1
\$\begingroup\$

I have a power managent IC that has features of Under and Over Voltage Protection through the EN and OVP pins. These pins are enabled by manipulating the voltages on the pin via a 3-resistor voltage divider.

Now my circuit takes in two configurations for the Li-ion battery for power, either at 3.7 volts (using 1 18650) or at 7.4 volts (using 2 18350). This makes it tricky for the power management IC. It is possible to use two of the PMIC for the different configuration making sure that only one is always active. But it take a lot more components and not much more GPIO pins for monitoring.

I was wondering if there is any way to switch between two different voltage dividers depending on the input voltage. Without using timers, I do not want to keep adjusting it.

enter image description here

The input voltage share the same line, but if the voltage is let's say lower than 5V, it will use the voltage divider on the left and if it's higher than 6V it will use the divider on the right.

\$\endgroup\$
6
  • \$\begingroup\$ It would help greatly if you tell us the model of your PMIC. Knowing more about it give us more options in helping you. \$\endgroup\$ – Edin Fifić Apr 2 '20 at 18:16
  • \$\begingroup\$ @EdinFifić The model of the PMIC im using is the TPS25944 \$\endgroup\$ – Jake quin Apr 2 '20 at 18:26
  • \$\begingroup\$ "The input voltage share the same line, but if the voltage lets say lower than 5v it will use the voltage divider on the left and if its higher than 6v it will use the divider on the right" is that safe? \$\endgroup\$ – Jasen Apr 3 '20 at 0:16
  • \$\begingroup\$ why not connect the 2 18350 in parallel? that seems much safwer, ans easier to handle too. \$\endgroup\$ – Jasen Apr 3 '20 at 10:14
  • \$\begingroup\$ @Jasen the battery holder is for a 18650, but you can actually fit 2 18350 with that holder, using a higher voltage is necessary for my application because there is a buck-boost converter that can only handle a certain input current. Increasing the input voltage means lower the input current. \$\endgroup\$ – Jake quin Apr 3 '20 at 12:57
1
\$\begingroup\$

You could use a 4053 multiplexer with a comparator and voltage reference for selecting. Be sure to carefully analyze this idea to see if it makes sense in all possible situations. The reference has to work from the lowest possible voltage with one cell, so you'll have to divide the battery voltage down.

A better solution might be to add a couple of jumper blocks or a DIP switch to change from 1 to two cells. You could use a single chain of 5 resistors.

\$\endgroup\$
6
  • \$\begingroup\$ I would prefer that i/user would not have to adjust anything to switch between the batteries configurations. the CD405 seems to do eveything in my posted answer, but the 125 ohms ON resistance may pose a problem later on(?) but it does solves the logic voltages \$\endgroup\$ – Jake quin Apr 2 '20 at 18:23
  • \$\begingroup\$ Maximum input leakage is +/-100nA so the resistance is of no practical consequence. \$\endgroup\$ – Spehro Pefhany Apr 2 '20 at 18:54
  • \$\begingroup\$ how about the voltage drop ? so the pin is receiving slightly less than expected. Although that can be mitigated by adjusting the voltage divider to accommodate that resistance \$\endgroup\$ – Jake quin Apr 2 '20 at 19:05
  • \$\begingroup\$ 100nA * 125 ohms is 12.5uV, which is totally insignificant. \$\endgroup\$ – Spehro Pefhany Apr 2 '20 at 19:08
  • \$\begingroup\$ My bad, i did not think that properly. The only thing discouraging me now into using this chip is driving it. Since this circuit is the first circuit receiving power before the mCU ill have to add lots more of components \$\endgroup\$ – Jake quin Apr 2 '20 at 19:21
1
\$\begingroup\$

I had an idea of using a analog switch SPDT IC to switch the pins, I could only find one where it can more than 5V TS12A12511

enter image description here

This circuit is not finished though since the voltage divider for enabling/disabling the switches is proving quire hard, below 5V the IN pin must read <0.8V and above 6V the IN pin must read >5V

EDIT:

I think I found the way to do it, is to use a power supervisor IC:

enter image description here

As to if I actually use this on my circuit I will still asses it.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ You got the levels wrong: IN has to be below 0.8V for logic LOW, and above 2.4V for logic HIGH. Also, you don't need 2 sets of voltage dividers for the IN pin, just connect the middle of one divider to both IN pins since they're both switching at the same time! I do see the difficulty for the switching levels. Give me a little bit of time, and I will find a solution. \$\endgroup\$ – Edin Fifić Apr 2 '20 at 18:40
  • \$\begingroup\$ You are right i should use only 1 voltage divider.The IC has different characteristics at 12v, i do not know if it the 5v high will apply at 12volts input or somewhere inbetween so its best to keep safe. I have edited it to fix the interchanged < and > and edited the schematics too \$\endgroup\$ – Jake quin Apr 2 '20 at 18:43
  • \$\begingroup\$ My idea right now is replacing the voltage divider with a 5v linear regulator as that will give me a 5v if input is 6v+ and 0v if input is lower than 5. currently looking for an IC \$\endgroup\$ – Jake quin Apr 2 '20 at 18:49
  • 2
    \$\begingroup\$ @Jakequin A 5V linear regulator will not do that (give you 0V for < 6V in). \$\endgroup\$ – Spehro Pefhany Apr 2 '20 at 18:51
  • \$\begingroup\$ @SpehroPefhany i think you are right, i thought the ones with Undervoltage Protection does this, it seems not \$\endgroup\$ – Jake quin Apr 2 '20 at 19:02
0
\$\begingroup\$

maybe you can get the ratios you need by only adjusting the upper resistor in the divider.

schematic

simulate this circuit – Schematic created using CircuitLab

here the switch could be facilitated by a shorting link on the battery connector, or a switch in the battery compartment

\$\endgroup\$
1
  • \$\begingroup\$ Thank you for idea, but i would really like it that the user doesnt have to do anything. It should toggle passively. \$\endgroup\$ – Jake quin Apr 3 '20 at 6:56

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.