I shorted the R31, will this still work? what does the r31 do anyways? what about replacing with a 220ohm resistor?!

I shorted the R31, will this still work? what does the r31 do anyways? what about replacing with a 220ohm resistor?!

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  1. 2 months ago
    Anonymous

    No, not as intended, could though.
    The r31 performs the function of being an r31, which makes r31 = r31.
    Replacing an r31 with a resistor could have benefits and drawbacks, analgous to other things in life so don't think and just frickin send it, BE STRONG, use 1 ohm. Real advice: Getting started in Electronics, Forest Mims. Check voltages with your meter, solve ohm's law and try to compare measurements...make guesses, and think of reasons, you may figure it out if you haven't lost the ability to use a pen and stare at the wall thinking until someone, invariably no less, distracts you.

    • 2 months ago
      Anonymous

      >in my mind I'm a very clever guy

      • 2 months ago
        Anonymous

        What the frick does clever mean, are you implying I'm an idiot. Buttpirate...

  2. 2 months ago
    Anonymous

    Looks like R31 and R8 do the same thing: they set the base current which determines the current through Q4 which is the charge current. I think you can safely remove either R31 or R8. When Q9 is open, the voltage drop on it is close to zero so you essentially have R31 and R8 in series. What's the point?

    • 2 months ago
      Anonymous

      The point might be to avoid overcurrent on Q4. Depends on the specifications for whatever transistors OP is using.

      You can probably safely replace the resistor with a 220R but charging will be slower probably.

  3. 2 months ago
    Anonymous

    >but charging will be slower probably

    nonsense.
    both Q4 and Q9 act as switches, to turn on/off charging current to the battery
    they dont vary the amount of current in any way
    why do you need 2 switches instead of 1?
    Q9 is a voltage/current amplifier coz the BBIC box cant handle that much power.
    as for R31, it's a traditional-design safety thing, it can be eliminated but it's not good practice.

    • 2 months ago
      Anonymous

      >both Q4 and Q9 act as switches
      How do you know both transistors will saturate? Genuine question transistor circuits aren't my favorite.

      • 2 months ago
        Anonymous

        >both transistors will saturate?

        you can look at the base resistors and intuit the answer
        or you can simulate it to confirm both are saturated

        • 2 months ago
          Anonymous

          Ah I see - I made a mistake in some working out I scribbled down and got Q4 as being forward active. Oh well! Thanks for the help anon.

        • 2 months ago
          Anonymous

          Who determined the voltages?
          They aren't specified in the OP.
          Also: OP has a 2-cell battery with two + terminals.
          No indication of cell type is given.

          • 2 months ago
            Anonymous

            >No indication of cell type is given
            better not be frickin lithium with a ghetto ass charge controller like that...then again...this is PrepHole...

          • 2 months ago
            Anonymous

            >Who determined the voltages?

            common sense and his albino brother, experience

        • 2 months ago
          Anonymous

          Why is R4 100 ohm? Of course Q2 will saturate since most voltage will be dropped on R4. But it is not the way to simulate the charge. You should replace R4 with a V source with a low internal resistance of like 0.1 ohm and you will see that Q2 will not saturate. In fact it will have to dissipate a silly amount of power. It has to be a beefy transistor. Q1 will most certainly saturate. Also I am not sure why you powered Q1 from a separate power supply.

          • 2 months ago
            Anonymous

            Never mind Q1 is powered separately since it is the "control line".

          • 2 months ago
            Anonymous

            To be precise, Q2 should be able to eat the difference in voltages between the charger and the battery. So if the battery is almost fully charged, it won't matter much. But it is fully discharged, it would have to dissipate like 8V times the charge current.

    • 2 months ago
      Anonymous

      Post the IC part name/number.

      That implies CHG+/- is a constant current source, which I can accept, but if they were just switching Q4 on hard then why would they:
      >make Q4 a PNP instead of a P-ch MOSFET
      >add R8 to limit the emitter current of Q9
      The existence of R8 especially lends me to believe that they're eliminating the variability of Q9's hFE and relying on Q4's hFE being sufficiently constant to provide a regulated charging current. Assuming IC1 outputs a 5V level to Q9, then that's ~4.3/100 = 43mA of current being pulled out of Q4's base. If Q4 is something like a TIP42, the hFE will be no more than 75, which suggests a charging current of 3.2A, which is quite believable if it's a moderately sized lipo.

      There's also voltage feedback, but no current feedback. I imagine it's cheaper to have the user set the desired charging current by just changing resistor values than to have a current sense amplifier.

      If this is the case, shorting out R31 will significantly increase the battery charging current, replacing it with a 220Ω resistor will significantly decrease it. Using two 220Ω resistors in parallel might be satisfactory.

  4. 2 months ago
    Anonymous

    >I shorted the R31
    why?

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