capacitor bank for surge currents - Page 1
Forum section: electronics / general

author: Simon
author_rank: Unknown
timestamp: October 06, 2011, 05:53:13 pm
content: I want to measure the startup current of some fans (or DC motors in general). noe to get an accurate reading and actually measure the current I guess i need to be able to supply it (the surge current). So of course not having a power supply of up to 100 amps I think my best solution is to use a capacitor "bank" on the output. So my first question:How do I calculate the capacitance to provide a certain amount of amps for a certain amount of time ?Am i assuming correctly that I'll need like 0 ESR caps which could be very exspensive and that a better solution is lots of smaller capacitors of say 470-2000 uF in parallel, I mean I can get caps of a few hundred uF for pennies and the cost per uF is less than the larger caps and i get lower ESR in the fist place.I'll also need to look carefully at charging them, I guess some sort of soft start circuit will be required

author: jahonen
author_rank: Super Contributor
timestamp: October 06, 2011, 06:00:55 pm
content: Note also that inductance might also be a factor if you want very fast current rise time. That is however easily minimized by putting the capacitor "bus bars" on top of each other, for example as thin as possible two-sided pcb where other side is dedicated to positive and another side to negative rail. This yields to very low inductance.Instantaneous current is only limited by the ESR. Even small capacitors can deliver easily 100 A for some microseconds or so. However, you can't take any current from a capacitor bank without voltage drop, so you must decide how much voltage drop for given ampere-seconds you can tolerate, and work out the capacitance from there. Pulse current in amperes times pulse duration in seconds divided by the allowable voltage drop in volts gives the needed capacitance.Regards,Janne.

author: Simon
author_rank: Unknown
timestamp: October 06, 2011, 06:06:06 pm
content: I'd probably use some thick wires with the capacitors soldered to them or some very wide PCB tracks, the idea of many capacitors is that I don't have a single point carrying huge currents except for the main bus and output, I am after all simulating automotive startups so haing ideal conditions is not neccesary in a vehicle the fan will have some feet of wire from the battery and it won't be huge wire so anything better than that is fine, I can get 470 uF caps for like 17 pence if I buy 50 so 50 x 470 = 23'000 uF with am ESR of 0.008 ohms

author: IanB
author_rank: Super Contributor
timestamp: October 06, 2011, 06:12:34 pm
content: You might be coming at this from the wrong direction? I think that large start-up currents are generally an undesirable consequence of switching on a motor. Rather than accommodating the surge with fat wires and low resistance it would be better to limit the surge with appropriate current limiting resistance in the circuit.Which is another way of saying you should measure the start-up current under typical installed conditions with normal wiring and power supply, rather than under artificial conditions with huge capacitor banks. You don'twantto maximize the surge, it does nobody any good.

author: Simon
author_rank: Unknown
timestamp: October 06, 2011, 06:14:52 pm
content: ok I'l start from the begining, i want to make a soft start circuit for the fans under test, so i want to demonstrate that there is a problem in the first place and that it is cured. my power supply is 25A and if I'm expecting currents of up to 100A on a fan that usually runs at 12A I need that temporary capacity in order to actually see what is going on.

author: IanB
author_rank: Super Contributor
timestamp: October 06, 2011, 06:39:32 pm
content: Can you get hold of a 12 V car battery to test with? That basically is going to act like a power supply with close to zero internal resistance and unlimited current.

author: Simon
author_rank: Unknown
timestamp: October 06, 2011, 06:40:35 pm
content: yes I could, I suppose that would do it.

author: jimmc
author_rank: Frequent Contributor
timestamp: October 06, 2011, 07:57:34 pm
content: For a shunt wound or permanent magnet motor, the initial surge current is limited by the resistance of the armature,the current will only reduce as the motor begins to turn and generate a back EMF.Neglecting losses, once the motor is running at full speed (without a load), the back EMF is equal to the supply voltage and no current flows.eg If the stall current is I then at half (no load) speed current = I/2 etc.I suspect that this surge will last in the order of ten milliseconds or more.If you assume that the current is constant then the formula is simplyCV = ITwhere C is in farads, V is in volts, I is in amps and T in seconds.eg What capacitor is required to supply 100A for 10mS with 1volt drop over this time?Now C = IT/V = 100*(10*10-3)/1 = 1Farad.200+ of your 470uF caps for only 10mS surge, 2000+ to hold up for 100mS.I know I neglected ESR, inductance and the fact that the current is not constant but it does show that capacitors are not the easy way to go.On the other hand, if you use a battery then the capacity required in ampere-hours is quite small.100A for 1 sec = 100*1/3600 =0.028AhAs has been suggested a small car (or motorcycle) battery is probably the cheapest option.Jim

author: Simon
author_rank: Unknown
timestamp: October 06, 2011, 08:00:20 pm
content: hm looks like I need quite a few caps if i go that route

author: Zero999
author_rank: Super Contributor
timestamp: October 06, 2011, 08:35:57 pm
content: There should be no need to measure the surge current. It can be calculated very easily. Measure the resistance of the motor and use Ohm's law to calculate the surge current.The duration of the surge is a bit more difficult as it's dependant on the load. You're probably better off measuring it with a current probe connected to an oscilloscope. As the frequency is low, a standard hall effect sensor will probably do or you could use a very low value current sense resistor but bear in mind its resistance needs to be low compared to the motor.You'll need a low impedance power supply. As mentioned above, a SLA is probably the best option.

author: Simon
author_rank: Unknown
timestamp: October 07, 2011, 05:49:24 am
content: Quote from: Hero999 on October 06, 2011, 08:35:57 pmThat's true, my idea was to be able to demonstrate that the soft-start works by showing before and after startup current graphs. Yes I was going to use a small series resistor

author: Simon
author_rank: Unknown
timestamp: October 07, 2011, 03:04:01 pm
content: does anyone have any idea of what ESR a capacitor that does not have it specified will be roughly ? the highest they bother to specify is 0.4 ohms, are there any worse than this ?