Getting more current from power supply

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Edd
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Re: Getting more current from power supply

Post by Edd » Tue Mar 18, 2003 4:41 pm

Sir Grant:
Confirmed that you had received the heat sink insulative items that I had sent you for the motor driver cktry, . some of the larger insulators……...but not the silicone units……should also be compatible with your TO-3 plastic casings or just custom cut from the mylar sheeting.
I see that you are now in a quandary with its PS. Following down on the comments on the PS troubleshooting I didn’t ever see reference to the voltage rating of your filter cap. Also its not at all uncommon to see Sprague reforming their units…..a pristine unit with good seal integrity and with an inert gas fill can last one long time. Also you can check for code date of manufacture on the unit you have, if curious, I see one unit at hand marked as 7945 then followed by 8212 which would translate to a manuf date of the 45th week of 1979 and then a reform at the 12th week of 1982.
I experience a minimal ESR problem on electrolytics at LINE FREQ operation in linear supplies but is of utmost importance in switch mode supplies where they’re getting hammered by those high current square waves at ten’s to 100’s of khz rates. One quick spot check I do on equipment is to check the compression indentations
on the end cap for any bulging, a delayed forewarning of the unit having pressure buildup internally. Also a
simultaneous low tech /high info touching of a fingertip to that end cap to make a temperature check. High
ESR and hi temp go hand in hand , so time to cull out those units before rupture of that expansion crimp or the worst case…..explosion of a unit and dispersion of all its confetti foil and electrolyte (IF any is left) into the internal electronics.<p>Relevant to the transformers rating, which I am uncertain what it was actually rated at, I believe that and the filtering are your last hang-ups as Ron cued you in on the current sharing and its parts parameters in the reg circuitry. If you were in loss of the odd value of pot just put a smaller fixed res in series to get on up into the higher voltage range required and
thereby leaving the pots adjustment range deficient in the lowest voltage area which is typically of no interest to U.
On older Mil/Aero/Ind surplus transformers I would use the same technique that you finally used in dumping the raw filtered DC voltage from the Xformer/FWB rect/Filter(s) into a dummy load. Initially measuring the DC voltage unloaded I would expect it to drop 10% when it was loaded down to the specs on the transformers sec winding.
Lately I’ve seen transformers that drop down up to 20%. The voltage you experienced is shy of your desired amount considering the pass element losses in your regulatorcircuitry. If I were in the same situation and all of the transformers I have wound/modified….but primaries are NO fun……If this was an open exposed core design I would surgically cut across the heavy paper/fishpaper outer insulative covering of the bobbin and expose the sec winding which I would expect to be ~#18-20 ga (*) wire for those current specs and see if there was clearance in the window to accommodate one more winding layer which should just bring you up to that required extra voltage/power you need.Premeasuring your wire length and interleaving in the winding without the necc of core breakdown. I feel the core area mass would be tolerant of the few extra watts that you were shy of.Typically your application is not for continuous max pwr use. On the ripple I usually bring the primary voltage up on a Variac and just watch the output on a scope at the dummy load and watch to see if the filtering was adequate at that loading by seeing the percentage of that quasi-sawtooth present. If no scope, use the AC scale of a meter, but some cheaper meters act squirrelly with a DC component also present , so just read thru the isolation provided by a 1(+) ufd paper/poly/mylar of adequate voltage rating.
If you find your filtering to be deficient I see a prime 18,000 @ 75VDC cap with a ’97 code date in my bench... gratis if U need.
Another easier fix, if U happen to have the transformer on hand, is to take a unit with a 6.3 VAC secondary with a current rating of the same or greater current than you need and place its secondary….. series aiding…. with your main transformers secondary.
Addenda (*): Enamel/Formvar wire..... available from an electrical motor rebuild/rewind shop.<p>73's de Edd
[email protected]..............(Interstellar~~~~Warp~~~~Speed)
[email protected] (Firewalled-Spam*Cookies*Crumbs)
;)<p>[ March 19, 2003: Message edited by: Edd Whatley ]</p>

rshayes
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Re: Getting more current from power supply

Post by rshayes » Thu Mar 20, 2003 12:44 am

Watch out for ripple current in the transformer secondary. The current capability of a transformer is determined by the total temperature rise. Part of this, usually about half, of this is core loss, which is determined by the maximum magnetic flux density. This in turn is determined by the maximum input voltage, the supply frequency, and the number of primary turns. This is usually called the "core loss". The loss due to current flowing through the primary and secondary resistance is called the "copper loss", and can be calculated by the square of the current in the winding times the winding resistance. The limit on these sum of these losses is the temperature rise that the transformer can withstand. The current rating of the transformer is set to limit the maximum temperature rise to a value the insulation can safely withstand.<p>The catch is that this rating is root-mean-square (RMS) current and not average current. If the conduction angle of the rectifiers is 10%, a load current of 1 Amp will require that an average current of 10 Amps flow during the conduction time of the rectifiers. The RMS value of this current will probably be well above 10 Amps, and it is this current which heats the transformer and filter capacitor.<p>Adding capacitance to the filter reduces the ripple, but it also shortens the conduction angle and increases the RMS current. Tens of amperes of load current can result in hundreds of amperes of ripple current. I have seen power supplies where a 1 inch wide copper trace literally burst into flame due to this current.<p>Power supplies delivering heavy currents often use choke input filters instead of capacitor input filters. The output voltage tends to be about 70% that of a capacitive filter (average voltage instead of peak voltage), but the lower ripple current (since rectifier conduction may be continuous) may allow several times as much power to be drawn from the same transformer. Bridge rectifiers also help make better use of the transformer capacity compared to full wave rectifiers.

