Help! What size Capacitor & Bridge

[Trainhound]

I just bought a toriod transformer input 120v, output 48vac/6vac 750va. So I'd like to know what size cap and bridge rectifier would I need? I'm trying to get about 68vdc to run 640oz motors and G202 Geckos. What dc current should I expect out of the 68vdc supply?

BTW I'm using a campbell designs combo board that has 5vdc to power the gecko.

[ViperTX]

Well kinda depends on the toroid, what are the current specification's for each winding? I can guess 15 amps for the 48VAC and 5 amps for the 6VAC....but then that is just a guess.

[Evodyne]

Originally Posted by Trainhound I just bought a toriod transformer input 120v, output 48vac/6vac 750va. So I'd like to know what size cap and bridge rectifier would I need? I'm trying to get about 68vdc to run 640oz motors and G202 Geckos. What dc current should I expect out of the 68vdc supply? BTW I'm using a campbell designs combo board that has 5vdc to power the gecko.

Hi. The 5VDC to power the drives will see minimal current draw-just enough to power the logic circuitry. So let's assume that all 750 VA can be used by the drives. In that case that transformer can deliver at most 750 VA / 48 VAC or a smidge over 15 Amps. You'll lose about 0.7 volts across each diode in the bridge (or 1.4 volts for the two that are conducting at one time), so your DC out should be close to (48 VAC-1.4V) * 1.414 = 66 VDC.

The diode bridge is chosen by two parameters-the forward current rating and the peak inverse voltage rating. We know the most current you could get is 15 Amps, so any bridge that handle more than that is fine. The most the reverse biased diodes will see is the 66 volts, so choose a bridge whose PIV rating exceeds those values. Thus a 25Amp, 100 PIV bridge would work just fine. Bigger would work too.

The capacitors are rated by capacitance and a maximum voltage rating. Again, the voltage rating must exceed what your feeding into them-so you would want to pick caps that are rated in excess of 66 volts by 20% or so (a safety margin). Say 66 * 1.2 or 79.2 V. Of course you will never find a 79.2 volt cap-use the next standard size higher-80 VDC or higher.

The capacitance can be estimated from C = (80000 * I) / V, so C = (80000 * 15) / 66 = 18,000 uF. Thats a minimum-more is OK. Capacitance adds up when you wire units in parallel. So two 10,000 uF caps in parallel would yield 20,000 uF total.

Make sure to observe the polarity on the electrolytic caps too-they pop (explode) quire nicely if you wire them in backwards.

Hope this helps...

Evodyne

[Trainhound]

Thanks guys! I greatly appreciated the help.

[gar]

051114-0743 EST USA

Steady-state continuous load current rating under normal conditions for a bridge rectifier with a capacitor input filter may be approximated by TRANSFORMER VA RATING divided by DC OUTPUT VOLTAGE. See technical discussion at Signal Transformer. This would be approximately 750/68 or 11 amperes for your transformer. Therefore your average load should be less than approximately 11 amps.

Under short circuit load conditions this current will be much greater.

The peak diode current is much greater than 11 amps for an 11 amp dc load. The peak diode current may be much higher with a toroid transformer than a standard transformer because of the lower leakage inductance of the toroid.

You possibly should consider a peak inrush limiter such as an inductor in series with the primary to limit peak diode current and also a negative temperature coefficient thermistor to limit turn-on inrush current.

Relative to load current the primary considerations are temperature rise in the various components.

.

[MrWild]

Thanks for the great info. It's good to know.

I'm running three 12v 20a normal transformers for 36v 20a. (36v - 1.4) * 1.414 = 49v Safety factor 1.2 * 49. = 59v. for a safety factor. Sad shame too as I have 3 x 10,000uf 50v salvaged caps. 20 * 80000/49v = 33,000uf.

Did I nail it?

[Trainhound]

Guys I have another problem, I just found out I may have to limit the current to the steppers. my steppers are 640oz 4 wire bipolar 2.3v rated voltage current is 5.5amp dc per phase. So what do I need to do to make my transformer work with these 3 steppers? current limiting resistors or what? sorry for my ignorance.

[tctech]

...the important thing is the current rating ...the voltage rating is a static rating based on ohms law ...

...get a bipolar microstepping drive capable of driving upto, say 6 Amps or more per phase, connect the four wires of the motor to the drive by working out the two pairs by checking continuity with a multimeter. Connect a pair showiing continuity to the A/A- and the other to the B/B- connections on the drive. Other connections will be given with the documentation for the drive.

A supply of 48V should be OK to support almost any drive on the market (but pls check the documentation ...check that your power supply can supply the total current you require for your motors. Bipolar stepper motor drives use PWM current regulation and maintain a constant current drive to the motor. This obviously implies that you should set your drive to deliver a little below 6 Amps. If your motors are warm (up to say 40-50 deg C) to the touch after some running, you have set the current correctly...

I hope this helps ...

[Trainhound]

The drives that I'm using are Gecko G202 will these work and shouldn't I still need to limit the current?

[2muchstuff]

You don't need to limit the current going into the drive. The drive will take what it wants/needs. The Gecko 202 will still need a current limit resistor to limit the amount of current going out to the motor, in your case 5.5 amps.

[Trainhound]

What size resistor should I use?

[Evodyne]

Originally Posted by Trainhound What size resistor should I use?

Trainhound,

From the manual (available online at www.geckodrive.com under support):

"This input programs the G202’s current output to the motor windings. The G202 will accommodate motor winding currents from 1 to 7A and 0.3 to 2A. Use the following equation to calculate the value, (in kilo-ohms) of the current set resistor: R (in kilo-ohms) = 47 * I / (7 – I) or for the low current range, R = 47 * I / (2 – I)."

For 5.5 amps I'm coming up with 172.333. That 172.3 kohms. Nearest standard 5% resistors are 160 kohm and a 13 kohm: wire these in series for a total of 173 kohms. You would rather be a bit higher than 172.3 than lower.

Take care.

Evodyne