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I want to charge a battery my power supply is 6V , I Need to add a battery to my circuit and need to charge it for a long time. should i add a switch in series or in parallel to the diode? this is my schematic : dc/dc converter

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    \$\begingroup\$ That's only part of a schematic. And a very confusing part at that. Is the 7.4V component supposed to be your battery? And the 6V thing is your power supply? What do you think will happen when you connect a 7.4V battery to a 6V power supply? Please explain more clearly what the circuit is supposed to do and where you plan to connect you battery. Also provide a more useful schematic. \$\endgroup\$
    – StarCat
    Commented Sep 7, 2020 at 15:45
  • \$\begingroup\$ @StarCat in parallel to the power supply(6V) i connect the battery (7.4V). \$\endgroup\$
    – hos
    Commented Sep 7, 2020 at 15:50
  • \$\begingroup\$ The voltage of the charging source must be higher than the battery voltage in order to charge the battery - but you can't simply connect a charging source and battery in parallel without some means of controlling the charge current. \$\endgroup\$ Commented Sep 7, 2020 at 17:51
  • \$\begingroup\$ hos: We need to know 1: What is the battery chemistry? (LiIon/LiPo, NiMH, ...) Q2 What is the 6V power source? (Another battery. a mains power supply, ...) Q3: You say " ... for a long time ." -> What does that mean? Do you mean it must be able to be left connected without damage, or you want the charge rate to be slow, or ...? You need to provide more detail for us to be able to answer this question well. \$\endgroup\$
    – Russell McMahon
    Commented Sep 10, 2020 at 0:38

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To charge a battery the charging voltage MUST be at least as high as the maximum voltage that you wish the battery to be charged to.
In this case, if the "7.4 V" battery consists of 2 x LiIon or LiPo cells then the minimum voltage for normal charging is 2 x 4.2V = 8.4V as a LiIon or LiPo cell is usually charged to a maximum of 4.2V/cell.

Connecting a 6V supply to a 2 cell LiIon battery will discharge the battery to 3V/cell which is about fully discharged. Adding a diode as shown will discharge the battery to ABOUT 6.6V and not charge it at all. Reversing the diode polarity will not charge the battery as the available voltage is then 6V-Vdiode = 6V-0.6V =5.4V = 2.7V/cell.
This is at the very bottom end of the cells' discharge range.

IF you wish to charge a 2 cell LiIon or LiPo battery you must have at least 8V, and usually more like 9V available AND MUST have a proper charge control circuit.

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