Multiple Power Supplies - Page 1
Forum section: electronics / beginners

author: johndon2000
author_rank: Contributor
timestamp: June 12, 2013, 02:05:45 pm
content: Hi,I have 2 power supplies but I need to obtain 3 voltage levels (+5V, +12V, -12V) to bench test a circuit board… is there a simple solution or do I have to use 3 power supplies?!Many thanks,John.

author: madires
author_rank: Super Contributor
timestamp: June 12, 2013, 02:11:06 pm
content: Quote from: johndon2000 on June 12, 2013, 02:05:45 pmIf the +5V rail doesn't need to be isolated you could add a 7805 (1A max, for 2A use a 78S05 ...) to the +12V rail.

author: Dave
author_rank: Super Contributor
timestamp: June 12, 2013, 02:11:47 pm
content: How much current do you need on those supply lines and what are your 2 power supplies capable of?

author: johndon2000
author_rank: Contributor
timestamp: June 12, 2013, 02:42:00 pm
content: Thanks guys- I need approx 3A per voltage rail; have a 10A/30V and a 3A/60V power supply...

author: Unknown
author_rank: Unknown
timestamp: June 12, 2013, 02:49:46 pm
content: A linear regulator would dissipate (12 V - 5 V) * 3 A = 21 W. That's a fairly hefty heat sink, assuming you get a regulator that can supply 3 A. The simplest solution I can think of would be a buck converter with one of those simple switcher ICs. Either that or an LM350 strapped to a CPU heat sink.

author: madires
author_rank: Super Contributor
timestamp: June 12, 2013, 02:54:21 pm
content: Quote from: johndon2000 on June 12, 2013, 02:42:00 pmThen you could beef up a 7805 with an additional power transistor or go for a buck converter like the LM2576-5.0 (3A) or LM2678-5.0 (5A)..

author: johndon2000
author_rank: Contributor
timestamp: June 12, 2013, 04:09:45 pm
content: Thanks guys, I'll look into the boost converter then

author: mariush
author_rank: Super Contributor
timestamp: June 12, 2013, 05:05:10 pm
content: Not a boost converter, a buck converter. Buck means from higher voltage down to lower. Boost means to change lower voltage into higher voltage.If your two power supply outputs are floating, you could set both to 12v and connect the positive of one to the negative of the other and that connection would basically become the ground and you have -12v - 0 - 12v . You can get the 5v using a 7805 (or a couple) from the +12v.if you don't want to keep your two power supplies busy, get a 24v AC 80-100VA transformer with center tap or two independent 12v secondary windings, a couple of diodes (or a bridge rectifier) and two large capacitors and you have unregulated 12-14v dc and -14 - -12v DC ... then use 7812 and 7912 (two-three of each or using external transistors) to have smooth 12v and -12v.5v you can get from 12v with a 7805.

author: Dave
author_rank: Super Contributor
timestamp: June 12, 2013, 05:05:40 pm
content: Buck converter, not boost.----Edit: Ha, mariush beat me to it.The most commonly used topologies of non-isolated switch mode power supplies:Buck (also called step-down) - reduces voltageBoost (step-up) - increases voltageBuck-boost (inverting) - increases or reduces input voltage, whatever is needed (this one does have the downside of inverting the input voltage)SEPIC - offers the capability to increase or reduce the input voltage without inverting (it does, however, need a more complex circuit)There are also other types, you can read about them onWikipedia.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 12:06:45 am
content: I would use one of your bench supplies for the -12 at 3A, parallel 3 LM7805's for the +5V on a fat heatsink and set the second supply to+12V which also powers the 3 7805 regulators. Unless you have a LM2576 or LM2678 laying around and the right 5-amp inductor, you might as well heat up the room a little with the 7805's for the time to do your testing.

author: Dave
author_rank: Super Contributor
timestamp: June 13, 2013, 02:12:07 am
content: Sorry, but paralleling 7805s is an awful idea.Two of them are just going to sit there, while one will be working overtime - until the thermal protection kicks in.You can increase the ampacity by adding one resistor and one power transistor into the circuit of a 7805. It does increase the effective dropout voltage, but you have plenty of headroom at 12V, so you don't really have to worry about it. You can find the schematic inLM7805 datasheet (page 24, figure 14 - High Current Voltage Regulator).

