capacitors and rectifiers

[Runner4404spd]

i'm looking at building my own power supply for my cnc mill conversion. i already bought (2) 24 volt 10 amp transformers that run on 110VAC. i need to know how to size the capacitor and also how to size the bridge rectifier for this application. if you have part numbers or suggestions that would be appreciated but i'd like to know the equations behind this so i know what i'm doing in case i need to modify something.

i will be running gecko 320 drives and am running 3 servo motors that run at 6 amps each. this mill/drill is an RF-25 clone. its not as big as the RF-31 but it should be fine for what i want to do. i am running a 3.6:1 reduction on the gearing and i will be using ball screws (5/8)

[ViperTX]

I personally prefer new capacitors and would go to mouser.com and look at the Sprague Electrolytics....50 volt 40 to 47,000 mfd. The bridges should be at least 100 volt 10 to 20 amp units...which you can find on ebay or at mouser.com. Half the fun is looking for P/N's...isn't it??

[Al_The_Man]

The rectifier voltage rating or PIV (Peak Inverse Voltage) on a Full wave bridge should be minimum of the RMS (ac voltage) x 1.414 or if using a centre tap full wave supply the PIV should be AC x 2.282.
Al.

[Runner4404spd]

what does full wave center tap mean?

[ViperTX]

Means when the transformer secondary has a center tap which is being used as the neg. and the other 2 ends are feeding the bridge rectifier ~ (AC in).

[Runner4404spd]

is there a benefit to using the fullwave input?

[ViperTX]

yes, requires smaller capacitors (mfd wise)....but you will most likely need the extra current from paralleling the outputs.

[gar]

50720-2126 EST USA

Runner4404spd:

There are several items that need clairification.

First, you can idealize a diode rectifier as a non-linear resistor that has zero resistance when forward biased, and infinite resistance when reversed biased.

The conventional symbol for a semiconductor diode is a triangle with a straight bar in contact with one corner. This straight bar is called the cathode (the name cathode derives from vacuum tube days). The side of the triangle opposite the cathode is called the anode (again from tube days). The triangle forms a pointer indicating the direction of positive current flow. This is the low resistance direction.

The term rectifier is primarily applied to a diode when it is used to convert ac to dc. Diodes can be used for other purposes than converting ac to dc. For example as a mixer in a superhetrodyne receiver. Some special diodes are voltage variable capacitors, and others are light emitting devices, and so on.

Second, if we are dealing with a single phase circuit, then there are typically two broad classifications of rectification, full wave and half wave. In a three phase system there are 3 phase, 6 phase, etc, rectification. The classification is determined by the number of peaks in the output.

AC power is distributed with a voltage waveform that is ideally a sine wave, and therefore is composed of a single frequency component. As soon as you distort a sine wave it then contains other frequency components that are harmonics of the fundamental.

If you do not know what a sine wave looks like, then make a plot on graph paper of the sine function. Here are some points --- 0 deg 0 v, 45 deg +0.707 of Vpeak, 90 deg +Vpeak, 135 deg +0.707 Vpeak, 180 deg 0 v, 225 deg -0.707 Vpeak, 270 deg -Vpeak, 315 deg -0.707 Vpeak, 360 deg 0v. Plotting these points will give you an idea of a sine wave.

(edit 050721-0640) Some additional points --- 30 deg 0.5 Vpeak, 60 deg 0.866 Vpeak. You figure out how to mirror and map these base angles to others. A scientific calculator will provide you a means to get the sine function at any desired angle. See the Handbook of Chemistry and Physics under SERIES for the infinite series defintion of the sine function.
(end edit)

If you have a sine wave voltage source and connect the anode of a single diode to the source, the cathode to a resistor, and the other end of the resistor to the other side of the voltage source, call this other side common, then only one half of the sine wave appears across the resistor. This is classified as half wave rectification because only during one half of the cycle is current flowing. This is not a very smooth waveform, but it is pulsating dc and has an average value of 0.318 of Vpeak.

Next create a second voltage source that is exactly the same as the first, but shift it in time by 180 deg. Connect its common to the common of the first source. Now when one peak is positive the other is negative, and vice versa. Connect a second diode to the second source, and the cathode to the cathode of the other diode. Now we fill in the gaps between the half wave rectifier pulses and its a full wave rectifier. When you have a center tapped transformer this is the result you get. This is a circuit that generates the absolute value of a sine wave. Here the dc average value is 0.636 of Vpeak and the waveform is much more smooth. The constants 0.318 and 0.636 are obtained by using integral calculus to determine their average values.

