0
\$\begingroup\$

I have a simple DC circuit (Arduino related) with a USB power supply that's giving me 4.9V.

Is there a way, with minimal components and without using an inductor, to boost that voltage to 5.2V? I have a sensor that starts being stable right around 5.1v.

I'm looking for a current of around 300mA.

\$\endgroup\$
  • 5
    \$\begingroup\$ Yes there is. I'm thinking of a chargepump like circuit. If we could get a square wave signal out of the Arduino it would not require many components. Some Arduino output pins can generate a 400 Hz PWM signal, that would do the trick. But you haven't told us how much current you need at 5.2 V. You should explain why you need that 5.2 V. That is, if you want a usable answer. \$\endgroup\$ – Bimpelrekkie Feb 13 at 7:54
  • \$\begingroup\$ Charge pump will do the job without an inductor. But why can't you use an inductor? What current are you after at 5.2V? What ripple can you have on the 5.2V? \$\endgroup\$ – Puffafish Feb 13 at 8:59
  • 2
    \$\begingroup\$ X/Y Problem. \$\endgroup\$ – Rev1.0 Feb 13 at 13:19
  • \$\begingroup\$ I've added some additional information. \$\endgroup\$ – user1142433 Feb 13 at 15:24
  • \$\begingroup\$ There are many reasons to want to boost voltage a little bit. An so many cause for voltage drops... \$\endgroup\$ – Fredled Feb 13 at 17:13
1
\$\begingroup\$

300 mA is a lot to ask for in a "minimal components" charge pump circuit. The driver needs a complimentary pair of transistors rated for at least 1 A. At 1 kHz, the pump capacitors would be over 300 uF each. The output would need a linear regulator rated for at least 0.5 A, dissipating over 1 W. Overall, this seems like a cumbersome workaround to an inductor.

\$\endgroup\$
  • \$\begingroup\$ "300 mA is a lot to ask." One time I asked the same question but for 10A. My question was closed after two days. I think i was asking too much. \$\endgroup\$ – Fredled Feb 13 at 17:16
  • \$\begingroup\$ I'm not saying it can't be done, but you've really got to hate inductors. Capacitors are much less efficient storers of energy. \$\endgroup\$ – AnalogKid Feb 13 at 18:34

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.