Bench Power Supply

[chopnhack]

I think its time I built a bench power supply to continue my learning of electronics. I have a few thoughts of what I might need, but would like to hear other more experienced voices in this area. I am looking to build not buy a basic psu - and lets not get into the whole linear vs. switching debate: I am a novice without an oscilloscope so I can't see the difference nor do I think I would be building circuits so sensitive that it should matter ;-)

Some specs that I thought would be good to have: 0-24VDC both +/- no more than 5A.

I came across this schematic and thought it was pretty cool:
http://www.eleccircuit.com/dual-power-supply-3v5v6v9v1215v-with-lm317lm337/

I would change the rotary switches out with perhaps two coarse and fine control multiturn pots, a pair each for the +/- rails.

What do you guys think? What do you use?

Thanks!

 

[KrisBlueNZ]

That looks reasonable. I like the idea of using trimpots to set the output voltages.

BTW I fixed your link so it points to the article. Previously it didn't work at all - the forum software included the trailing >> in the URL! You need to put the URL by itself.

The diodes in the bridge rectifier are under-rated at 1A. I would use 1N5404 (3A) or an integrated bridge (http://www.digikey.com/product-detail/en/GBU4D/GBU4DFS-ND/1055186 USD 1.00) instead of separate diodes. Personally I would use more smoothing - 3300 µF or 4700 µF for C1 and C2.

The circuit has the weakness that if there's a break in the switch and the voltage setting resistors, the output voltage will go to maximum, which could damage your circuit. To avoid this, you would need to use a "shorting" switch, aka "make before break", where the moving part makes contact with the next contact before it breaks the connection with the previous contact, as you rotate it. These are a lot less common than the "non-shorting" aka "break before make" type, where one contact disconnects before the next one connects.

Actually there's a simple design change that allows you to use either type of switch and will improve performance and reduce the parts list! Connect your six voltage-setting trimpots between the output and 0V, and connect their wipers to the contacts on the rotary switch. Connect the common pole of the rotary switch to the Adj pin on the regulator, as in the current circuit, and change R2 to about 22k or so. Repeat the changes for the negative regulator. The voltage setting trimpots should be multi-turn ones with a resistance of about 2k: http://www.digikey.com/product-detail/en/PV36W202C01B00/490-2880-ND/666507 (USD 1.51)

Edit: WRONG! Change R2 to about 22k and move the top end from OUT to the 0V rail, i.e. put it across C5. Same for the negative side.

The "non-shorting" aka "break before make" is the commonest type of rotary switch - see http://www.digikey.com/product-detail/en/KC26A30.001NLS/EG1954-ND/101797 (Digikey's cheapest suitable rotary switch at USD 5.50). This type has two poles, so you could switch both the positive and the negative regulators from the one switch, assuming you wanted the rails to always be symmetrical.

Also, the fuse and the switch should both be in the phase line! Phase wire from plug --> fuse --> switch --> transformer.

The regulators will need heatsinking. Either separately or on the same heatsink. You'll need TO-220 insulating kits unless you use separate heatsinks and put them inside the case (you'll need to drill ventilation holes in this case).

Apart from that, it looks good. Here's a suitable transformer: http://www.digikey.com/product-detail/en/VPS36-2200/237-1285-ND/666171 (USD 18.48). This transformer is rated for 80 VA which is a bit more than you need. The secondary voltage is 18-0-18 instead of 15-0-15 which I think is a good idea if you want to draw 1.5A out of each output. The smoothing capacitors (C1, C2) will have to be rated for at least 30V though.

 

[chopnhack]

Excellent analysis Kris, thank you for weighing in!

I don't like the idea of a rotary switch because it forces a set number of voltages, I would rather have a smooth range. I was thinking about removing the switch and associated resistors and replacing with 10 turn pots. I had thought perhaps having two pots in series for each rail, one for coarse adjust and one for fine adjust - I found a few that start at 100 ohm or 500 ohm and run up to 2.5k or 5k ohm: http://www.digikey.com/product-detail/en/7286R100L.25/987-1131-ND/2408709
They get kinda pricey at $13.60 ea. !! Perhaps I can use a combination of a 10 turn for the larger range and a standard pot for the smaller range?