Whether the common lead, center tap, is negative or positive is dependent upon the orientation of the diodes. Also in a full wave center tapped rectifier you do not refer to the rectifiers as a bridge.

Third, a bridge rectifier is a group of four diodes configured in a square such that when an ac signal is applied to opposite corners that full wave rectified dc is produced from the other opposite corners. The term bridge derives from the very early days of electrical science and relates to a circuit of the square shape. The Wheatstone bridge is the original basic four arm resistance bridge.

Fourth, diodes have limitations. Major ones are --- One is peak inverse voltage. This is the maximum reverse bias on the diode without damage. Another is forward current which is power dissipation related.

Fifth, a capacitor is an energy storage device and is used to smooth out the ripple from the rectifier. For the same ripple a half wave rectifier requires more capacitance than for full wave, a three phase less, and even lower for six phase.

Out of time for now.

.

[Runner4404spd]

i appreciate the the run through on the electrical stuff. i do have a masters in mechanical engineering and i have take a couple EE courses but that was a while ago. i am handling all the mechanical stuff on the conversion, my dad is helping with the electrical side of things since he has a PhD in electrical engineering. i just like to know the basics so we can at least have a somewhat educated conversation.

thanks again.

[gar]

050721-0854 EST USA

Continuing my comments.

Sixth, When you add a capacitor in parallel with the previously mentioned resistor you have simple single RC network. The voltage discharge curve of the capacitor with respect to time is an exponential with a time constant of RC where time is in seconds C is in Farads and resistance is in Ohms. The voltage change in one time constant is 63% of its initial value. As a rough approximation you can treat this portion of the curve as a straight line. A closer approximation to a straight line is at 1/10 time constant where the change is 9.6%.

In a 60 Hz system (USA) the time between positive peaks of the sine wave is 1/60 second or 16.67 milliseconds.

In a half wave rectifier the capacitor gets charged to the peak voltage once every 1/60 second. In a full wave rectifier it is every 1/120 second ( 8.33 ms ). In a three phase rectifier it is every 1/180 second ( 5.556 ms ) and a six phase is 1/360 second ( 2.778 ms ).

Without a capacitor both the half wave and full wave rectifiers have a peak ripple of Vpeak. A three phase has a peak ripple of ( 1-0.5 ) Vpeak. This comes from a 150 degree cross over point. A six phase cross over is at 120 deg and the voltage is 0.866 Vpeak at this point, thus peak ripple is ( 1-0.866 ) Vpeak or 0.134 Vpeak. Thus without any capacitor a six phase rectifier gives you 13 % ripple.

Back to a capacitor on a full wave rectified source. The time between charges is 8.3 ms. At 1/10 of an RC time constant we have a drop in voltage of 9.6%. Approximate this at 10% and assume that the curve is linear from 0 to 10%. So a ripple of 1% occurs at 1/100 of an RC time constant. Let K be the desired % ripple, then RC = 100 * 0.0083 = 0.83 . R = V/I so C = 0.83 * I / V for 1% ripple and C = ( 0.83/K ) * ( I/V ). For 5% ripple with 30 v and 10 amps this gives us 0.055 Farads, or 55,000 mfd. In actuality with ripple the time between charging shortens slightly, but this gives you the idea of what happens.

Seventh, on the PIV calculations assume the capacitor is charged to the peak of the source voltage and there is no ripple, then look at what maximum reverse voltage can be applied to the diode. Generally you want to use a diode with a much larger PIV rating than the peak you calculate because of transisents generated when switching a transformer off. Increased PIV is cheap.

Eighth, full wave center tapped vs full wave bridge. The bridge has twice the power dissipation in the rectifier as the full wave center tap, but makes better utilization of the transformer. For the same output power a slightly larger transformer is required for full wave center tap operation.

If you parallel transformers, then the phasing must be correct.

If you use a bridge rectifier and an output is connected to earth ground or equipment chassis, as distinguished from floating the output, then you must have an isolated ac source.

Ninth, capacitor input filters cause large pulse currents in the diode rectifiers, power dissipation is your primary concern here, but diodes are cheap so oversizing is not a problem. However, average dc current is still a good estimator of diode selection. As always running parts at lower temperatures improves reliability.

.

[ViperTX]

Tenth, Vpeak is not Vrms. AC VOMs measure Vrms. Just to add to gar's...

[DieGuy]

You know what is scary! I almost think I am starting to understand that!