supply ?

[mdreitzusa]

i got a good deal on some part and am trying to make a supply that will run my cnc.the parts that found are

transformer : 1-120VAC primary and 2 secondaries. One secondary is 24 VAC at 2amps and the other is 36VAC (center tapped) rated 750mA.
bridges are :
1-100v 2amp and 2-50v 4amp
diodes :
16-1n5400 and 25-1n4001
transistors : 12
tip31c
my control chips are 3 ucn5804b
can these parts be used to build a supply that will run my 3-pj55 steppers they are 5.6 ohm,1.2 amp,6.5v motors.i was hoping to run the motors at 24v.
i also have a 7805 regulater to run my chip and a 7812 to run my fans.got all the parts for about 30$.
what i am wondering is one transformer enough or do i need to run two in paralel.
thank for any help.
i need another kick in the rite direction.lol

[mdreitzusa]

after running some tests and building a power supply.i have finally got it nailed down.i now have 5v regulated,12v regulated those where a real pain to work out.to many years since basics in school.and i have a filtered 32.4v to drive my motors with.took a lot of picking my brain but i did figure out the bugs in my driver board to.thanks for letting me rack my brain and figure out what i should never have had to ask.

[gar]

050718-0625 EST USA

mdreitzusa:

You have not indicated where the source is for the 7805 and 7812.

The National Semiconductor data sheet sheet specifies an absolute maximum input voltage of 35 v. Use your judgement how close you want to work to this limit.

Your biggest problem from a high voltage source, other than the voltage limit, is power dissipation. Suppose you had 35 v input to a 7805 and loaded it to 1 amp, then the power dissipated in the regulator would be ( 35-5 ) * 1 = 30 watts.

It is more appropriate to source any 5 v series pass regulator from a source of 9 vdc at 120 vac input. When the input goes to 135 v then the 9 v goes to 10.1 v. At this overvoltage condition and a 1 amp load the 7805 dissipation is 5.1 watts, much more reasonable.

The dropout voltage for the 78xx series is 2 v. Thus at minimum supply voltage the the 7805 needs 7 v input. If we assume 95 v as our lowest ac input voltage, then the nominal 9 v supply will provide (95/120) * 9 = 7.125 v. Thus we are ok.

.

[mdreitzusa]

gar thanks for the responce.i need to start using more detail in my questions.the transformer has one primary and two secondarys.input is 120 , out is 24vac 2amp ,and a center tapped 36vac 750ma.i used the center tap and one leg of the 36v to feed my 12v,5v regulators.i had a problem with the 5v.could not get it below 15v so i put a 1k resistor inbetween the bridge and the reg.now i have 4.92v.the only thing it is for is to run the ucn5804 controllers.the 24vac witch tests at 31.9vdc.i also put a 10k bleeder across each of my caps.should i put a fuse between the supply and the boards just incase?it all look good with the meter but will know forsure when i connect the boards.here is my supply (zipped but in diptrace format .dch)

[gar]

050722-0750 EST USA

mdreitzusa:

The data sheet for the ucn5804 indicates a maximum Idd of 30 ma. Your 18 v will produce 25 vdc or more depending upon load. You are within the voltage and current limits of the 7805. The power dissipation would be about 20 * 0.04 = 0.8 w. Should not have been a problem with appropriate heat sink. Did you have bypass capacitors close to the 7805?

You could also feed the input to the 7805 from the output of the 7812, but this adds 30 ma load to the 7812.

If the load is actually 30 ma (note this is the max spec), then the voltage drop across 1k is 1000 * 0.03 = 30 v. Thus the load is not 30 ma. Does not look like the 7805 is doing much. Is it a good chip? Is it wired correctly?

Try a 300 or 330 ohm resistor. Does the 7805 regulate? If not something is wrong. Do this with no load.

On my Internet Explorer your posts have no space between period and next sentence, also no paragraphs.

I have no idea what a .dch file is, and my system has no software that responds directly to it.

.

[mdreitzusa]

sorry about the weird sentence stucture my writing never was my stong point. a .dch file is a schematic diagram for a program called diptrace. i found it as freeware.
i had a bad feeling that the 7805 may be bad after the first test run. the voltage tested to high to be normal. i think i still need to do some more work on it before i risk the rest of my cuircit.
sorry about the spelling not to good with that ether.LOL
thanks for the help i will do some more tweeking before the whole thing goes poof.

[gar]

060734-0722 EST USA

mdreitzusa:

If you hold a 7805 so you can read the part number and the three leads are toward you and the mounting tab is away, then the left lead is the positive input, the middle is common (negative) and also is connected to the mounting tab, and the right lead is the 5 volt positive output.

Put a 220 to 330 ohms resistor in series with the positive input lead, a capacitor from the input lead to common (see data sheet).

