at 11:56
  • \$\begingroup\$ @MarkoBuršič I'm unsure what you are asking or saying. It would work in CCM if the load resistance allowed it to get to a big enough current. \$\endgroup\$ – Andy aka Oct 1 '18 at 12:16
  • 2
    \$\begingroup\$ I'll go bring some popcorn. \$\endgroup\$ – Gregory Kornblum Oct 1 '18 at 12:17
  • \$\begingroup\$ I don't fully understand about the 15.6mA constraint. But I understand that large inductor has high ESR. But theoretically speaking, supposing the ESR=0, would this boost converter work or not? And would the one cylcle constraint (15.6mA) still matter or not? \$\endgroup\$ – Marko Buršič Oct 1 '18 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.