I have 5 identical NEW 10,000uf electrolitic caps rated at 63 voltes.
I have (45VAC * 1.44 full bridge rectifier) = 64.8VDC no smoothing.
I would like to use the caps I have for my new power supply feeding 1 x G203V and 1 x 6amp Nema34 1200oz stepper from Kelinginc.
I have seen a circuit using capacitors in series here Capacitors - Multiple Capacitors
My question, can I get away with using the caps I have or should I just order 2 x 10,000uf caps at 100VDC.
Basically I forgot about the smoothing caps before I started the upgrade and I have these other ones on hand.
Thanks for the help.
I have used electrolytic capacitors in series (with the equalizing resistors) in high voltage power supplies before. It does work as advertised. But it can be a pain to connect them up (in the past I have made a custom circuit board to hold them), and you will have to get the resistors. So it may be easier just to get capacitors with a suitable working voltage.
Thanks for the reply dk, I have quite a good assortment of resistors on hand but having the correct wattage might be a problem. I have to read up more to find the value and wattage I need if I actually proceed down this road.
I will give you a hand calculating the resistors. To start we need the spec sheet of the 10,000 uf caps. What is the full part number of the cap and mfg.
The short version is we need to pick resistors that allow 10x more current than the worst case leakage current of the caps.
Thank you very much for the help!
The caps are nippon, chemicon, 63v@10000uf(M)
diameter 35.17mm, length 51.69mm
I purchased them from digikey. I was hoping that I would not have to use all 5
As I could use one on another project.
From the cap spec sheet.
Leakage current I=0.02CV or 3mA, whichever is smaller.
C = 10,000uF
V = 63V
I is in micro amps.
.02 x 10,000uF x 63V = 12,600uA
12,600uA = .012600 A = 1.26mA
So 1.26mA is smaller than 3mA so you are supposed to use 1.26mA.
Personally i am not a real believer in what the mfg’s say. I think we should use the 3mA number to be safe and see were it leads us in values.
You have a 45 volt ac transformer.
45VAC x 1.414 = 63.63VDC and subtract 2 diode drops for 1.4 volts = 62.23VDC lets just call it 62VDC.
Our 2 resistors in series will carry 3mA across 62VDC. R = E/I with E in volts and I in amps, so 62VDC/.003A = 20,666 ohms or 20.67K.
We have to divide this by 2 because there are 2 resistors in series across the caps. 20.67K/2 = 10.36K and of coarse 10K is a standard and popular value so rounding down gives us slightly more current. We now know we need 10K ohm resistors.
Now we need to know the wattage each resistor will dissipate. P power in watts = I in amps x E in volts across the resistor. .003mA x 1/2 the total voltage 62VDC/2 = .093 watts an 1/8 watt is .125 and 1/4 watt is .250. You are close to 1/8 watt but never run a resistor near its full wattage rating for long life. Use the 1/4 watt at 10K ohms for long life.
That is a very reasonable resistor.
OK i made a mistake, i forgot that we needed 10X the current in the resistors $%^&.
so 3mA x 10 = 30ma
R=E/I 62VDC/.03A = 2,066 ohms divided by 2 = about 1000 ohms or 1K.
P=IxE .03A x 62VDC = 1.86 watts.
I would use 1K 3 watt resistors.
Last edited by cnc_4_me; 04-28-2012 at 02:08 AM.
Now to figure out if your cap values are OK.
The Gecko capacitor formula is
(80,000 x I in amps) / V in volts = C in uF
You have a 6 amp motor. Gecko tells us that you only draw 2/3 motor current with their drives. 2/3 x 6A = 4A
(80,000 x 4A) /62VDC = 5161uF
Two 10,000uF caps in series = 5000uF, close enough.
Your capacitors are rated at 4.40 amps max. So you can just get by with 2 caps in series since each cap will see the full current.
Or you can make up 2 sets of capacitors in series and then parallel them, you will then have 10,000uF and 8.8 amps capacity.
Last edited by cnc_4_me; 04-28-2012 at 02:10 AM.
Reason: minor edit
thanks again cnc_4_me, I didn't consider the idea of 2 sets of 2 in parallel. I have some 1/2 watt 10K resistors in my tray which should work well.
This has saved me a digikey order, I'm very happy.
We could also check to see if your transformer has enough capacity. Do you have the specs for it.
I think that you had a misplaced decimal point there, as 12,600uA would be 12.6mA, however since you did the calculation at 3mA anyway, it looks like you're good to go...
Originally Posted by cnc_4_me
Thanks for slogging through the detailed analysis. I'm sure that it will be helpful to lots of folks facing similar design decisions...
I may have to dig a little deeper for the larger resistors , if I remember correctly I might have a few 1K@5watt ceramic resistors in the tray, maybe a little closer to the back and collecting dust.
Here is the twist, I have a G540 and 1 x G203V.
I have about 10 of these:
Amp Transformer Mid West Co. 430-9001 (8) Class 130B
Model 430-9001, Transformer
Input: 125 VAC black/white - 115 VAC brown/white
Output: 14 VAC 15 Amp ( terminal .250” )
On the brown/white inputs the output is 15VAC while the black/white inputs gives you 14VAC output.
My plan is to use 3 of these in series, take a tap off the first 2, to provide the voltage for the G540 and use all 3 for the G203v.
In my current setup I have 2 of the 430-9001 transformers in series running the G540 with 3 nema23 steppers, XY-axis are using CNCRP ~firstname.lastname@example.org amd stepper and the Z-axis uses a Tormach 280oz@3amp stepper. The G540 current resistor is set at 2.8 amps.
The bridge rectifier I'm using is the MB251 100V@25amp. (I also have 10 of these)
In the new setup I'm upgrading the X-axis to a Kelinginc(automationtechnoligies) Nema34 1200oz@6amps with the G203v. The Y-axis will stay the same and the Z-axis will get the stepper from the X-axis. This leaves me 2 free drivers on the G540 to maybe use the new plastic extruder I seen in this thread. http://www.cnczone.com/forums/diy-cn...d_printer.html As well I have always had my mind set on a 4th axis.
Good catch, it is 12.6mA
Originally Posted by doorknob