# Thread: Electromagnet Control Wiring

1. gar
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171007-1724 EDT

The photo of post 17 very clearly shows three heat sinks, and they need to be isolated from each other. One heat sink has two diodes mounted to it. Each of the other heat sinks has one diode. This is why one can conclude that the diodes are wired as a bridge.

With a 240 V RMS sine wave applied to the bridge input and a resistive and/or inductive load the bridge average DC output voltage is 240*0.636/0.707 = 215.898 V. Then one must subtract diode drops. Or approximately 214 to 215 V DC output.

If you add sufficient capacitance at the bridge output (make it a capacitor input filter), then you can make the average DC output voltage approach the peak voltage of the sine wave, or about 339V.

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2. gar
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171007-1746 EDT

Two diodes and a center tapped transformer is called a "center tapped full wave rectifier", not a half wave rectifier.

A half wave rectifier has only one diode and with a resistive load the load current only flows during 1/2 cycle.

A half wave rectifier produces an average DC current in a source transformer secondary, and this unbalances the hysteresis loop. A full wave center tapped rectifier produces no unbalance flux in the transformer core integrated over one cycle.

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3. gar
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171008-0846 EDT

To determine the average value of a sine wave you calculate the area under the curve and divide by the base of that area.

If you take a half cycle of a sine wave (0 to Pi or 0 to 180 deg), and do the average calculation, then the result is approximately 0.636 (2/Pi) of the sine wave peak value.

If you have a full cycle of a sine wave, then the result is 0, because the positive half cycle cancels the negative half cycle.

A typical moving coil DC meter has an averaging time constant somewhere in the 5 to 20 Hz region. Connect this meter, such as a Simpson 260, to a sine wave source of variable frequency. At 60 Hz the meter needle will not visibly move from zero for any input voltage. Lower the input frequency to 0.1 Hz and the meter will quite closely track the instantaneous voltage of the input.

In the early 1950s there were some moving coil strip chart recorders that had a fast enough response that you could record a 60 Hz waveform.

If you use a half wave rectifier on a sine wave source, then the result is 0.318 of the sine wave peak. This is because the averaging time base is 2 times the half sine base, and 0.636 needs to be divided by 2.

For a full wave rectified sine wave the average DC output is the 0.636 of the sine peak.

Experimentally you can prove out this theory with a single diode, resistor, sine wave source, and both an AC and DC meter. Then do it with a full wave bridge rectifier.

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4. gar
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171008-1354 EDT

The integral of sin u du = - cos u
Evaluating this at 0 and Pi gives us an area under the curve of 2.
Divide by the base of the half sine wave, and the result is 2/Pi = 0.636 .

For a half wave rectifier, current only flows for 1/2 of the cycle. Thus, the average of the half wave rectified output is 1/Pi or 0.318 .

If we have a sine wave and measure its RMS value, then its peak value is RMS/0.707 (0.707 is (sq-root of 2)/2).

If one knows the RMS value of a sine wave, then the full wave rectified average DC value is RMS*0.636/0.707 = RMS*0.900. For a 240 V RMS sine wave the average DC output is 240*0.9 = 216 V.

This is true for either a resistor alone, or inductor, or both.

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5. gar
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Some component values you can use to run your own experiment.

A run of the mill 1 A diode with a 400 V PIV rating can be used. I used a 1N5060, a 1N4004 would be good.

For a load resistor I used 10,000 ohms 1/2 W.

You can use 1 or 2 meters. I used a Beckman true RMS 4.5 digit for the AC. A Simpson 260 could be used. This is an average reading calibrated for RMS on a sine wave in AC mode. I used both a Simpson 260 and a Fluke 27 for the DC measurement. At about 123 V input the DC output was about 55 V. This correlates well with the 0.9 ratio divided by 2 = 0.45 .

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6. Member
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Wow! Thanks very much all of the information is very helpful. I don't want to use the existing setup because the owner that originally had the smaller magnet got a slightly larger magnet and he is using this controller to drive that magnet. The smaller magnet is now at a different location and my friend needs a whole setup on the excavator. I also think I can do the job much more neatly than the person that put this controller together. I want my friend to have something that will last a very long time and that he can depend on for his business. I really can't thank you guys enough for all the help.

Jilmo your offer is very generous and if I did not have the means to machine the heat sink I would take you up on that. Before I was an electrician I was a machinist. If anyone here needs anything in terms of metal fabrication I can help. We have a large facility with lathes from 4 feet to 18 feet in length and some turret lathes for turning large diameter parts. Boring mills, radial arm drills, Blanchard grinders, bridge-ports and a nice welding shop. Everything is old school and most without digital readouts so its with with an indicator and micrometer. If there is anything that I can help anyone with in terms of metal work please let me know! We used to be a large operation servicing paper mills and the locks on the eerie canal but things have slowed down here a bit for a number of reasons.

So the consensus is that I will use a bridge rectifier good for 400v. Gar I really appreciate the explanation of the process of rectification and I am curious if this product would suffice to build the rectifier? Also if I were to raise the capacitance at the bridge output and bump the voltage up would it result in better performance of the magnet? Would I risk damaging the magnet coil?

https://www.digikey.com/product-deta...N1188-ND/77577

Thank you!

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Also where do you guys like to order contactors and heavy duty switches?

8. gar
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crispysonofa:

Measure the DC resistance of the magnet. Tell us what it is.

As a wild guess you might consider a Comchip GBPC 3506 G from Mouser as an integrated package for your bridge rectifier. Something over \$3.

The heat sink size you required is a function of the actual load current, ambient air temperature, type of cooling, and desired junction temperature.

The above diode bridge can be directly mounted on the heat sink without an insulator.

Run an experiment with the bridge output directly connected to the magnet. Just use a cord and plug or equivalent to switch power to the input of the bridge.

Measure DC voltage and current. See if you can pick up a load and dump it simply by removing AC input. After this experiment you decide what to do next.

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Ok I will get a coil resistance ASAP. Thank you for that information Gar.

10. Originally Posted by jimo144
A Full Wave Bridge would produce right at 170 Volts DC from 120 VAC (RMS) and may be a little easier to set up.
You seem to be assuming peak voltage for the DC. Average, which is likely what you would get with an electromagnet, would be around 110Vdc.

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