0
\$\begingroup\$

I recently built step up converter which work quite good. Now I would like to build a circuit which would enable this step up converter when battery voltage drops below 1.8V (Arduino BOD).

  • I need to keep in mind power consumption because of battery power

Now, I do not know what is the best method to do that.

I guess that couple of transistors would do the job but I do not know where to start. Also I dont know if it is optimal solution bacouse of battery power.

The EN pin is a logic-level input used to enable or disable device switching and lower quiescent current while disabled. A logic high (>70%of VIN) will enable the regulator output. A logic low (<20% of VIN) will ensure that the regulator is disabled.

So if I understand it correctly, I need "circuit" that will "bring" VIN on EN pin when VIN < 1.8 V.

Please give me some direction to start with this. Thank you.

EDIT 2017-03-07 I went search in direction mentioned in comments and found this video. Which I guess is good start.

p.s. Iam using MCP1625 which has enable pin.

\$\endgroup\$
4
  • \$\begingroup\$ Basically its factory recommended design of MCP1625 discussed here electronics.stackexchange.com/questions/236184/…. \$\endgroup\$
    – MightyPolo
    Mar 6, 2017 at 14:23
  • \$\begingroup\$ The solution will include a reference, a divisor, and some type of comparator to drive the boost enable. If cost is not a concern, you can find all of this in a single IC which is targeted at this sort of application. \$\endgroup\$
    – M D
    Mar 6, 2017 at 14:24
  • \$\begingroup\$ How many cells in series are you using for this? 1 alkaline? \$\endgroup\$
    – Dejvid_no1
    Mar 6, 2017 at 14:58
  • \$\begingroup\$ @Dejvid_no1 right now in my prototype I went for CR123A because of its size. Alternatively I would use two AA bateries. \$\endgroup\$
    – MightyPolo
    Mar 7, 2017 at 9:21

2 Answers 2

0
\$\begingroup\$

I'm guessing you won't be able to squeese out so much energy of the battery when it is almost empty.

Anyhow you could either use a low voltage comparator with a reference, like this:

enter image description here

Change the resistors accordingly for your application. You might want to invert the output with an NPN transistor and a pullup.

Or you could build something with transistors. I don't know if this actually works in the "real" world but it works in my LTSpice simulation:

enter image description here

Q1 holds EN low until VBAT reaches approximately 1.8 V when it releases EN to VBAT. The trip point can be adjusted by changing R5 and R4. R7 provides a bit of hysteresis around the trip point. I'm mediocre at designing transistor circuits so I would definitively go with the IC version.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you sir, this is what I was looking for. I use 3V battery and BOD is around 1.8 V. That is 60% of battery would be vasted. Transistor circuit is impressive, but I will go with IC. Thank you. \$\endgroup\$
    – MightyPolo
    Mar 7, 2017 at 14:12
0
\$\begingroup\$

Could you not just use a comparator circuit to measure when the battery voltage drops below 1.8V, then have this toggle the EN pin?

\$\endgroup\$
2
  • \$\begingroup\$ But how do I get reference for OA when voltage is changing in time.. Divider would return slightly different values each time. \$\endgroup\$
    – MightyPolo
    Mar 7, 2017 at 11:40
  • \$\begingroup\$ You can use a reference voltage chip, and have the divider set against that, or you can get some type of comparator chip with a built in low reference voltage. Seems the guy above me answered pretty extensively though, and that IC he mentioned will do the trick nicely! \$\endgroup\$
    – MCG
    Mar 8, 2017 at 8:35

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.