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 02:20:56 am
content: Dave, you might not quite understand how the paralleling of 7805 works. If you tried it, you would see it does. The answer is obvious, they must all work together...they are 1-amp regulators.

author: Dave
author_rank: Super Contributor
timestamp: June 13, 2013, 02:30:43 am
content: Please take the time to study the datasheet I linked. If you look at the block diagram, you will see that there is no current monitoring whatsoever, so these regulators will not switch to constant current once the maximum output current is reached. Therefore they will not equally share the load.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 03:29:38 am
content: Dave, I don't think you have this right.Assume that one regulator takes the lead in supplying current and its current output rises towards the current limit of the 7805, which is little more than an Amp, to 1.5A, depending on input voltage and device temperature. But well before it gets anywhere near to this current limit, it's output will drop a few mV and one or more of the other 7805 will kick in and no one would notice. There is current limiting in the 7805, that's one of the characteristics of this versatile regulator that makes it such a rugged device and makes them work as a team.Take three 7805's and try it.There is no need for them to equally share the load at low currents, but at the 3-amp requirement they will share to meet the load requirement.

author: mariush
author_rank: Super Contributor
timestamp: June 13, 2013, 04:16:02 am
content: Paul, what you highlighted only says that the regulator can do more than 1A, which is already established as true. Most can do up to 1.5A.However, you won't be able to easily make two or more 7805 regulators to both output the same current, one will go in current limit or something like that.LM317 datasheets have some high current examples, for example see "Typical Applications" in this datasheet:http://www.ti.com/lit/ds/symlink/lm117.pdfHave a look for example at page 24 where there's 3 lm317 working together with 0.2 ohm balancing resistors at output and an opamp controlling the adjustment of the three regulators.

author: Dave
author_rank: Super Contributor
timestamp: June 13, 2013, 04:24:10 am
content: I am still not convinced. A few millivolts drop when the regulator is loaded (datasheet states 9mV typical drop from 5mA to 1.5A) is nothing compared to the tolerance of the output voltage (couple of percent, depends on which regulator you select).I did, however, find a schematic of the LM78xx internals (attached below), and it does shine some light on your argument. It does actually have current limiting inside (I outlined it in red). Ha!I will test your theory this weekend, I'll report the results.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 04:27:32 am
content: Mariush, I think you are wrong about this. Take three 7805's and try it.One of the three regulators, if they are not closely matched, one of them might attempt to carry all the load and it will do so until the current limiting begins to kick in and it's output voltage drops. The other 7805s will increase their current share to regulate to the 5V output voltage.Because they are regulators, they will and must work as a team.There is no need for them to equally share the load at low currents, No one will know the difference. Even at 2 amps, maybe the third 7805 might keep its cool. But at the 3-amp requirement they will share to meet the load requirement because each will try to regulate to 5V. If any one tries to take more than it can handle, its output voltage will drop. The other 7805[s must sense the 5V is not 5V and kick in.

author: ptricks
author_rank: Frequent Contributor
timestamp: June 13, 2013, 10:31:39 am
content: It is never a good idea to connect the output of two regulators without something to isolate between the feedback input. Like was said the regulators each will try to regulate with a back and forth type switching. If you need to increase the current capability of a 78xx part you really should do it with a pass transistor.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 03:03:19 pm
content: PTricks, I think you don't quite understand how the 7805 works. There is no oscillation or shifting out of regulation. You simple get good regulation. In the situation of three paralleled regulators of the 7805 type, when 3-amps are needed, and 3 7805's are available, it is the optimal simple solution, a quick solution to the -12 +12 +5 at 3A problem. When 3-amps are needed, the 3 7805s will balance the load between them.Try it, get 3 7805's and see for yourself.It can be a good idea to not isolate the feedback inputsso as to create a mult-amp regulator with commonly found parts,There is no back and forth switching, (I would not notice any oscillation on the output) and the long term trend should be towards equal sharing because of 3A loading and temperature.

