Multiple Power Supplies

[johndon2000]

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.

[madires]

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?!

If 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.

[Dave]

How much current do you need on those supply lines and what are your 2 power supplies capable of?

[johndon2000]

Thanks guys - I need approx 3A per voltage rail; have a 10A/30V and a 3A/60V power supply...

[alm]

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.

[madires]

Thanks guys - I need approx 3A per voltage rail; have a 10A/30V and a 3A/60V power supply...

Then 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).
.

[johndon2000]

Thanks guys, I'll look into the boost converter then

[mariush]

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.

[Dave]

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 voltage
• Boost (step-up) - increases voltage
• Buck-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 on Wikipedia.

[Paul Price]

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.

[Dave]

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 in LM7805 datasheet (page 24, figure 14 - High Current Voltage Regulator).

[Paul Price]

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.

[Dave]

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.

[Paul Price]

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.

[mariush]

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.pdf

Have 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.

[Dave]

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.

[Paul Price]

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.

[ptricks]

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.

[Paul Price]

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 inputs so 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.

[ptricks]

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.   

A 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. 

[Paul Price]

 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 are regulating the 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?"

[Paul Price]

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.

[c4757p]

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.

So did you use these shunts? How well did they share the current?

[Paul Price]

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 +-.01V

The resistors have an arbitrary unknown very 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 perfectly

The 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

[fpliuzzi]

The only experience I have with using linear regulators in parallel comes from this article at edn.com

www.edn.com/design/power-management/4331884/High-current-supply-uses-standard-three-terminal-regulator

I 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