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Posted by Don Foreman on March 15, 2006, 4:03 pm
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On Wed, 15 Mar 2006 19:15:02 GMT, Bruce L. Bergman
>On Tue, 14 Mar 2006 11:38:54 -0600, 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.
>
> 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.
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.
>
> Been there, bitten by that, sometimes hundreds of times in a day.
If you open the circuit between the bridge and the coil, yes.
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Posted by jim rozen on March 15, 2006, 4:24 pm
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>If you open the circuit between the bridge and the coil, yes.
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
--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================
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Posted by on March 14, 2006, 11:00 am
Please log in for more thread options On Mon, 13 Mar 2006 21:59:09 -0500, Spehro Pefhany
>On Mon, 13 Mar 2006 20:00:58 -0500, the renowned Spehro Pefhany
>
>>On Tue, 14 Mar 2006 00:31:26 GMT, the renowned "Jerry Foster"
>>>
>>>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 <snip>
>
>Sorry, forget it, the bridge clamps it to two diode drops on top of
>the input supply anyway. <dunh>
>
>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
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
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Posted by Spehro Pefhany on March 14, 2006, 12:26 pm
Please log in for more thread options On Tue, 14 Mar 2006 16:00:28 +0000, the renowned pentagrid@yahoo.com
wrote:
>On Mon, 13 Mar 2006 21:59:09 -0500, Spehro Pefhany
>
>>On Mon, 13 Mar 2006 20:00:58 -0500, the renowned Spehro Pefhany
>>
>>>On Tue, 14 Mar 2006 00:31:26 GMT, the renowned "Jerry Foster"
>>>>
>>>>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 <snip>
>>
>>Sorry, forget it, the bridge clamps it to two diode drops on top of
>>the input supply anyway. <dunh>
>>
>>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
>
> 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.
Yup, agreed. Two pairs of diodes in series/parallel across the coil.
8-(
> 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.
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.
> With a large output capacitor the output approaches the
>PEAK value of the input voltage, again less the diode drops.
>
> Jim
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
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
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Posted by Don Foreman on March 14, 2006, 12:35 pm
Please log in for more thread options On Tue, 14 Mar 2006 16:00:28 +0000, pentagrid@yahoo.com wrote:
>On Mon, 13 Mar 2006 21:59:09 -0500, Spehro Pefhany
>
>>On Mon, 13 Mar 2006 20:00:58 -0500, the renowned Spehro Pefhany
>>
>>>On Tue, 14 Mar 2006 00:31:26 GMT, the renowned "Jerry Foster"
>>>>
>>>>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 <snip>
>>
>>Sorry, forget it, the bridge clamps it to two diode drops on top of
>>the input supply anyway. <dunh>
>>
>>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
>
> 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
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.
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