author: ptricks
author_rank: Frequent Contributor
timestamp: July 02, 2013, 01:11:15 am
content: Quote from: Paul Price on June 13, 2013, 03:03:19 pmA couple things you are not mentioning. 7805 parts are not equal, even from the same production run .If someone wants to parallel them they should try to find ones that match the closest and then should have resistors keeping the outputs separate.Monitoring the output for faults is also harder because if one 7805 fails you have no way of knowing that now two 7805 are regulating the load.This is one of those things in electronics like using a transistor in place of a diode, it can work but it is best to use a diode and do it right .I would not do this to generate a 5V 3A supply, a pass transistor is the way to do it right and way cheaper than (3) 7805.

author: Paul Price
author_rank: Super Contributor
timestamp: July 02, 2013, 02:04:26 am
content: ptricks, you are not quite yet getting the idea yet. You can always tell if one had failed, (which is very unlikely because each has a robust protection circuit) because the regulator would under full load fall out of regulation. But this is not going to happen. In any case, the current does not have to be exactly shared by these 7805's but they will tend to do that anyway. There is no need for an electron egalitarian utopia for 7805's circuit to work well.They do not have to be matched, the circuit current draw will match them. And if two regulators are regulating the load, then two regulators areregulatingthe load, but this is not how it will work, they will share the load.Where I come from 7805's are almost as cheap as transistors, and a lot more easy to find off the shelf.You guys seem to make every situation more complicated that it is, seems to think that every test setup it has to match the conditions for a 100000 unit production run of an aircraft altimeter.Hello..this is a simple test setup.Try connecting three 7805's and prove me wrong!Remember what the OP asked for, "is there a simple solution?"

author: Paul Price
author_rank: Super Contributor
timestamp: July 02, 2013, 04:07:41 am
content: Just to repeat a test I made many years ago, I just took 3 random choice On-Semi 7805 T0220 regulators purchased in a lot of 10 from Farnell and mounted them all on the same heatsink, no heatsink thermal compound, no extra time finding an insulating washer and insulating spacing material required if I was trying to make things hard for myself and need a TO-220 PNP power transistor or when the other suggestive alternative, sharing the real estate in combo on a heatsink with an LM317, (that combo of parts takes some time more to assemble that I didn't have to spend).Used for input an 18.5V 4 amp laptop P/S, )didn't have 12V or 15V handy), but this is even more stress on the 7805's.Used a 220uF/25V cap to bypass the input V to the 3 musketeer 7805's, loaded them with a 2-ohm 25W iwirewound in parallel with some misc 2W resistors metal film to trim the combo to the approx 1.67 +- 5% ohm odd value,Measured the unloaded output of each reg before assembly and the results were 5.00V, 4.99V,4.98V. (These <.3% from 5V are typical and what I always expect to get from any modern production of 7805's)Attached a 2-in #22 wire form each 7805 reg output to a common +5V output point so as to function as a current measuring shunt to check results.Result: 4.98V at +/- .01mV(limits of my Fluke DVM) at timed test intervals, T=1Min, T=15Min T=40Min when I decided it was enough to prove the point, beautiful 7805 5V clean output on my 20MHz scope and some of the chill somehow disappeared from my work room.

author: c4757p
author_rank: Super Contributor
timestamp: July 02, 2013, 04:28:18 am
content: Quote from: Paul Price on July 02, 2013, 04:07:41 amSo did you use these shunts? How well did they share the current?

