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Electronics Forum / Basics / July 200624Vdc Power supply
Could somebody please help me with a very simple 24Vdc power supply, capable of delivering 20amps. Does not need to be a very clean supply, Just to drive 3 of 6amp dc motors. Hoping to just use 240ac to 24ac transformers, some single phase rectifiers, and capacitors. My problem is not knowing what size components I require. Ideally, something that I can build on at a later date for smoothing out the output without reducing its load capability. Do need to achieve 23.5 to 24Vdc at 20amps plus Thanks in anticipation "Ian Tedridge" > Could somebody please help me with a very simple 24Vdc power supply, > capable [quoted text clipped - 6 lines] > rectifiers, > and capacitors. ** No need for capacitors. > My problem is not knowing what size components I require. ** Really ? So you don't imagine a 24 V ( ideally 26.5 V ) 20 amp tranny would do? Just add one 40 amp bridge rectifier on a modest heatsink. If there is any chance one of the motors could stall, then add a 20 amp breaker in series as well. ...... Phil Thanks Phil, My thoughts on the capacitors, was just to clean up and stabilise the output a little. Something I've seen used before. Must admit, had not thought about a breaker for protection. Think I need to slow down a little and take stock of what I am trying to do. Cheers > "Ian Tedridge" > > [quoted text clipped - 23 lines] > > ...... Phil > Thanks Phil, > [quoted text clipped - 33 lines] > > > > ...... Phil Hi, Ian. Ditto on everything Phil said. If you'd like some control of the motor speed, though, you can drive the primary of the transformer with the output of a Variac like this (view in fixed font or M$ Notepad): | _ T1 | o--o_/ \o--. [quoted text clipped - 9 lines] | | | | | | '--------' '----' (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) But plain ol' rectified 24VAC should do the trick just fine. Cheers Chris >Could somebody please help me with a very simple 24Vdc power supply, capable >of delivering 20amps. [quoted text clipped - 12 lines] > >Thanks in anticipation Some more to think about . . . if you filter it you raise the voltage because with raw RMS DC 24V=24V, with filtered DC you approach the peak voltage which is 1.4142 X the RMS value, or about 36 volts (unloaded - the transformer will drop a little voltage as will the rectifier diodes) Transformer ratings are for some specified current and at some nominal voltage - Voltage can be plus or minus 10%. off nominal. And, you are using 20 amps. each diode in a four diode bridge will drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24 watts wasted power and requires a good heatsink. Use a center-tapped transformer and you cut the waste power in half - ditto the heatsink and size, and likely as not cost as well. I use an adjustable 24 VDC 10 amp to run some fans - the SCR's doing the controlling also rectify the AC to save on wasted power and keep the heatsink small. "default" > Use a center-tapped > transformer and you cut the waste power in half - ditto the heatsink > and size, and likely as not cost as well. ** Shame how you then *waste* the output capacity of the transformer instead - by loading only half of the secondary copper at a time. Very inefficient . ........ Phil >"default" > [quoted text clipped - 6 lines] > > Very inefficient . Correct. John >>"default" >> [quoted text clipped - 10 lines] > >John I see the errors of my thinking. Intrinsic lossses in the transformer cause the I sq R loss ot increase faster than the diode drop losses in that (24V) voltage range. At lower voltages the diodes waste more power than the transformer and a center tap makes more sense. "default" >>>** Shame how you then *waste* the output capacity of the transformer >>>instead - by loading only half of the secondary copper at a time. >>> >>> Very inefficient . > I see the errors of my thinking. ** Not really. > Intrinsic lossses in the transformer > cause the I sq R loss ot increase faster than the diode drop losses > in that (24V) voltage range. At lower voltages the diodes waste more > power than the transformer and a center tap makes more sense. ** Drivel. The "full wave, center tap" ( two diode) arrangement REQUIRES that a transformer's VA be DE- RATED compared to using full wave bridge.The reason is as I posted earlier - only HALF the secondary copper is being used at any moment. This applies to a transformer of any size. ....... Phil >The "full wave, center tap" ( two diode) arrangement REQUIRES that a >transformer's VA be DE- RATED compared to using full wave bridge.The reason >is as I posted earlier - only HALF the secondary copper is being used at >any moment. --- I disagree. For a resistive load and the same output voltage, current, and regulation, both transformers must use the same gauge of wire on their secondaries. However, since the CT arrangement requires that each half of its secondary conduct only half the time, it will _inherently_ have a higher VA rating than the FWB transformer, but some of its capacity will be unused. That is, it will run cooler than the FWB transformer for the same load. The same is true if the load includes a reservoir capacitor, as indicated by example 1 at: http://www.mcitransformer.com/i_notes.html  SignatureJohn Fields Professional Circuit Designer "John Fields" >>The "full wave, center tap" ( two diode) arrangement REQUIRES that a >>transformer's VA be DE- RATED compared to using full wave bridge.The [quoted text clipped - 4 lines] > --- > I disagree. ** Dig you own grave deep as you like if you wish ..... > For a resistive load and