I agree with the smoothing caps, I read somewhere about 2-3k uF per 1A of power. With more smoothing comes higher retained charge, do I need to be concerned with a load resistor to drain them when powered off? I assume I would need additional circuitry to put the resistor in series only when power is disconnected otherwise I would have parasitic loss and extra heat whenever the unit was running.

Should I include a secondary switch to allow me to connect the psu to test circuits while on but with the leads unpowered? I.E. - psu live, but test leads are switched off until desired.

Naughty, naughty, fusing the neutral ;-)

I can tell that its an old schematic, do you think the technology is adequate or am I building from the obsolete pile? Should I be looking at other designs?

From a user perspective, should I include a voltmeter/ammeter LCD for each rail or just rely on a second standalone multimeter?

TIA

 

[OLIVE2222]

You can made a cheap fine adjustment with two potentiometers in series. One close to the nominal value and one with 1/10, 1/20 of the nominal value for the fine adjustment.
You can also look at a so called tracking power supply, who both plus and minus voltages are tuned simultaneously, like here:

http://members.shaw.ca/roma/supply-2.html

Olivier

 

[KrisBlueNZ]

I don't like the idea of a rotary switch because it forces a set number of voltages, I would rather have a smooth range. I was thinking about removing the switch and associated resistors and replacing with 10 turn pots. I had thought perhaps having two pots in series for each rail, one for coarse adjust and one for fine adjust - I found a few that start at 100 ohm or 500 ohm and run up to 2.5k or 5k ohm: http://www.digikey.com/product-detail/en/7286R100L.25/987-1131-ND/2408709
They get kinda pricey at $13.60 ea. !! Perhaps I can use a combination of a 10 turn for the larger range and a standard pot for the smaller range?

Yes, as Olivier says, you can probably get away with two single-turn pots with resistances in a ratio around 30:1. The higher resistance one is the coarse adjustment. If you decide to go with that idea, I'll draw up a circuit. It's pretty simple. Also let me know if you want the tracking arrangement. I think it's a good idea.

With that circuit, the output voltage isn't linearly proportional to the pot position, but you'll probably find it acceptable.

I agree with the smoothing caps, I read somewhere about 2-3k uF per 1A of power. With more smoothing comes higher retained charge, do I need to be concerned with a load resistor to drain them when powered off? I assume I would need additional circuitry to put the resistor in series only when power is disconnected otherwise I would have parasitic loss and extra heat whenever the unit was running. Should I include a secondary switch to allow me to connect the psu to test circuits while on but with the leads unpowered? I.E. - psu live, but test leads are switched off until desired.

Normally, bleed resistors are permanently connected across the smoothing capacitors. The value is chosen so they don't get too hot and waste a lot of power while the supply is running. So they take a little while to discharge the caps. I think that's enough, personally, especially if you have the output switch. Actually you could combine them into a single three-position switch if you didn't mind switching mains voltage and output voltages with the same switch: http://www.digikey.com/product-detail/en/G-368S-0000/SW336-ND/164483

I can tell that its an old schematic, do you think the technology is adequate or am I building from the obsolete pile? Should I be looking at other designs?

Linear regulators are old technology, and they require plenty of heatsinking, but they're cheap and easy to work with. Replacements will probably be available for another 20 years or so. You don't need high efficiency, or small size, or particularly high output power. So I think it's a reasonable choice.

From a user perspective, should I include a voltmeter/ammeter LCD for each rail or just rely on a second standalone multimeter?

I would include them. They're available on eBay for very low prices. But there could be a problem with the ones on the negative rail because some of them use their negative supply rail as the negative reference for voltage measurement, and very few of them can work with either of their inputs outside their power supply rails. You might need a small extra mains transformer with one or two secondaries to supply the meters. It's probably worth it though. Or you could put a PTC resettable fuse (http://www.digikey.com/product-detail/en/E53493-000/E53493-000-ND/1045805) in the negative output (probably should have one in each output anyway) and not bother metering the negative side at all.

