1
\$\begingroup\$

I don't know why but i'm not getting 5v on the input of the Arduino, the only thing I've done is to add a switch in order to interrupt the circuit. Here it's my schematic: enter image description here

I have another issue with the battery charger, I don't know why but in order to output tension it needs to be plugged to the main power supply (220v AC) so that it leaves the "stand by" mode. Is there any way to bypass this? Or how can I simulate this behavior without doing so? I'll leave a picture of the controller too.

enter image description here enter image description here

\$\endgroup\$
3
  • 1
    \$\begingroup\$ That battery / board combination looks a lot more complicated than the battery symbol on your schematic. Why are you cutting negative? The photos aren't great at explaining what you've done. \$\endgroup\$
    – Transistor
    Oct 18, 2017 at 11:37
  • \$\begingroup\$ @Transistor isn't it the same cutting negative or positive? As soon as the circuit is interrupted there should be no problem. Or am I wrong? I used the usb A output/gnd pins and that's what it's being represented on the schematic. \$\endgroup\$
    – Fjallbacka
    Oct 18, 2017 at 11:41
  • 1
    \$\begingroup\$ The shield may also be grounded. You need a more detailed schematic to be able to understand what's happening. \$\endgroup\$
    – Transistor
    Oct 18, 2017 at 12:10

2 Answers 2

Reset to default
2
\$\begingroup\$

The big component marked 2R2 is an inductor. That's a tell tale sign of a boost (or buck) converter. This allows the power bank to output higher voltage (or more current in case of a buckconverter) than the batteries are able to provide.

Boost converters work by shorting the inductor to ground and then using the stored energy to boost the voltage.

Most power banks using such a boost circuit will have a sensing circuit to turn it off if there is no load.

The simplest method of keeping the circuit enabled is by adding more load to increase the current.

\$\endgroup\$
2
  • \$\begingroup\$ Oh I see, I'll try adding a 10k ohm resistor to VIN of the arduino that might do the trick I hope. \$\endgroup\$
    – Fjallbacka
    Oct 18, 2017 at 11:48
  • \$\begingroup\$ apparently there's something going on inside the switch... If I measure tension after the switch I get 5v, but If I do it before it goes down to 1v... \$\endgroup\$
    – Fjallbacka
    Oct 18, 2017 at 12:15
2
\$\begingroup\$

Nano operates at 5v, but its required input voltage is 7-12v, because there is a 5v regulator in the nano board, and this 2v gap is required to make the onboard regulator work.

But if you're feeding only 5v to Nano Vin, the Nano's regulator will not output the 5v required and that's the reason the Nano is not working. If the Nano doesn't leak any current, the power-bank regulator shuts down its output.

You could either:

  • adjust your regulator to feed 7v to Nano's Vin; or
  • feed the 5v from your power bank directly to the Nano 5V pin (this second option bypass the onboard regulator).

If you power-bank regulator has only an USB output (5v), there's only the second alternative.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Thanks, I don't know why but I worked this around just by trial and error as suggested by ratchet freak I added a 100ohm load to the switch and now it is working. Ps. I'm still using the VIN pin and my power bank generates barely 5.02V \$\endgroup\$
    – Fjallbacka
    Oct 19, 2017 at 11:14
  • 1
    \$\begingroup\$ It´s a curious thing! The Arduino Nano's website in "tech specs" part, says that: "Operating Voltage 5 V", and "Input Voltage 7-12 V". But the "documentation part" says that "The Arduino Nano can be powered via the Mini-B USB connection, 6-20V unregulated external power supply (pin 30), or 5V regulated external power supply (pin 27). The power source is automatically selected to the highest voltage source." So, there should be three options: a) USB; b) pin 30, 6-20v; c) pin27, 5v. But if it is working now, good, and do this way! ;-) \$\endgroup\$
    – mguima
    Oct 19, 2017 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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