Stupid Electronics question

Yes, you do need a capacitor. Size depends on how much current you are drawing, it can be calculated (depends on how much ripple you can take) but something like 5,000 uF per amp is probably about right. Make sure you get one with the correct voltage rating. 4,700 uF should be fine with 1 A.

The AC should demagnetise it OK.

Leon

Should have said: the DC voltage across the capacitor will be about 1.4 x 110 V. I'd get a 200 V one.

Leon

Yes of course. That much I knew...

Ive got some big assed caps in cans...600 vots, 15,000 mfd if I recall, but Ill have to check. I found out they were too high a value for balancing caps for an RPC..so they are sitting on the shelf.

Too much?

Gunner

"A prudent man foresees the difficulties ahead and prepares for them; the simpleton goes blindly on and suffers the consequences."

- Proverbs 22:3

A motor start capacitor around 100uF should work just fine, also consider putting a current limiting resistor in series with the chuck to cut down the inrush current and don't forget to put a bleeder resistor across the cap, i usually just use a incadesent pilot lamp for 110v bleeders.

Best Regards Tom.

You're getting a full wave recitified DC outout, not half wave with a bridge rectifier.

120 times a second with a full wave rectification, 60 with half.

That should work.

Gunner are the caps bipolar (ie for AC)? I'm note sure but I think you may have to derate the max voltage value for DC use. Someone over at sci.elec.design will know.

No warranties but after checking the voltage rating I'd try the caps. The average voltage output may increase. The higher capacitance value may also result in more inrush current when you first turn it on and charge the caps. A slow blow fuse may help with this.

rob

Gunner:

Try the caps out of an old PC supply. They're usually rated something like 330uF 200V (they are used as a voltage doubler in the 120VAC input mode, so they are in series). One of two of those should suffice

330uF o(r 660uF @ 200V with two in parallel). While your in there scavenging, grab the input surge supressor (NTC thermistor) and put that in series with the input (looks like a disk capacitor and gets warm). That'll keep from killing your bridge with the surge to charge the caps.

Best regards, Spehro Pefhany

A bridge rectifier will give you full wave DC pulsed at 120 cps (htz). Each bottom half wave is inverted to the top, so to speak. You just need a bit of fill between the peaks, as you said. If the chuck gets a bit warm because the feed is a bit hot from the direct DC-ed 120vac, I'll send you a little variac that will easily allow adjustment. That would also allow a safer degauss cycle. Respectfully, Ron Moore

The problem with using AC to demagnitize is that when you turn it off you may turn it off when the voltage is high and you remagnitize it. You need to slowly reduce the AC current.

Bill K7NOM

I had a thing like that once that was intended for running portable power tools. The rectified/boosted power really made portable AC/DC tools run strong. It was intended to compensate for the drop in a long extension cord. After you get this thing built, you might try plugging in a couple of power tools, just for fun. Mine was stolen, and as far as I know they are no longer made.

A bridge rectifier (if connected correctly.... ) produces full-wave rectified DC. The RMS value of this DC will be about the same as the applied AC voltage. If the chuck was rated at 110 VDC I don't think you want any caps, just fullwave rectification. The inductance of the chuck winding may well smooth out the current ripple quite a bit. Using guesses of 100 ohms for coil resistance and 500 mH as inductance, I get 1 amp of avg current with 0.127 amps RMS of 120 Hz ripple current.

Adding capacitance can raise the DC voltage closer to the peak voltage (165 volts) which may overdrive (overheat) the chuck.

Surprised no one pointed this out. 110 VAC is rms voltage. The output of a bridge rectifier/capacitor combination will be about that times the square root of two, practically somewhere around 150 volts.

You probably need a variac in front of the rectifier if you really want 110 VDC.

Jerry

Act-ually, the first thing Gunner should try is really E-Z, and that's to put an inverse parallel rectifier across the coil* (REVERSE-biased). As it sits now, there's no path for the current when the current drops below the peak, until the diodes avalanche, which means a whole bunch of energy is likely going into the diodes and not staying in the electromagnet.

• care to polarity or all 5 diodes will release their smoke

Best regards, Spehro Pefhany

Sorry, forget it, the bridge clamps it to two diode drops on top of the input supply anyway.

But there is a potential problem when the input connection is broken with no capacitor on the input or output of the bridge.

Best regards, Spehro Pefhany

Blink blink....huh??

Gunner

"A prudent man foresees the difficulties ahead and prepares for them; the simpleton goes blindly on and suffers the consequences."

- Proverbs 22:3

Pls. see correction. ;-)

Best regards, Spehro Pefhany

Ah...I didnt understand the correction either.......

Now ask me about Welding, or ballistics, or skinning and curing hides, or other Stuff.....

Gunner

"A prudent man foresees the difficulties ahead and prepares for them; the simpleton goes blindly on and suffers the consequences."

- Proverbs 22:3

The problem is that the coil of the magnetic chuck stores energy in the magnetic field. How much energy is stored per amp of coil current is proportional to the inductance of the coil.

The problem is that when the coil current is abruptly interrupted, the energy in this magnetic field comes back out, attempting to keep the coil current flowing in the same direction as the original current flowed. This is seen as a high-voltage spike, reaching to hundreds or thousands of volts, and will puncture any attached semiconductors (like the rectifier diodes).

The classic solution is to put a rectifier diode in parallel with the coil, oriented so that under normal conditions no current flows through the diode. When the coil current lead is opened, the magnetic field energy goes into this rectifier diode, preventing the voltage spike.

If one mis-orients this rectifier diode, it will be a dead short across the power supply, and all five diodes (4 in the bridge, 1 across the coil) will blow, unless the fuse manages to blow first.

Joe Gwinn