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High dc pulse current

  1. Nov 18, 2015 #1
    hi dears
    i need your advice. i wand to produce high pulse dc current in below described circuit:
    - power source is lithium battery 3.6 v (limited output current, maybe 1A)
    - load is resistor 0.01 ohm
    - current by value of around 50 A during 0.001 second need to be passed from the load.
    - frequency of repeating this is 1 second.

    it would be highly appreciated if you can help me how can i produce this?
     
  2. jcsd
  3. Nov 18, 2015 #2

    Svein

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    Science Advisor

    You need a "Flying capacitor" circuit - load the capacitor to 3,6V and then connect the capacitor to the load. The trouble is that you need a very large capacitor with a very low equivalent resistance - my first approximation gives a value for C = 100 000μF and I do not know if the 3.6V battery can charge that capacitor in 1s.
     
  4. Nov 18, 2015 #3

    davenn

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    Gold Member

    I was thinking the same ... as I considered how long it takes for a camera flash to charge up

    EDIT---- in fact, stripping the electronics out of an old camera flash unit and experimenting with that
    may be the best way to go for a start. At least it's going to give you some indications of circuit requirements

    Dave
     
    Last edited: Nov 18, 2015
  5. Nov 18, 2015 #4

    donpacino

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    Gold Member

    some notes:


    to get 50A on that 0.01 ohm load you will need a voltage of 0.5 V

    I made a quick and dirty matlab program (seen below) that calculates the charge time of the capacitor in seconds. i did it in 15 mins so forgive me if it seems sloppy
    Then it assumes the capacitor is charged, computes the dissipation of the capacitor when it gets hooked up to the load. You can see what the load current will be for the 1 ms period.

    It then compensates for the voltage sag in the capacitor and automatically boosts the voltage by a small amount such that the average current is 50A.

    it will spit out
    v_charge_ideal: the ideal cap voltage. you can change this if your load or desired current changes
    v_charge_new: the voltage you should charge the capacitor to
    i_ripple_final: the amount the current will decrease of th

    some output voltage and current graphs

    you can tweak the inputs to change the values. based on your current ripple requirements you might have to tweak some values. the larger you make your capacitor, the smaller the ripple. But the longer it will take to charge.
    c: capacitance
    rcap: capacitance esr
    rbatt: battery esr

    Code (Text):
    clc,clear,close all

    %% setup variables
    t=linspace(0,0.01,1000);

    r_load=0.01;
    vbatt=3.6;
    rbatt=0.1
    rcap=0.1
    pulse_time=0.001;
    c=1

    %% initial setup
    charge_time=5*(rbatt+rcap)*c

    v_charge_ideal=r_load*50
    [null index] = min(abs(t-pulse_time));

    %% first pass
    v_load=v_charge_ideal*exp(-t/(r_load*c));


    i_load=v_load/r_load;
    % figure()
    % plot(t,v_load)
    % hold on
    % area(t,v_load.*[ones(1,index) zeros(1,length(t)-index)])
    % figure()
    % plot(t,i_load)
    %
    % hold on
    % area(t,i_load.*[ones(1,index) zeros(1,length(t)-index)])


    i_ripple = i_load(1)-i_load(index);

    %% compensate for voltage drop
    v_charge_new=v_charge_ideal+i_ripple*r_load/2


    v_load_new=v_charge_new*exp(-t/(r_load*c));


    i_load_new=v_load_new/r_load;

    figure()
    plot(t,v_load_new)
    hold on
    title('load voltage')
    area(t,v_load_new.*[ones(1,index) zeros(1,length(t)-index)])
    figure()

    plot(t,i_load_new)
    title('load current')
    hold on
    area(t,i_load_new.*[ones(1,index) zeros(1,length(t)-index)])


    i_ripple_final = i_load_new(1)-i_load_new(index)


    rms_current=rms(i_load_new(1:index))
     
    a capacitance of 0.5F looks good
    i dont think 0.1F will work. too much current sag for my tastes. it might be fine depending on your requirements.

    __________________________________
    some further notes....

    esr of lithium batteries typically will be 0.1 ohms, maybe as high as 0.2 ohms
    the capacitor esr might be anywhere between 0.05 and 0.4 depending on the capacitor type and size

    the time to charge the cap will be 5 time constants of the RC filter

    charge_time=5*R_battery*R_cap*C
    as long as that time is below 1 second you are good

    but that assumes an unlimited supply current
    peak supply current will be battery_voltage/(R_battery+R_cap), which is much higher than your rating of 1A.

    if it is an actual current limited battery, then the charging will be very nonlinear and i doubt it will charge in 1 second. it might be close
    __________________________________________________________

    but we need to change our charging model, because the capacitor will be charged to battery voltage. this will slow down the charging significantly, especially if the battery is current limited

    I don't think it is possible to meet the requirements you are asking without
    1. getting a battery with larger current rating
    1.A this will allow you to possible use active circuitry and drive the capacitor harder to charge (boost the voltage, which will increase charging speed, then regulate the output and cut it off when the voltage gets to an acceptable amount (our 0.5V))
    2. change the load current from 50A to 350 amps, which will allow you to charge to capacitor to the battery voltage. This still will be marginal in terms of charge time
     
  6. Nov 18, 2015 #5

    donpacino

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    Gold Member

    that model also assumes ideal switching
     
  7. Nov 18, 2015 #6

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Can you say what the application is? As you can see from the replies, this may not be possible given the constraints you have listed.
     
  8. Nov 19, 2015 #7

    meBigGuy

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    Gold Member

    Sounds like an igniter
     
  9. Nov 19, 2015 #8

    berkeman

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    Staff: Mentor

    Well that sounds ominous... But why the repetition frequency of 1Hz?
     
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