the same output voltage, current, and > regulation, both transformers must use the same gauge of wire on > their secondaries. ** Your SILLY false assumption is now plain. The **context** here, is where the SAME transformer is used in one or the other rectifier mode. Do try to improve your reading skills - John. Unconscious context SHIFTING is such a GLARING indicator of your congenital AUTISM. ....... love, Phil >"John Fields" >> [quoted text clipped - 22 lines] >Unconscious context SHIFTING is such a GLARING indicator of your >congenital AUTISM. --- Well, let's see... If we want to use the _same_ transformer to feed a full-wave bridge or as a center-tapped transformer feeding a 2 diode full-wave rectifier, then we need to have a transformer with two secondaries which can be paralleled or connected in series. Just to nail things down, let's say that we want to feed the rectified output of the transformer into a 12 ohm resistive load and that we need 1 amp through that load. That means that we need a 12VA transformer, and in the case of the FWB, the entire secondary needs to pass 1ARMS continuously. For a regulation of 10%, that means that the no-load secondary voltage will be 13.2V, the secondary's resistance will be: 13.2V - 12V R = ------------- = 1.2 ohm 1A and it'll be dissipating: P = I²R = 1A² * 1.2R = 1.2W Since both secondaries will be wired in parallel, that means that each one will have a resistance of 2.4 ohms and will be dissipating 0.6 watts. Now, if we wire the secondaries in series in order to use them center-tapped, when either side of the secondary is delivering current to the load its resistance will be 2.4 ohms, So, instead of: 13.2V | 1.2R | +-->12V | 12R | 0V the circuit now looks like this: 13.2V | 2.4R | +-->11V | 12R | 0V Notice that now, instead of 12V into the load, we'll have 11V, with one half of the secondary dropping the other volt. That means that we're now looking at about 18% regulation instead of 10%, and to get the regulation down to 10% we'll have to double the area of the wire in the secondary, which means _increasing_ the VA rating of the transformer using the center-tapped full-wave arrangement. No only that, each half of the secondary will be dissipating: (13.2V - 11V)² 4.84 P = ---------------- = ------- ~ 2 watts, 2.4R 2.4R But for only 50% of the time, so that's about 1 watt. Since the paralleled secondaries will only be dissipating about 0.6 watts each, that means the series-connected secondaries will be running hotter that the parallel connected ones, requiring the series connected arrangement to be rated for greater, not less, than 12VA. That is, unless I missed something... SignatureJohn Fields Professional Circuit Designer "John Fields" = desperately needs a remedial reading class >>>>The "full wave, center tap" ( two diode) arrangement REQUIRES that a >>>>transformer's VA be DE- RATED compared to using full wave bridge.The [quoted text clipped - 4 lines] >>> --- >>> I disagree. > Since the paralleled secondaries will only be dissipating about 0.6 > watts each, that means the series-connected secondaries will be > running hotter that the parallel connected ones, requiring the > series connected arrangement to be rated for greater, not less, > than 12VA. ** Which is IDENTICAL to my original comments. > That is, unless I missed something... ** English class at school ?? ....... Phil >>>>>The "full wave, center tap" ( two diode) arrangement REQUIRES that a >>>>>transformer's VA be DE- RATED compared to using full wave bridge.The [quoted text clipped - 12 lines] > > ** Which is IDENTICAL to my original comments. --- If you meant that one would have to run the transformer at a lower VA rating than it was rated for if the center-tapped arrangement was used, then I agree with you. However, the argument you gave, that the reason for that was that only half the copper was being used at any given instant doesn't seem to support that, since half the copper is being used, but only for half the time. Had you delved into it a little deeper, as I did, (since this is seb) your meaning would have been clearer and this exchange would not have been necessary. SignatureJohn Fields Professional Circuit Designer "John Fields" > "Phil Allison" > [quoted text clipped - 21 lines] > VA rating than it was rated for if the center-tapped arrangement was > used, then I agree with you. ** That IS what the term " DE -RATED " means !!!!!!! > However, the argument you gave, that the reason for that was that > only half the copper was being used at any given instant doesn't > seem to support that, since half the copper is being used, but only > for half the time. ** 50% duty cycle at DOUBLE current = a net doubling of copper loss. I squared R is not that hard to understand is it ? > Had you delved into it a little deeper, as I did, ** The only thing YOU are deep in is your own excreta. ....... Phil >> --- >> If you meant that one would have to run the transformer at a lower [quoted text clipped - 9 lines] > >** 50% duty cycle at DOUBLE current = a net doubling of copper loss. --- That's true, but you can't achieve that by using the same transformer in a series or parallel configuration because of the effect of the load resistance. Consider: E1 | R1 | R2 | GND Where E1 is the open-circuit output voltage of the transformer, R1 is the resistance of the secondary, and R2 is the load resistance Let's set conditions up such that the open-circuit voltage is 10 volts, RMS, the secondary resistance (both windings in parallel) is 1 ohm, and the load resistance is 10 ohms. In that case, the current out of the transformer would be: E1 10V I = --------- = ----- = 