author: Paul Price
author_rank: Super Contributor
timestamp: July 02, 2013, 12:54:19 pm
content: Do the math. Those of you who have not taken algebra in school will remain in the dark on the individual current magnitudes.Total current 3Amp +-5% at 4.98V +-.01VThe resistors have an arbitraryunknownvery small resistance and I used the 2mv range on my Fluke.The readings were 1.4mv, 1.6mv 2.4mv and all readings +- 1 LSD digit.Remember, they have to share, but there is no reason they have to share equally for this test power supply circuit to work perfectlyThe point is, I was able to construct this whole, simple, test power supply, with only one hole drilled in less than 10-mins, including finding all the parts and drilling one hole.No messy heatsink washers and insulating spacers, no trying to find a suitable PNP transistor and an insulator for it, no terminal strips for wiring isolated parts or areas of the circuit and supporting potentiometers I did not need.The 7805's all worked together as a team in their simple and same bed, without even thermal paste as a blanket.I used small heatsink from a dead ASUS PC motherboard that once lived over the North Bridge chip and I drafted my 4-in desktop PC fan always sitting on the bench to blow away soldering fumes to cooling duty.The heatsink already had two holes drilled on it for attaching itself to the PC motherboard and I used these two holes for mounting two of the 7805's

author: fpliuzzi
author_rank: Regular Contributor
timestamp: July 02, 2013, 02:45:05 pm
content: The only experience I have with using linear regulators in parallel comes from this article at edn.comwww.edn.com/design/power-management/4331884/High-current-supply-uses-standard-three-terminal-regulatorI find that this technique comes in handy when I need something whipped up in a hurry for a quick test jig, etc (usually temporary). I always have a large quantity of low cost surplus linear regulators on hand.The article infers that hundreds of LM7812 linear regulators can be paralleled; although I'm not sure that this is really a practical solution if you need hundreds of amperes of output capability.Interesting, nonetheless.Regards,Frank

author: Paul Price
author_rank: Super Contributor
timestamp: July 02, 2013, 08:55:16 pm
content: fpliuzzi,This guy hasn't got it right either, there is no need for wasting parts and power with those extra diodes on the output and common point, these diodes just serve to add unneeded complexity as well as ensuring poor regulation..a poorly thought out circuit design.

author: fpliuzzi
author_rank: Regular Contributor
timestamp: July 02, 2013, 10:15:23 pm
content: I was always curious about the extra 1N4007 isolation diodes in EDN's implementation.Regards,Frank

author: Unknown
author_rank: Unknown
timestamp: July 03, 2013, 09:50:25 am
content: Given the huge amounts of bulk capacitance that might be present at these current levels, I would add the datasheet-recommended reverse-biased diode from output to input instead. The 7812 might be unhappy when a huge cap gets discharged through the regulator if the input gets shorted.

author: Paul Price
author_rank: Super Contributor
timestamp: July 03, 2013, 12:05:01 pm
content: alm, I don't think you quite understand how the 7805 works or the EDN circuit functions.There would not be any great discharge current through any individual 7805 regulator and the diodes would not help to protect if there was such an event.The diodes are a poor design idea.

author: mrflibble
author_rank: Super Contributor
timestamp: July 03, 2013, 12:20:45 pm
content: I should hope all the 7805 opinions are rather academic, since the OP will of course have taken the pragmatic path and use buck regulation. I mean ... he has 30V and 60V supplies available and needs 12V and 5V... Seems buck is a better match than linear.

author: Paul Price
author_rank: Super Contributor
timestamp: July 03, 2013, 01:29:02 pm
content: mrflibble, You are quite right, buck is better than linear,but it is not simpler, it is not the simple, "I need to get this job done and get to the beach!" solution.Remember what the OP asked for, "..a simple solution?"

author: Unknown
author_rank: Unknown
timestamp: July 03, 2013, 01:41:48 pm
content: Quote from: Paul Price on July 03, 2013, 12:05:01 pmDo you think that claiming everyone who disagrees with you doesn't understand the rather trivial circuit is an intelligent response?Assume you are designing the 100A supply the EDN article suggests. You might have local bypassing after every regulator and maybe some bulk capacitance to improve transient response. Say you have 1000 uF of capacitance in total, charged up to 12 V. Now the input gets shorted. Each 7812 contains an internal diode from output to input. The TI datasheet recommends a reverse-biased diode with over 10 uF of output capacitance. So we have a cap charged to 12 V, then 100 diodes in parallel, and then a short. Silicon diodes have a negative tempco, so the load sharing will be extremely poor. That's not a robust design.The series diodes also prevent this problem, but will affect regulation, as you stated.

