0
\$\begingroup\$

I am designing a small bidirectional circuit consisting of two batteries in series as a power source. Part of the circuit will be a charger, to charge the batteries when they are not in use, which is triggered by external circuitry.

The problem I am having is that each battery has its own charging circuit, therefore I think that the must be disconnected from each other in order to facilitate charging.

I had the idea of using a MOSFET which joins the positive of one battery to the negative of the other battery when the circuit is discharging, and then disconnect the two when it is charging. But I am having difficulty implementing this. I have tried simple simulation with PSPICE using only voltage sources however I could not get it to work. Is this possible? If so how can I do it?

\$\endgroup\$
  • 1
    \$\begingroup\$ Please post the schematic you have simulated. \$\endgroup\$ – Dmitry Grigoryev Sep 6 '16 at 14:13
  • 1
    \$\begingroup\$ What MOSFET do you use? How do you drive it? \$\endgroup\$ – Edesign Sep 6 '16 at 14:52
  • 1
    \$\begingroup\$ If the two chargers are isolated from each other, you shouldn't need to disconnect the batteries from each other. I have done this with two 12V off-the-shelf chargers to charge automotive batteries in series. \$\endgroup\$ – Tut Sep 6 '16 at 16:08
  • 1
    \$\begingroup\$ Even if you used a MOSFET to break the series connection, your two battery chargers need to be electrically isolated. That is, they cannot share a common ground. Usually, it is just easier to design a charger that can charge the two in series. This can have the advantage of allowing both charging and use of the system at the same time. \$\endgroup\$ – Vince Patron Sep 6 '16 at 19:33
1
\$\begingroup\$

While thinking that could be used relays the desired operation, I provided an outline of a possible circuit using MOSFETs. Be especially careful that the control signals must be complementary. The intrinsic diode of M1 does not impose a problem for charger 2, since M3 is conducting in this case. I leave to you the determination of resistors and proper part-numbers.

Series battery -connect/disconnect

\$\endgroup\$

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.