Hey guys i need to run my arduino for +- 1/2 second, the current draw of the arduino and sensors are about 150mah.
So i need to work out what capacitance i need to accomplish this.
The supply voltage at the moment is 5v but I will change that later to 12v.
So if someone could please explain to me how to work it out that would be great.
Thanks for the help
M......
I think you have to figure out the minimum voltage your circuit requires. Are you using the built-in linear regulator? An external boost-buck regulator is more efficient but more expensive.
Do you mean that you want it to continue working for 1/2 S after the power goes down?
If so you can work out the capacitance needed from the equation I = C dV/dt
Let's say that from 12 V you can let the voltage drop to 7 V before the regulator.
Then dV = 5 V, dT = 0.5 S, and i = 0.150 A
So C = 0.15 * 0.5 / 5 = 0.015 F or 15,000 uF
Russell.
Current drawn might be 150mA, it won't be 150mAh as that is a measure of charge.
Assuming your capacitor is charged to 12V and feeds a 5V regulator with 1.5V
drop out. That means there will be adequate supply as the capacitor discharges
from 12V to 6.5V.
Charge = capacitance x voltage, so that's 5.5C coulombs where C is the capacitance
in farads.
150mA for 0.5seconds is 75 milli-coulombs, so we want
5.5C >= 0.075, or C >= 0.014F, ie 15,000uF would be the next standard value.
However electrolytic caps are usually +100/-50% rated, so 33,000uF (16V) would be
right value to choose - quite large in fact.
Don't even think about trying to hold the 5V steady directly with a capacitor bank
unless you use supercapacitors.
Thanks for the explanations guys.
Do you mean that you want it to continue working for 1/2 S after the power goes down?
Yes exactly thanks Russell
Assuming your capacitor is charged to 12V and feeds a 5V regulator with 1.5V
drop out.
Well yes when I change my circuit it will charge to 12v and feed my arduinos regulator.
Does the arduino regulator have such a high drop out?
However electrolytic caps are usually +100/-50% rated, so 33,000uF (16V) would be
right value to choose - quite large in fact.
That is big and expensive even the 15000uF cap is very expensive here, I was hoping to get away with 5000uF, so if I have a 15000uF it should run the arduino for 0.25seconds?
Could I have 15 * 1000uF caps in parallel, would that work?
Im running my arduino and sensors off my motor power supply so the problem is when these motors kick in they draw the max current of 33amp that the battery can supply then my arduino resets.
Edit: Ok I found a 10000uF 25v cap which is not expensive if I connect 2 of them up parallel that should work right?
Could I have 15 * 1000uF caps in parallel, would that work?
Yes
if I connect 2 of them up parallel that should work right?
It will supply the current but:-
so the problem is when these motors kick in they draw the max current of 33amp that the battery can supply then my arduino resets.
You have to track down that problem first. It could be caused by a drop of voltage on the power supply or it could be the interference caused by your motors switching on.
You have to track down that problem first. It could be caused by a drop of voltage on the power supply or it could be the interference caused by your motors switching on.
Hmmm I think it could be a voltage drop because when I power my arduino with my laptop via usb then it does not reset and everything runs fine.
The way I know or think it resets is I have a counter(++) running in my code and serial printing the counter, when on motor power supply the counter will just stop counting when motors kick in and start up again randomly but when I power my arduino with my laptop(I disconnect the vout wire) that does not happen the counter runs smoothly even when motors kick in,
But let me explain ive got a Pololu - 4.a. Assembly for Use as a General-Purpose Motor Driver motor driver as a stand alone driver which has a pinout(vout) to power my arduino with reverse voltage protection so I was thinking of adding the caps between vout and arduino vin unless you say otherwise.
Thanks
I was thinking of adding the caps between vout and arduino vin unless you say otherwise.
That is fine but it might not be enough. Look at the Pi circuit the last one on this page:-
http://www.thebox.myzen.co.uk/Tutorial/De-coupling.html
However you supply might not be capable of supplying the current that is needed.
Hi,
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?
It will help us work out how to solve your problem.
I understand that all you want to do is keep the arduino going during a power supply dip due to a with high start current.
A diode and an electro should fix it, unless you have some physical wiring problems.
A picture of your project would help also if possible.
Tom.... 