0.909 amperes R1 + R2 11R and the power dissipated in both secondaries would be: P = I²R = 0.826 * 1R = 0.826 watt or, 0.413 watt per secondary. Now, if we connect the secondaries in series and use each one of them half the time, we wind up with this for each secondary: 10V | 2R | 10R | GND so we now have a current in the "hot" winding of: 10V I = ----- = 0.833 ampere 12R and a dissipation in the winding of: 0.833A² * 2R P = -------------- = 0.694 watts 2 Note that 0.694 watts is _not_ twice 0.413 watts. :-( --- > I squared R is not that hard to understand is it ? --- Apparently, it is... --- >> Had you delved into it a little deeper, as I did, > > ** The only thing YOU are deep in is your own excreta. --- Could be, but at least I can see through yours! ;) SignatureJohn Fields Professional Circuit Designer "John Fields" >>** 50% duty cycle at DOUBLE current = a net doubling of copper loss. > > --- > That's true, but you can't achieve that by using the same > transformer in a series or parallel configuration because of the.. ** GIANT Yawn ....... Wot a tedious w.nker is this Septic Tank prick from Austin Texas. " He's a Long Tall Texan ... and he thinks lot of crap .... He's a Texan and he's full of Crap ..... " . ....... Phil >"John Fields" >>> [quoted text clipped - 11 lines] > > He's a Texan and he's full of Crap ..... " . --- Well, it looks like the grave was for you! :-) Thanks for the easy win. SignatureJohn Fields Professional Circuit Designer "John Fields" ** GIANT Yawn ....... Wot a tedious f.cking LYING TURD is this ASD f.cked Septic Tank *c.nt* from Austin, Texas. " He's a Long Tall Texan ... and he thinks a lot of crap .... He's a Texan and he's soooo full of Crap ... Oh Roy, oh Roy - is that your horse ? " ....... Phil >"John Fields" > [quoted text clipped - 8 lines] > > Oh Roy, oh Roy - is that your horse ? " --- There once was a chap from down under who thought "I squared R" was a wonder. but try as he might from morning 'til night could not work it out, he'd just blunder. SignatureJohn Fields Professional Circuit Designer "John Fields" ** GIANT Yawn ....... Wot a tedious f.cking LYING TURD is this ASD f.cked Septic Tank *c.nt* from Austin, Texas. " He's a Long Tall Texan ... and he thinks a lot of crap .... He's a Texan and he's soooo full of Crap ... Oh Roy, oh Roy - is that your horse's arse ? " ....... Phil >"John Fields" > [quoted text clipped - 8 lines] > > Oh Roy, oh Roy - is that your horse's arse ? " --- No Dale, It's just Phil... SignatureJohn Fields Professional Circuit Designer >"John Fields" > [quoted text clipped - 8 lines] > > Oh Roy, oh Roy - is that your horse's arse ? " --- There once was a retard from Oz, whose rhet'ric brought nothing but bahs. So he stopped all the pain with a round through his brain, to a round of applause and hurrahs. SignatureJohn Fields Professional Circuit Designer >For a resistive load and the same output voltage, current, and >regulation, both transformers must use the same gauge of wire on [quoted text clipped - 3 lines] >some of its capacity will be unused. That is, it will run cooler >than the FWB transformer for the same load. Why would a designer use the same size wire in the CT secondary? Each winding is working at a 50% duty cycle so the required ampacity of the wire is cut by half - things being what they are, the transformer maker would save copper by using smaller wire. AND Looking at transformer ratings where they provide two independant secondaries the output will be rated 12V @ 2A , and 24V @ 1A (same game I would think) >>For a resistive load and the same output voltage, current, and >>regulation, both transformers must use the same gauge of wire on [quoted text clipped - 8 lines] >wire is cut by half - things being what they are, the transformer >maker would save copper by using smaller wire. --- And pay the penalty for the greater I²R losses in poorer regulation. --- >AND Looking at transformer ratings where they provide two independant >secondaries the output will be rated 12V @ 2A , and 24V @ 1A (same >game I would think) --- Except when you want to get 12V 2A by connecting the secondaries in series and using each of the 12V 1A windings half the time. See my reply to Phil's post for more detail. SignatureJohn Fields Professional Circuit Designer >>AND Looking at transformer ratings where they provide two independant >>secondaries the output will be rated 12V @ 2A , and 24V @ 1A (same [quoted text clipped - 3 lines] >Except when you want to get 12V 2A by connecting the secondaries in >series and using each of the 12V 1A windings half the time. --- That is, by trying to get 2A out of each secondary for half the time. SignatureJohn Fields Professional Circuit Designer > And, you are using 20 amps. each diode in a four diode bridge will > drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24 > watts wasted power and requires a good heatsink. Use a center-tapped > transformer and you cut the waste power in half - ditto the heatsink > and size, and likely as not cost as well. Incorrect. A half bridge configuration wastes transfomer utilisation. The I^2R losses increase far more than anything you'll save in a bridge. Graham >> And, you are using 20 amps. each diode in a four diode bridge will >> drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24 [quoted text clipped - 6 lines] > >Graham No way. > >> And, you are using 20 amps. each diode in a four diode bridge will > >> drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24 [quoted text clipped - 7 lines] > >Graham > No way. I'm afraid you're utterly wrong. Graham |
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