 

[Arouse1973]

Isn't the 317T like 1.5A. Wouldnt you need the 317K?
Adam

 

[KrisBlueNZ]

I think he's expecting 1.5A on each output.

 

[Arouse1973]

Oh ok.

 

[Harald Kapp]

The op wrote "not more than 5A", so probably a bit more than 1.5A are expected. I don't see where the LM318K should deliver more than 1.5A (typ), at least not in the datasheet. A boost circuit can be employed to deliver >1.5A.

You will need a very good heatsink: The voltage across the electrolytic capacitors will be ~20V. At a minimum output of 3V this is equivalent to a voltage drop of 17V across the regulator. At 5A this means 85W of power dissipation!
Even at 1.5A the dissipation is already 25.5W. Using a TO220 case you have Rth=3°C/W. Max. junction temperature is 125 °C. Assuming a ,ax. ambient temperature of 40 °C this leaves 125 ° - 40 ° = 85 ° for the case+heatsink. 85 °/25.5W = 3.3 °/W so the heatsink would need 3.3 °/W-3°/W = 0.3 °/W thermal resistance. That's going to be quite big (and this calculation neglects the Rth case-heatsink!).
Therefore the power supply should be derated (<< 1.5A) for low output voltages or the dissipated power needs to be distributed across more than one component by using e.g. a boost circuit (see above).

Also the electrolytic capacitors are a bit underrated at 25V. For better endurance I suggest using 35V rated capacitors.

 

[chopnhack]

Yes, as Olivier says, you can probably get away with two single-turn pots with resistances in a ratio around 30:1. The higher resistance one is the coarse adjustment. If you decide to go with that idea, I'll draw up a circuit. It's pretty simple. Also let me know if you want the tracking arrangement. I think it's a good idea.

I will look for some less expensive pots, do you think the 300 degree pots will have enough range to easily control the volt setting? The 10 turn are nice, but to sink ~$52 for 4 seems silly to me. Perhaps three of the less expensive in series? This is a 5k one that looks to be of good construction - http://www.digikey.com/product-detail/en/P160KNP-0FC25B5K/987-1311-ND/2408888 Only 84 cents! This is a 100 ohm pot at $5.34 http://www.digikey.com/product-detail/en/026TB32R101B1A1/CT2154-ND/203775

With that circuit, the output voltage isn't linearly proportional to the pot position, but you'll probably find it acceptable.

I noticed the non-linearity with the resistor to labelled voltage range and I agree that its not an issue. If I wanted to establish a scale on the faceplate, would I need a separate trimmer to adjust or "zero" the scale?

Normally, bleed resistors are permanently connected across the smoothing capacitors. The value is chosen so they don't get too hot and waste a lot of power while the supply is running. So they take a little while to discharge the caps. I think that's enough, personally, especially if you have the output switch. Actually you could combine them into a single three-position switch if you didn't mind switching mains voltage and output voltages with the same switch: http://www.digikey.com/product-detail/en/G-368S-0000/SW336-ND/164483

Excellent point - a high value resistor to gradually drain would be acceptable so long as there is an isolating switch. I would prefer a separate mains switch and two separate switches, one for each rail.

I would include them. They're available on eBay for very low prices. But there could be a problem with the ones on the negative rail because some of them use their negative supply rail as the negative reference for voltage measurement, and very few of them can work with either of their inputs outside their power supply rails. You might need a small extra mains transformer with one or two secondaries to supply the meters. It's probably worth it though. Or you could put a PTC resettable fuse (http://www.digikey.com/product-detail/en/E53493-000/E53493-000-ND/1045805) in the negative output (probably should have one in each output anyway) and not bother metering the negative side at all.

Good pickup on the voltmeter not working on the negative rail - is there a way to invert the voltage in to the meter? This very first one I found only uses <30mA. Working input V ranges from 4.5-30VD - why would we need a separate transformer? To isolate meters from fluctuations? At 80VA the transformer you selected should have enough room to power two of these devices as well as the rails, no? The PTC would be a nice feature in case of accidental shorting - I saw some of the commercial units had a LED warning lamp come on when a short was detected, would there be a simple way of implementing that when the PTC opens?
Here is one am/voltmeter I found that was reasonable:
http://www.ebay.com/itm/New-Dual-LE...928?pt=LH_DefaultDomain_0&hash=item2a3cfbf618

Should I include an AC voltage supply as well in the same box? So long as the transformer is adequately sized, it seems like a good idea to me.

