C = Capacitance
I = Amperage
the * means to multiply
C=(80,000 * 7)/65
Is this 7amps for the entire system? and the Vdc is this the max rated or is it your nom. voltage. I am guessing that you are using Ametek servos????
Iv'e seen this equation for calculating the size cap I need to use On my power supply. I don't know enough math to use this equation, could someone tell me how to use it?
C=(80,000*I)/Vdc
My Vdc is 65V
My apms need to be 7A
What is the "*" in the equation?
What is the "I"?
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C = Capacitance
I = Amperage
the * means to multiply
C=(80,000 * 7)/65
Is this 7amps for the entire system? and the Vdc is this the max rated or is it your nom. voltage. I am guessing that you are using Ametek servos????
Industrial automation ????
www.challengermechtech.ca
65V and 7 amps is the end output of the supply. I'm using Geko g201a drives.
Ok so you have a PS that can potential put 7amps max on your sys. @ 65vdc after rectification.
So the capacitance you need in theory is.
C=(80,000*7)/65vdc
C=8615 round up to 9000uF of capacitance to be safe.
Industrial automation ????
www.challengermechtech.ca
I also forgot to add, you may find it cheaper to buy more capacitors rated less then 1 large capacitor. look into 5 @ 2000uf instead might be cheaper.
hope this helps
Industrial automation ????
www.challengermechtech.ca
so is it 65/7x80,000?
or 7x80,000/65?
I still don't understand how to do the equation. My math skills only go to about 8th grade.
ok, no prob.
math inside the brakets () is always done first. (80,000 multiplied by 7amps) = 560000
56000 divided by 65vdc = 8615 which is the capacitance needed to help smooth the ripple on your power supply.
8615 uf is an odd number, therefore you would be better off to round it up to the next whole number, which in this case is 9000uf.
My math teacher use to beat into my head "BEDMAS" which is an acronym for doing math equations in the right order, otherwise your awnser will come out wrong.
B= brackets- anything in brackets will always be done first
E= Exponents-anything with exponents is done second
D= divide... this or multiply 3rd
m= multiply
a= add...lastly adding and subtracting is done to the equation
s= subtract
Industrial automation ????
www.challengermechtech.ca
Remember that your actual current draw will depend on what motors you're using, not the drive.
Also, make sure your capacitors have a voltage rating higher than 65
Awesome!!! thanx so much for explaining this to me. I did not know the capacitor needed to be able to handle more than the 65V, now I know.
The 2 steppers I'm going to buy use, 5V and 2.6A per phase,they are 2 phase motors so the actual amperage drawn from both motors would be,,,2.6x2(phase)=5.2 x2(motors)=10.4 - 67%= 6.93A this sound right?
I understand the power supply needs to put out4 to 20 times the motors voltage so 65V is 13 times more than 5V and within the rated voltage of the gecko drives. This is how I came up with my numbers for the power supply.
I dont know where I came up with 2.6A per phase on my motors? its actually 4Aper phase. I just went and checked the website that has them. So now were looking @ 11A 65V
So I need about a 14000uf 70V capacitor right?
Last edited by shortbus; 03-25-2005 at 12:25 AM.
That looks correct, although my gut feel is that the power supply don't need to be nearly as large as what the calculations suggest, but I could be wrong.
A 14,000µfd will give approx 3.5% ripple at max. load (=6.4vpp).
A 20,000µfd will give 2.5% ripple at max.load (=4.6vpp)
You will need 47vac secondary voltage.
Also the larger capacitor, the larger VA for the transformer.
Al
CNC, Mechatronics Integration and Custom Machine Design
“Logic will get you from A to B. Imagination will take you everywhere.”
Albert E.