Stupid Electronics question

the power supply, and all five diodes (4 in the bridge, 1 across the coil) will blow, unless the fuse manages to blow first.

Come on now, we all know that the third law of electronics, states that, the $300.00 picture tube is supposed to blow first to protect the $0.20 fuse!

Peter

Reply to
Bushy Pete
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With the coil fed by a bridge rectifier, when the AC input is disconnected, the inductive overshoot drives all four diodes into forward conduction and this safely dissipates the stored energy in the resistance of the magnet coil. A protective capacitor is not necessary on the input or output of the bridge.

With no output capacitor the DC output voltage will be the MEAN value of the input voltage - 0.9 x RMS. less the diode drops.

With a large output capacitor the output approaches the PEAK value of the input voltage, again less the diode drops.

Jim

Reply to
pentagrid

Ah!! Ok. Makes perfect sense.

Gunner

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

- Proverbs 22:3

Reply to
Gunner

Addendnum. Im using a 50 amp (or bigger) bridge rectifier out of a Hardinge HC control box.

Gunner

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

- Proverbs 22:3

Reply to
Gunner

After reading your post, I started wondering about the mag on my Arter Grinder. It has a clockwork rotary switch that one has to turn 270 degrees then it locks on. To turn it off, one turns the switch a few degrees and it unlocks and turns off slowly. I think it pulses the mag.

Reply to
Tom Gardner

Yup, agreed. Two pairs of diodes in series/parallel across the coil.

8-(

Yes, & the heating value will be the RMS input voltage (ignoring diode drops), the valley magnetizing current will depend on the coil inductance. If something held near (or with a piece at an angle to) the chuck buzzes, the inductance is too low to be ignored.

What is normally used to power these things? Just a bridge or something else? From using bridges to power DC relays and solenoids from AC, I know the pull-in strength can be significantly compromised by not filtering the supply. I recall this issue has come up before here.

Specifically Vavg ~= sqrt(2)*Vin - (0.5* Iout*(1/120)/C + 1.2)

So with 660uF, 120V in, 1A out, about 162V. With 330uF, about 156V.

That will cause about 75% more temperature rise in the coil than with unfiltered full-wave rectified power. If that's too much, a suitable series resistor could be used, preferably between the bridge and the capacitors (or a variac, of course).

Best regards, Spehro Pefhany

Reply to
Spehro Pefhany

Yup, the diodes will snub the coil on turnoff.

Most DC voltmeters do respond to average value. However, neglecting diode drops, the RMS value of the rectified DC must be the same as the RMS value of the AC by definition. The rectified voltage waveform is the same as the AC wave except that alternate half-cycles are reversed in polarity. RMS means root mean squared, and the squaring operation makes polarity irrelevant. Therefore, the RMS values (neglecting diode drops) are identical.

Reply to
Don Foreman

Draw the circuit, Joe. There is a path for the inductive current thru the diode bridge. There won't be any spike on turnoff.

Reply to
Don Foreman

If you dont have a transformer for isolation I would check for a short to ground in the magnet. As far as needing a cap or ballast resistor, it is not necessary. Most of the magnetic chucks Ive repaired dont have any. Your diode bridge may be the problem. I would never reuse a bridge diode unless you have a very good method of testing it.

This method wil work.

Put a light bulb across it and see if you get full brillance on the bulb. Then put a diode in series with the light bulb and see if you get half brillance. Reverse the diode and do the same thing. This will check the complete bridge diode. Many diodes get "weak" they get an abnormal voltage drop across them under any load but test good with a diode tester. I've changed more of them than I can remember that had abnormal voltage drops. In avionics the KN-65 DME was notorious for bad diode bridges.

Reply to
john

According to Gunner :

O.K. The mag chuck, I presume you mean, is rated 110 VDC?

Huh? How do you get half-wave with a bridge rectifier? That should give you full-wave rectification.

Yes.

