DIY power supply project going bad. Shorted caps.

[erxeto]

Hi,

I'm in the process of building a bench power supply, nothing fancy. It will be a small (in size as I don't have much space) and in power (I don't really need much, just 0-15V 0-1A).
Trying to cut some cents here and there I split the design in 3 boards. My idea was (is) take advantage of the ridiculously low prices on pcbs (usually less than 100x100 mm) for the variable regulator and the control board and use a perfboard for the power board, as it is a lot simpler.

I've attached the schematic for the power board (the other boards are not done yet, I was just trying to put the first part together).
The actual board is in my trash bin as the "debug" process went mad an I ended up desoldering all the caps just to realise that all 4 were shorted (they were brand new before soldering).

I've also blown 3 fuses in the process until I understood something was totally wrong ! How did I do that ? No idea, I'm totally puzzled, I followed the "tracks" 100 times and did not see any flaw, so I'm asking because I think it should be in the design itself ... (first time I play with transformer + bridge rectifier and all that). But I'm so blind by now that I see nothing.

If somebody sees something totally wrong with this, I would appreciate the help.
Thanks.

[Kalvin]

Did you check the cap's polarity?

[erxeto]

Yes I did, polarity was ok.

Just to give a little bit more info, the caps were those exactly:

https://www.tme.eu/en/details/sd1v478m1835mbb/85c-tht-electrolytic-capacitors/samwha/

[Twoflower]

Are you absolutely sure that the caps are shorted? What type of fuse did you use (fast or slow blowing)?

I'm thinking your oversized capacitor bank has a huge inrush current. That is blowing the fuse. Just use one 4700µF is sufficient. And you might use a slow rated fuse.

[Belrmar]

Right from the top op my head it could be a problem of the soldering procedure, maybe stress on the leads?

Did you check the caps right before soldering? maybe it could be shipping damage...

[Atom]

Well that's very strange  the voltage isn't the problem 17*sqr2 - 0.7*2 = 22.6V so you caps are fine.

i would reccomend to use a 12VAC transformer that will put out less voltage because right now the power disspiations on the ic is very large and you will have a lot of problems evan at 500mA.

for the caps 1 4700uF should be ok, and for the shorted capacitor that's again really strange, where did you buy them?, and also remember polarity and check how you connected you diode bridge
 
bye

[Delta]

How exactly did you test the caps to find they are shorted? A multimeter set to Ohms will see a large capacitor as a short. 

If your soldering and construction is OK, I would say your caps are fine and the inrush current is popping your fuse.  If this is the case, use one or more of the following:
Fewer caps.
A soft-start circuit.
A slow-blow fuse.

[abraxa]

Are you absolutely sure that the caps are shorted? What type of fuse did you use (fast or slow blowing)?

I'm thinking your oversized capacitor bank has a huge inrush current. That is blowing the fuse. Just use one 4700µF is sufficient. And you might use a slow rated fuse.

That's what I'd bet on, too.

Furthermore, I'd like to suggest you add another voltage regulator stage to spread the thermal load. You see, at 17VAC from the transformer and full wave rectification, you end up with 17V * sqrt(2) = 24V. Minus the 2 * 0.7V for the rectifier, you end up with 22.6V at the filter caps. Feeding that into the 5V regulator means that at maximum current, it'll have to dissipate 1A * (22.6V - 5V) = 17.6W in thermal power. Now sure what your thermal solution looks like but personally, I'd feel more comfortable breaking that up into more manageable pieces. One way to do this is to add, say, a 7815 that feeds the 7805. That way, the 7815 would dissipate 1A * (22.6V - 15V) = 7.6W and the 7805 would dissipate 1A * (15V - 5V) = 10W.

The prefered way to solve this would be to change the transformer for a lower voltage one but I'm assuming you are building the PSU around it

[Twoflower]

The voltage rating is a bit tight to my taste. Depending on the unloaded voltage of your transformer that 23V can be reasonably higher.

Not sure if this was right, but I might remember a rule of thumb: 1000µF per Amprere load. Unless there's some special need this sounds reasonable to me.

[erxeto]

Hi, thank you all for the replies.

The fuses: https://www.tme.eu/en/details/zcs-1.6a/fuses-5x20mm-fast/eska/520519/

I say the caps are shorted, because the multimeter says so in continuity ...

I thought about the inrush current thing (I'm a noob and know little about all this, but I've heard about it), that would explain the blown fuse ... but not the shorted caps. I assume they were ok before soldering (I just checked one that I replaced in the "debugging" process), but the fact is that they are all shorted after powering the board a couple of times (no load).

Does that mean that nobody sees a flaw on the design ? (a part from having too many capacitance). If that's the case I might try to order the board and take out of the equation the most likely problem (my hands !).

Cheers.

[Twoflower]

That fuse is a quick blow. And attach the multimeter several seconds to the uncharged(!) cap and you'l see that the resistance starts ton increase slowly as the measurement current will charge the cap.

THere's no flaw in your circuit beside the already mentioned problems.

[Kalvin]

Try with only one 4700 uF capacitor or 1000 uF capacitor (if you have one).

[The Soulman]

Can't see why your design wouldn't work, it is always good practice to take a multimeter and measure for continuity before powering up the board.
i.e make sure everything is connected correctly and there are no shorts.