Bernius1
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Re: Getting more current from power supply

Post by Bernius1 » Fri Mar 21, 2003 4:31 am

Ron & Steven;
Will a large inductor AFTER the bridge-rec. store enough current to prevent peak current burn-out? 70% seems low. As the cap bleeds down, it would still have current to push (through the coil). No?
Can't we end all posts with a comical quip?

rshayes
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Re: Getting more current from power supply

Post by rshayes » Fri Mar 21, 2003 6:25 am

Both capacitors and inductors store energy. The energy in a capacitor is proportional to the square of the voltage across the capacitor and the energy in an inductor is proportional to the square of the current through the inductor. Capacitors resist sudden changes in voltage while inductors resist sudden changes in current. If a series of identical cycles are assumed (steady state conditions), the average current through a capacitor must be zero and the average voltage across an inductor must be zero.<p>In a choke input filter, the ripple voltage across the output capacitor is usually low. After all, that is what the filter is for. For part of the cycle, the rectified AC voltage is higher than the capacitor voltage, and the inductor increases steadily during this period. Assume that the capacitor voltage does not change during the cycle. If a full wave rectifier is being used, there will be two points during each half cycle when the rectified voltage is equal to the capacitor voltage. Assume that the rectified voltage is rising at the first point and falling at the second point. Between these two points, the input voltage is greater than the capacitor voltage, and the inductor current increases steadily during this period. At the end of this period, the inductor current is at its maximum value and the greatest amount of energy is stored in the inductor. At this second point, the input voltage starts to fall below the capacitor voltage, ant the inductor current starts to decrease. The current will decrease until the current reaches zero, or until the point on the next half cycle where the input voltage again equals the capacitor voltage. If the inductor is large enough, greater than a value called the "critical inductance", the inductor current never reaches zero, but continues to flow throughout the cycle. It varies between a maximum and minimum value. This continuous current flow greatly reduces the RMS ripple current and thus the stress on the power transformer and the filter capacitor. The penalty is that often the filter choke is nearly as large as the power transformer. This is compensated to some extent by the smaller size of the power transformer.<p>Since the current flow is smoother, the size of the the filter capacitor may also be smaller for the same ripple voltage.<p>If the inductor current is continuous, the average voltage of the input waveform must equal the average output voltage. The average voltage of a rectified sine wave is 2 over pi times the peak value of the sine wave (about .637 times the peak value).<p>As the load current decreases, the output voltage will remain at this value until the inductor current falls to zero during the cycle. For lower load currents, the voltage starts rising, until it equals the peak input voltage for zero load current.<p>A properly designed choke input filter can reduce the RMS current in the transformer to a value of slightly above to possibly twice the load current. A capacitor input filter may require an RMS current of five to ten times the load current (or more) which substantially increases the size, weight and cost of the transformer.

grant fair
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Re: Getting more current from power supply

Post by grant fair » Wed Aug 13, 2003 7:44 pm

Edd, stephen and no_vice - I was rereading this and discovered I had not read your posts, which were on page 2, in March. So a belated thanks to you all.<p>Grant
Grant

k7elp60
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Re: Getting more current from power supply

Post by k7elp60 » Thu Aug 14, 2003 7:36 pm

Grant,
I hope I read the spec's of your transformer
right. It is an 8A transformer. I have built a lot of power supplies of various currents up to 30 amps. A number of years ago a high current power supply was working, but the transformer was getting real hot. Contacting the manufacturer(Triad) I was informed that with a brute force filter the maximum current one can expect to get out of the transformer without overheating is rated currentX.566. In your case 8.0X.566=4.528 amps.
I have Hammond manufacturing power transformer spec guide and they show I D.C.=0.62X Sec. I A.C.
As soon as I used these guidelines the transformers always run cooler. It is my understanding the reasons for these limits is that the charging current is much higher. Another
rule of thumb I use is to figure 3000uf per amp. In doing so you will be able to get the dc voltage equal to the rms of the transformer.
Ned :p

grant fair
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Re: Getting more current from power supply

Post by grant fair » Fri Aug 15, 2003 11:52 am

Thanks Ned - yes, the transformer is rated at 8 amps. If I take higher currents out of it, it does get warm. Now I understand that the limit on current I am seeing with this transformer is normal; it is nice to know.<p>Grant
Grant

N9AOK
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Re: Getting more current from power supply

Post by N9AOK » Mon Aug 18, 2003 9:42 am

Grant, I tried the same thing with a 4amp rated transformer and could only get about 1.5amps out before its voltage dropped too low. <p>I assume you tapped your feedback from the base of your drive transistor? I made the mistake of trying to feedback from the output of the drive tansistor...the 317 does'nt go for that.<p>Todd Snyder

N9AOK
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Re: Getting more current from power supply

Post by N9AOK » Mon Aug 18, 2003 9:47 am

Grant, did you use the circuit on page 15 of the PDF?<p>Todd

grant fair
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Re: Getting more current from power supply

Post by grant fair » Tue Aug 19, 2003 8:47 pm

Hi Todd-<p>I used the circuit at the top of page 16, "five amp constant current/constant voltage power supply". The schematic is a bit incomplete, the details of this are in earlier posts on this list which helped me make the necessary changes.<p>Grant
Grant

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