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7th March 2011, 12:01 AM #31
AC relays will have a laminated core to reduce the eddy currents. DC relay cores can be solid.
The coils are wound differently too. The DC coils depend upon the resistance to limit the current (thin wire and lots of it to increase the resistance), whereas AC coils depend upon the inductance to limit the current (so may have less turns of thicker wire as they don't need - or want - the high resistance).
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7th March 2011, 12:36 AM #32GOLD MEMBER
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Doh eddy currents, I so should have remembered that they would have something to do with it. Its a pitty I've forgotten a lot of what I know about AC and transformers, as I didnt know that much to start with lol
Stuart
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7th March 2011, 09:56 AM #33
Hi .RC,
Continuing on from previous post...
If you put an electrolytic large enough to supply the inrush current downstream of the resistor, that would work.
I think the contactor coil is 3w, ( check the data sheet, or measure the current), that would be a holding current of 3/24 = 125 mA, so to drop 10v 10/.125 = 80 ohms
wattage required would be 1.25watts, So look for a 75ohm 5 watt resistor.
For the cap to supply the inrush current, 4700uF 63v should be more than enough.
Regards
Ray
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7th March 2011, 03:01 PM #34
If you are determined to build this thing, get yourself some 1000pf capacitors. fit one across each leg of the bridge. The 75ohm 5W is fine but not the 4700uF, that is overkill and a half. Maximum needed would be about 100uF. It it was me I would feed that to a 3 pin regulator. (input 36V, centre to neutral of system and output 24V) that needs a 0.001uF on input and output legs as near to it as possible, Also fit that on a plate of metal to loose heat.
So head to Jaykar or similar get a piece of fibreglass strip board with holes drilled. ($4.00)
Get 6 X 0.001uf (66c) about 250VDC capacitors a 7824T ($6.50) regulator A silicon washer ($1.50) for it or heat paste. some solder and advise some solder wick when you stuff up, and a 100uF63V low ESR105deg electrolytic (0.52c) and use a 10ohm 5W resistor (47c)
The stripboard has lines of copper so you have to cut and remove bits if you dont want the line to continue. Personally I would fit a 2.2 M ohm half watt resistor across the 100uF as a bleader when the power is off but thats not so important You will also need some insulated jumper wire to jump from track to track. If you have never soldered the only trick is keep the iron clean and heat the joint and with the joint getting heated add solder. If the board is dull polish it with oo steel wool.
Please note Prices are from another catalogue and were trade +GST
Other thing is I suspect you have some huge bridge block, If you have, get a 1.5A bridge (55c)
Put that together and you have a safe smooth supply and no issues.
Thinking on it I would add a small chofe (coil) on the output with another 0.001uF to neutral to reduce splash on switch off hitting the regulator.
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7th March 2011, 08:31 PM #35Pink 10EE owner
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I worked out to install a jog button will require a second contactor... Space is running out in the box as I have just ordered 3 fuse holders that hold 10 X 38 fuses.... Which I will use some 6 amp fuses (it is only a 3hp motor) That will also take care of the fusing of the transformer...
Also bought some components from Jaycar this morning... Did some experimenting.... Interesting results... Will install an inline fuse on the output side as well..
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7th March 2011, 09:04 PM #36GOLD MEMBER
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8th March 2011, 08:00 AM #37Pink 10EE owner
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I will have a go at drawing this evening........ I have it drawn on paper, but I am probably the only one who could understand it..
It came to be last night in bed that I would not need a second contactor, but it would depend on a switch where the NC side would break before the NO side engaged... I will explain in my diagram..
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8th March 2011, 09:35 PM #38Pink 10EE owner
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Here is a diagram, sans capacitors etc.. Just a basic stop, star,t jog.
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8th March 2011, 09:55 PM #39GOLD MEMBER
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Yeap thats it
Why do you have a diode across the output of the bridge?(should it be a cap?)
"sans"?
Stuart
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8th March 2011, 09:56 PM #40
Two suggested changes to get the basics right.
Remove D2 - it is just shorting out the bridge rectifier.
Place D2 to be directly across the coil of the relay - and make sure it is reverse biased under normal operation (i.e. not shorting out the coil).
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8th March 2011, 10:20 PM #41Pink 10EE owner
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OK Fixed..... Yes I was told to put it across the coil... With the cathode pointing towards the positive..
sans means without.. As in this case without capacitors and resistors..
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8th March 2011, 10:44 PM #42GOLD MEMBER
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Thanks .RC., I should have googled better.
What's D2 for?(in its new spot). Cutting down arching of the switch contacts? or collapsing the coil faster? or maybe something else lol
Stuart
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8th March 2011, 10:47 PM #43Pink 10EE owner
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It is to it short out any kickback or back EMF from the collapsing magnetic field of the coil when the power to the coil is cut.
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8th March 2011, 10:54 PM #44
Its called a free-wheel diode. The coil is inductive and therefore it is impossible to instantaneously interrupt the current flow. The free wheel diode provides a current path for the decaying current as the magnetic field collapses. The alternative current path (if a diode is not fitted) is arcing across the switch contacts.
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9th March 2011, 01:05 AM #45
Hi .RC
Nice to see you're making progress, What about e-stops?
I'm just back from a few days interstate, I'll have a closer look tomorrow..
Regards
Ray
PS Never mind, I see what you've done, the start switch latches the contactor in and the stop switch is the e-stop.
The jog unlatches the contactor for momentary operation... Looks ok at first glance.
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