As a comment on your design, using a 17V ac (assuming rms) transformer will give you close to 23 volts on your capacitors so use 35V ones or better.
Also remember that when using linear regulators the excess power is transferred into heat, 1A output means 1A input to the regulator,
so 23V at 1A equals 23W of power into the regulator, out comes 5V at 1A that is 5W and the difference of 18W is transferred into heat so you need a good size heat-sink for that.
Instead of connecting the input of the 7805 directly across the rectifier caps you could connect it at the output of the 7812 to spread the 18W.
Either way you need a big heat-sink.

As a general purpose power supply I would suggest you implement some form of current limiting to save your supply, your project and possible your house from damage.

Edit: good points made above, I took to long to post......

[The Soulman]

The voltage rating is a bit tight to my taste. Depending on the unloaded voltage of your transformer that 23V can be reasonably higher.

Not sure if this was right, but I might remember a rule of thumb: 1000µF per Amprere load. Unless there's some special need this sounds reasonable to me.

Yes that is the ballpark rule of thumb I remember as well, it would leave some ripple but is perfectly sufficient for this application (with the linear regulators and the more than adequate "headroom").

[Delta]

I bet you a pint your caps are not shorted.

I repeat: A multimeter on continuity will see a large (discharged) capacitor as a short!

Try powering a load through them (say 12vDC @ 1 amp, a car sidelight bulb or something) if you think they're short circuit.
Them thar caps be fine I says!

[Zero999]

Well that's very strange  the voltage isn't the problem 17*sqr2 - 0.7*2 = 22.6V so you caps are fine.

i would reccomend to use a 12VAC transformer that will put out less voltage because right now the power disspiations on the ic is very large and you will have a lot of problems evan at 500mA.

for the caps 1 4700uF should be ok, and for the shorted capacitor that's again really strange, where did you buy them?, and also remember polarity and check how you connected you diode bridge
 
bye

The trouble is, a 12V transformer would be too low for the LM7812.

If the LM7805 is getting too hot, add a series resistor and 100µF capacitor before it.

The voltage rating is a bit tight to my taste. Depending on the unloaded voltage of your transformer that 23V can be reasonably higher.

Not sure if this was right, but I might remember a rule of thumb: 1000µF per Amprere load. Unless there's some special need this sounds reasonable to me.

Yes that is the ballpark rule of thumb I remember as well, it would leave some ripple but is perfectly sufficient for this application (with the linear regulators and the more than adequate "headroom").

The trouble with that is it doesn't deal with what level of ripple is acceptable. The higher the voltage, from the transformer, with respect to the regulator, the more ripple is tolerable.

The approximate formula for 50 to 60Hz, with a fullwave rectifier is:
C = 1000*I_LOAD/V_RIPPLE

Where C is in µF, I_LOAD is in Amps and V_RIPPLE, the peak AC voltage less two diode drops, minus the minimum voltage the regulator will work from, before going into drop-out.

Yes I did, polarity was ok.

Just to give a little bit more info, the caps were those exactly:

https://www.tme.eu/en/details/sd1v478m1835mbb/85c-tht-electrolytic-capacitors/samwha/

Those capacitors do look too cheap. Try a decent brand, from a well-known distributor.

[Kalvin]

The smoothing capacitor size should be as follows:

C = 10000 * I_LOAD / V_RIPPLE [uF]

For example, if you want to have 1V ripple with 1A load, the required capacitance is:

C = 10000 * 1A / 1V uF = 10000 uF.

You can derive this easily from the capacitor voltage change dV when charged/discharged by a constant current I for given time t:

dV = I * t / C

For a 50 Hz mains frequency and 100 Hz rectified frequency time t is 0.01 s. When the discharge current is 1A and the maximum ripple is 1V we can calculate the required capacitance C :

C = I * 0.01 / dV = 1A * 0.01s / 1V = 10 000 uF.

The simulation is shown below. Note the high peak current of the capacitor / rectifier diodes!

[erxeto]

First, thank you all for your comments.

I'm now pretty convinced now that my caps are ok and the inrush current is the main factor here. I tested the resistance (not continuity) as Delta said, and sure, the resistance ramps up quickly as they charge.
If I have the time this week to put my hands on the soldering iron again, I'll do some tests and see how it goes with just one cap.

In the mean time, I've run a couple of simulations (I based them on my circuit, following the process described here: as this fellow has a similar problem ans explains it beautifully).

It turns out that, in my case, with the 4 caps in place and no load I get a melting integral value of 1.1077 A^2/s in the first 180ms this is too close to my taste to the rated 1.28 A^2/s on the fuse. And of course it goes over the chart with 1A load ...

I ran the same simulation with just one cap. The value is 0.265 A^2/s with no load and 0.724 A^2/s with 1A load ... which looks pretty good to me. Even the first 70ms which seem the worst ones are ok.

Live to learn !

[erxeto]

I had the chance to test it further.

I just soldered back in place one of my "shorted caps" (just one), and everything works nicely.

Now I'll have to take a look at the ripple, but I guess it will be ok for the final 15V.

For those of you who were concerned about power dissipation on the 7805 and the 7812, those won't be delivering much current (100mA tops, in fact I expect no more than 50mA, which means less than 1W on each). I need those rails to power the micro and a couple more things. The one that will suffer is the variable regulator  . But I'll try to keep it as cool as possible, and really I don't go over 400mA for more than a few minutes in my projects anyway ...

Cheers.