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Flip Flop Charging - Can this be done?

Discussion in 'General Electronics Discussion' started by Mike Bookbinder, Aug 21, 2016.

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  1. Mike Bookbinder

    Mike Bookbinder

    2
    0
    Aug 21, 2016
    Hello all,

    I'm new to electronics and I am trying to learn as I go. What I am trying to do is see if it is possible to charge one supercap while another super cap is running a load. When the voltage in the driving cap drops to a certain level, swap caps and have the cycle repeat. I'm experimenting with recovering some energy in this case back EMF to see how long I can keep a pulse motor running. I'm interested in energy recovery ideas and not perpetual motion or over-unity. Below is a basic rough design of how this might work (or not).

    Values are not calculated - Just guessing on my part

    • Q1 is on and C2 is charging
    • Q2 is off and energy in C1 is under load
    • SW1 (reed switch) is CLOSED and pulses a motor
    • SW1 is OPEN and back EMF is sent back to Q1 and Q2 and flows only through the transistor that is on, in this case it would be Q1 that is now charging C2.
    • Once C2 reaches maximum charge, it turns on Q1 (flip flops) and now begins to charge C1
    • The cycle repeats.
    • L2 supplies current as magnets pass the air coils when the motor is spinning
    • L1 is the drive (PULSE) coil and back EMF is routed back to Q1,Q2




    flopcharger2.png
     
  2. AnalogKid

    AnalogKid

    2,515
    718
    Jun 10, 2015
    Bullet items 1, 2, 4, 5, 7, and 8 are incorrect, have serious errors, or do not describe the components in the schematic. For example:

    1. Q1 is on and C2 is charging - for a *very* long time. Even with a small supercap such as 0.1 F, the time constant is 2.78 hours

    2. Q2 is off and energy in C1 is under load - Not sure what you mean by "under load", but C1 is not powering anything because there is not a complete circuit even with SW1 closed. In terms of conventional current flow, current can't leave the + end of C2 because it can't go backwards through Q1 (it is NPN) and Q2 is off. Also, there is no path for the positive voltage at Q1's emitter (if there were any) to get to the right side of L1 (marked GND); D1 is pointed the wrong way.

    7. L2 supplies current as magnets pass the air coils when the motor is spinning - but only for milliseconds, not long enough to do anything.

    In order for L1 back EMF to do anything through LED1 and R4, it must be very large and positive going. That means that when SW1 is closed the left side of L1 must be very negative. Where does that voltage and current come from? Remember, 100% of the back EMF energy comes from the current in the inductor at the time the switch opens.

    ak
     
  3. Mike Bookbinder

    Mike Bookbinder

    2
    0
    Aug 21, 2016
    Thanks for for your time, I really value your feedback!

    A few things to mention... "C1 under load" I was referring to its discharge as the coil is pulsed.

    The transistor types are not NPN they are PNP the arrow is pointing in to the base therefor it's a PNP type.

    The L2 inductor is 3 in a series. 3 coils wound with 30 guage wire, more can be added. Each coil has the magnet pass 4 times per revolution per coil.

    Yes D1 is backwards the line is negative (CATHODE) side and have it to prevent back emf from going through L2.

    Not sure how you calculated a time constant of 2+ hrs for the caps without any values provided. I can charge a 10F 2.7v supercap to 1.3v in 20 seconds using one 1.3v AA battery.

    Anyway, I appreciate your help, I didnt think something like this would work the way I need it to work.
     
    Last edited: Aug 22, 2016
  4. AnalogKid

    AnalogKid

    2,515
    718
    Jun 10, 2015
    You are correct about the transistor designation; NPN was a typo. However, current still cannot go through it backwards.

    1. Q1 is on and C2 is charging - C2 is charging though a 100K resistor. As I said, assuming a 0.1F cap (since there is no value on the schematic, I chose the smallest one I've seen), this is a huge time constant. With a 10 F cap, it is 100 times huger.
     
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