Put your voltmeter in ac mode with appropriate range, and measure the dc output of your power supply. If the meter has an input capacitor to remove any dc input when in the ac mode, then you will see a reading of zero or a small value from the power supply ripple. If in this mode you get a large reading, then look for another terminal on the meter labeled "output" or something that would imply an input capacitor.

With power applied, but no load on the 7805, the dc output should be near 5 v and the ac reading should be virtually 0. If the ac reading is not near 0, then the 7805 is probably oscillating. Working correctly the 7805 should not oscillate.

Next apply a 1k load on the 7805 output. There should be nearly no change in the measurements. Add another 1k, still should be no appreciable change.

See what this tells you.

.

[mdreitzusa]

gar- thanks for the test.

i did so more work on it and have come to the conclution that i killed it on the first test. i will go and get a new one and see if this fixes my wondering voltage. i will hook it to the output of the 7812 so i don't kill it to. i added 10k resisters across all of my caps for bleeders. my other voltages are 31.4vdc steddy no other load and 10.4vdc steddy with 2 cpu fans running.will this voltage help my stepper speed.

i made a driver board but got to wondering since my motors are only 1.2amp. would it be better to wire them bi-polar.the spec sheet for the 5804 has a listing for uni-polar and bi-polar. but if i did that were would i put the current limiting resisters for the higher voltage.

and the schematic i posted has a lot of mistakes. will have to work on that to.

[gar]

050726-0855 EST USA

mdreitzusa:

When you get your new 7805 do the test with the 200 ohm series input resistor from the 31 volt source, remember the capacitor. With a 500 ohm load ( two 1k in parallel ) at 5 v ( output of the regulator ) the load current is 5/0.5 = 10 ma. This is 0.01 squared times 200 = 0.02 watts. So a 1/4 resistor would work. However, this is not what you want to design for in the test circuit. The worst possible case would be 200 ohms across 31 v or 31 squared divided by 200 = 4.8 watts.

At 10 ma the voltage drop across 200 ohms is 0.01 * 200 = 2 v. Thus, this same 200 ohm resistor can be used to experiment from 12 v. The maximum load current from 12 with the 200 ohms in series is about (12 -5 -2)/200 = 25 ma. The -2 is for the minimum dropout of the 7805. So near 25 ma load you will start to lose regulation from the 7805.

Let us assume you might want 100 ma maximum from your 7805 and you have a source of 31 v, then ( 31 -5 -2 ) / 0.1 = 240 ohms. You could put a 5 watt resistor of this value in series with the 7805 and off load some of the power dissipation from the regulator to the resistor. Note, if you put a 100 ma load on the 7805 with its input directly connected to the 31 v source, then the power dissipated in the 7805 is (31 -5 ) * 0.1 = 2.6 watts.

.

[mdreitzusa]

thanks gar i will pick up the parts tomorrow.

[gar]

050727-0710 EST USA

mdreitzusa:

An additional note. Suppose you used the previously mentioned 240 ohms to limit power dissipation in the 7805. It may not be obvious, but maximum power dissipation in the 7805 occurs without a short on the 7805 and at a current less than 100 ma.

Maximum power transfer occurs when the source resistance equals the load resistance. Write the equations and solve for maximum load power. (edit 050728-0640) Without using calculus you can do this by plotting power dissipation in the load vs load resistance and observe the maximum point. Note at zero load resistance the current in the load is Vsource / Rinternal, and this squared times zero is zero power in load. At infinite load resistance there is zero current, and the power in the load is zero current times Vsource and that also equals zero power. Somewhere in between is maximum power in the load. (end edit)

Thus, if the source resistance is 240 ohms and the regulator is regulating, then the source voltage for the maximum power transfer is ( 31 - 5 ) = 26 v. The regulator will look like a 240 ohm resistor and the power dissipated in it is ( 26/2 ) squared divided by 240 = 0.704 watts. The current at this maximum power in the regulator is 13/240 = 54.2 ma. Greater current will produce less dissipation in the regulator. At the 100 ma load the regulator dissipation is 2 * 0.1 = 0.2 watts.

Compare this without the 240 ohm resistor. At 54.2 ma the dissipation is ( 31 - 5 ) * 0.0542 = 1.4 watts, and at 100 ma it is 26 * 0.1 = 2.6 watts as mentioned in my previous post.

.

[gar]

050728-0652 EST USA

A note on incorrect use of the power equation P = V * I.

Consider a three phase system with a purely resistive load with equal phase voltages, currents, and all sine shape waveforms, and V and I are RMS values.

Total power in the load is NOT --- 3 times Vleg-to-leg * Ileg . This is incorrect because the current in one leg is not in phase with the leg-to-leg voltage.

.