 

[chopnhack]

Aside from the above questions, what would be the advantages of a dual tracking supply?

 

[SOLARWIND]

Good day,
Have a look at the LM350 (3A) and the LM338 (5A).
http://diyaudioprojects.com/Technical/Voltage-Regulator/

 

[chopnhack]

Thanks Solarwind, in retrospect, I believe for my first psu, 1.5A per channel should more than suffice. I have given a great deal of thought into what features I would like. I have read various information and I am now debating the merit of starting at 0V.

Do you think I should worry about being able to start at 0V or deal with the LM317's inherit limit of ~1.25V starting point?

 

[SOLARWIND]

HI,If it is for general use, I would not bother to start at 0V. However, in some instances you might want to. In that case, rather look at a power supply based on the old reliable and very stable 723 voltage regulator. I am on my way to work right now, otherwise I might have something in that line somewhere in my files. Meanwhile, google the 723 and see what pops up.

Good luck!

Johann.

P.S. Somewhere to start: http://www.circuitstoday.com/ic-723-voltage-regulators

 

[davenn]

I wouldn't go the 723 way unless you were absolutely desperate, they are not the easiest reg to use and are very old
And you are no where near that point yet

If you are not really worried about getting down to 0V and are happy with a supply that will go from ~ 1.7V and upwards then no hassles

If you are really looking at being able to draw up to 1.5A on a regular basis and sometimes continuously then you really need to use the
LM350 (3A) and the LM338 (5A) and their negative rail counterparts.
The last thing you want to be doing is running a LM317 to its max current. Its much better to have some headroom for the sake of long life of the regulator

cheers
Dave

 

[chopnhack]

Not really sure if I would need something that low, but you read various accounts and then it seems like a good idea!
Good point Dave, the LM350 would be a better choice, thanks for that thought!

 

[chopnhack]

I hope you guys can help me complete this project. I reviewed the original idea and many parts are still valid and necessary. I decided to look at more current chips, namely LDO regulators. Still not knowing enough to design a SMPS, I would rather stay with this simpler technology.

Here is a recent schematic that I will annotate. Can you guys review it and see if there are mistakes?

I went through the datasheet and gathered ideas. I hope that I implemented them correctly!
Please note that nothing has been spec'ed out yet, I wanted to see if I have the general layout correct before digging deeper into the values and other factors.


TR1 - thinking of using a 48VCT 2A 100VA transformer, that means I can have 2A max power per channel
TR2 - is a small transformer to power the dual voltmeter/ammeter.
R1/R2 - essentially bleed resistors, although R1 serves double duty for the PSU power 'on' light.
C1/C2 - most likely be about 50μF polarized AL electrolytics, literature states that they must be low ESR otherwise they will have to be tants
B1 - should this be rated for the 1/2 cycle peak of 24VAC or ?
R4/R5 - each of these will actually be two pots in series, a coarse and fine pot.
to be added - current adjustment pots - have to figure out how to add them into the circuit.

 

[Arouse1973]

Hey John
What's the second transformer for? The top one looks good, the only thing I can see is R2 is shorted out.
Cheers
Adam

 

[chopnhack]

Hey Adam! Thanks for the reply.

The second transformer is a small one to power two combination volt/ammeter LED displays. From what I read and as others have mentioned it needs to be powered separate from the power source. The black boxes at the bottom are the meters - 1 pair of wires for voltage sensing, other pair split the circuit to allow for measuring current, last pair, opposite is for powering the meter itself. One will meter the positive (top) rail and the other the lower negative rail.

I'm not sure I understand negative voltage very well. Where should R2 be placed if it is to be the bleed resistor for C2? Is current flowing from the common rail to the negative rail? If so, should the resistor be place between the bottom of C2 and the negative rail?

Thanks in advance

 

[Arouse1973]

Why do you need a bleed resistor? If you need a bleed resistor then place it across C2.
Adam