Well -- it could be calculated to determine what the minimum capacitance needed would be -- but I simply used a 250V 1000 uF computer grade electrolytic. (At least 200V is needed because the peak voltage from that bridge should be on the order of 170 V, and with minimum ripple, the voltage across the capacitor will be that pretty much all of the time.

You can use larger capacitance if you want, or significantly less. The 200V is a minimum, and higher won't hurt.

You also want a resistor to limit the surge current. Let's see, with 1 Amp current, a 5 ohm resistor would drop 5 volts, and will need to be capable of handling 5 Watts. Probably a 10 Watt wire-wound would do nicely.

Right.

Yes -- if you have the switching right, so you don't apply that while the capacitor and bridge are connected to the chuck.

I have a schematic of one which I designed to replace what was in the base of my Sanford grinder (about a bit smaller -- 6x4 inch, IIRC).

formatting link
an click on the link for the pdf version of the drawing. It is easier to read when printed. The .gif is probably not as good, and certainly the reduced size schematic tends to have lines vanish, depending on the browser in use.

The switch which I used has a center off position, and is stable in either on position, though a momentary to the "degauss" position could be used. I consider the switch which I have better, as you get better demagnitization of the workpiece if it is removed while the AC is on -- akin to moving a tape head demagnetizer away from the head before switching off the power, which could otherwise switch off during a peak and thus leave more residual magnetism.

The chuck had a plug which fit into the side wall of the base of the grinder, and that is what is shown in the schematic -- not the chuck itself.

Note that when using this design, you *don't* want to use coolant. If you want to do that, add an isolation transformer between the power line and the bridge rectifier.

Good Luck, DoN.

Reply to
DoN. Nichols

Just a bit more:-

With no output capacitor the DC output voltage will be the MEAN value of the input voltage - 0.9 x RMS. less the diode drops.

The RMS value of the output voltage is equal to the RMS value of the input voltage (less diode drops)

However it is the MEAN value of the applied voltage that determines the DC component of the output current and it is the value of this current that determines the current drawn from the supply and the power dissipated in the magnet.

If,as is usual with this type of electromagnet,the inductance is large, the magnet current with be almost pure DC with only a small ripple component. This means that (neglecting the effect of the residual ripple component) the waveform of the current drawn from the supply will be a square wave of amplitude equal to the DC output current.

Jim

Reply to
pentagrid

Sorry, but there is. The coil is running on rectified DC, even if it isn't filtered. And when you cut the power and the magnetic field collapses there is a Counter-EMF spike coming out of the coil in reverse polarity - a big pulse of negative energy coming out the positive connection.

It can't dissipate in the incoming diode bridge - the CEMF pulse is going the wrong way to go back through the bridge. If you don't provide a spark gap, diode, varistor or neon lamp to dissipate it, you have a lot of volts there.

Been there, bitten by that, sometimes hundreds of times in a day.

(Strapping the DTA on a Step phone switch, hot moves and changes on the mainframe, etc. You learn not to jump.)

-->--

Reply to
Bruce L. Bergman

The collapsing field will try to maintain current flowing in the same direction that it was, which was and will be thru the diodes. When mains power is cut, the polarity of the voltage on the coil will reverse. The polarity of this voltage induced by collapsing field is correct to make the diodes conduct. If you draw the circuit, you will see that there are two paths thru the bridge for this current to flow if the line side of the bridge is disconnected.

If you still disagree, before posting further argument please try this on the bench with an inductive load, a bridge rectifier and a scope to observe what happens.

If you open the circuit between the bridge and the coil, yes.

Reply to
Don Foreman

The real worry is that the first positive-going spike will exceed the prv ratings for the bridge. Then you do get smoke.

Hardinge found out about this on their potted feed motor drive units, they were smoking quite regularly. The newer versions have MOV snubbers on any inductive thing that gets switched, and I suspect they boosted up the ratings on their active devices as well.

Jim

Reply to
jim rozen

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