author: johndon2000
author_rank: Contributor
timestamp: July 03, 2013, 02:07:44 pm
content: Thanks all, ended up using a DELL PSU off an old PC for the -12V, and then 2 power supplies for the +5V and +12V... much appreciated

author: Paul Price
author_rank: Super Contributor
timestamp: July 03, 2013, 02:16:45 pm
content: alm,Firstly, I think that anyone that doesn't understand this rather complicated argument about this "trivial" circuit is welcome to seek clarity and correct apparent confusion about what is actually happening in this circuit.I don't think you quite understand the problem,If any single diode attempts to take some more of its share of the input-shorted discharge of an output capacitor, then it's forward voltage rises causing all other diodes to be biased forward to share the surge current.The forward voltage of the protection diode rises in a few nSecs in respect to current, while the thermal 2.2mv/degC effect takes some much more time, making all the other protection diodes the shortest path that electricity likes to take.

author: mrflibble
author_rank: Super Contributor
timestamp: July 03, 2013, 03:11:01 pm
content: Quote from: Paul Price on July 03, 2013, 01:29:02 pmSure it is. *click* [Buy It Now] on ebay, go to beach. Continue going to beach for a few weeks. On the way back from beach grab cheapo buck regulator from the mailbox and integrate into design. Problem solved. After solving said problem put the rest of those LM2596 modules you bought in the parts box for future projects.Although to be fair, if that 3A current is sustained that might be pushing it with those ebay LM2596 modules without some additional cooling.

author: Paul Price
author_rank: Super Contributor
timestamp: July 03, 2013, 03:19:07 pm
content: mrgibble, You seem to have somehow moved to an universe where ebay transactions take place along with delivery at someting near the speed of light.You might take notice that the OP's post on this page shows he had the wisdom to find a simple, quicker way to the beach.

author: Unknown
author_rank: Unknown
timestamp: July 03, 2013, 03:23:50 pm
content: I'm not sure how "the speed of light" and "a few weeks" got conflated in your head. Ah well.

author: mrflibble
author_rank: Super Contributor
timestamp: July 03, 2013, 03:36:56 pm
content: Quote from: Paul Price on July 03, 2013, 03:19:07 pmTime for a new keyboard?QuoteI reside in the universe where cheap electronics from China typically takes about 2-3 weeks to arrive, people have a clue how component tempco affects current sharing, and where people ground their scope.QuoteI took notice of all the posts by the OP in this thread. He uses ready made switchers, which sounds like a good plan.

author: Unknown
author_rank: Unknown
timestamp: July 03, 2013, 03:38:20 pm
content: Quote from: mrflibble on July 03, 2013, 03:36:56 pmReality is a pretty good place to live in.

author: c4757p
author_rank: Super Contributor
timestamp: July 03, 2013, 03:42:57 pm
content: I'm not sure how "simple" and "guessing whether it works contrary to how it's supposed to - gee, I hope there are no unforeseen problems with this" got conflated in his head. Oh well.Using a linear regulator to get 5V from 12V at 3A is retarded. You're not making a bench power supply here. Using them at the edge of their rating contrary to the way the datasheet instructs is also retarded. And going along with "just trust me, it will work, I tried it" when you present completely incoherent arguments as to why this is would be supremely retarded on the part of the OP.It's hard to beat the long-term simplicity of doing it right the first damn time, and it's hard to beat the short-term simplicity of an LM2576 switcher module or similar.

author: Unknown
author_rank: Unknown
timestamp: July 03, 2013, 03:48:47 pm
content: Well, this bad advice would have only resulted in the popping of a few regulators, and not in bodily injury, so thank goodness for small blessings?