High DC Pulse Current: Get Advice Now!

In summary, 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. 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
  • #1
Majidshadow
1
0
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?
 
Engineering news on Phys.org
  • #2
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.
 
  • Like
Likes davenn
  • #3
Svein said:
I do not know if the 3.6V battery can charge that capacitor in 1s.

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:
  • #4
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:
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/2v_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 don't 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
 
  • #5
that model also assumes ideal switching
 
  • #6
Majidshadow said:
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?
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.
 
  • #7
Sounds like an igniter
 
  • #8
meBigGuy said:
Sounds like an igniter
Well that sounds ominous... But why the repetition frequency of 1Hz?
 

FAQ: High DC Pulse Current: Get Advice Now!

What is high DC pulse current?

High DC pulse current refers to a type of electric current that flows in one direction and has a high amplitude, or strength, compared to other types of direct current (DC) electricity. It is typically used in specialized applications, such as in industrial processes or for scientific experiments.

How is high DC pulse current generated?

High DC pulse current can be generated in various ways, but it is most commonly produced using power electronic devices such as thyristors or MOSFETs. These devices are able to switch the current on and off rapidly, creating high voltage pulses that can be used for specific purposes.

What are the potential hazards of working with high DC pulse current?

Working with high DC pulse current can be dangerous, as it can cause severe electrical shock and burns. It is important to follow proper safety measures when working with this type of current, such as wearing appropriate protective gear and closely following instructions from experienced professionals.

What are the applications of high DC pulse current?

High DC pulse current has a variety of applications in different industries, including materials processing, welding, and electroplating. It is also used in scientific research, particularly in the field of high energy physics, for experiments such as particle acceleration and fusion.

Can you provide advice for working with high DC pulse current?

If you are planning to work with high DC pulse current, it is important to first gain a solid understanding of the principles and safety measures involved. Seek guidance from experienced professionals and always follow proper protocols to ensure your safety and the success of your project.

Similar threads

Back
Top