Linear power supply advice

[RogerThat]

Hi,

So, the winter is cold and I need something to play with which also heats me: Building a linear power supply 

This is my first venture into mains powered electronics so starting with something simple. I've been building electronics (up to 24v) for some years so I'm aware about the dangers playing with this stuff.

What I want to build is:
Mains powered linear power supply with fixed 5v and at least 4A. As high efficiency as possible with linear and ultra low noise. Cost comes low on the list....

I found this schematics which looks simple enough:
http://www.circuitstoday.com/regulated-dc-power-supply

I have a few questions:

How to dimension the capacitors C1 and C3?
Suggestions for diods in the bridge? I'm thinking the one with the lowest Vf, am I missing something?
What voltage should the transformer output give to be able to have a 5v output?(How do I calculate)

Thanks for any advice and education.


 

[Kleinstein]

The circuit shown is about 1960s  technology. It is neither high efficiency nor very accurate or stable. Depending on the zener, with lots of luck it may be low noise, but with typical parts it can also be quite noisy.

For low noise linear regulation is much easier, but this limits the efficiency. So high efficiency and lowest noise are kind of conflicting requirements.

A beginners project should avoid going for high power, high current, as there is more magic smoke to escape, but not much more to learn. So the first test should usually better be low power like 0.2 or 0.5 A.  Scaling up to higher current can than be some once the small version works.

P.S. was there any winter this year ?

[not1xor1]

I found this schematics which looks simple enough:
http://www.circuitstoday.com/regulated-dc-power-supply

forget it
-1) 2N3055 has really low gain. I'm unsure since I never used 2N3054, but I suspect that for 4A you'd probably need to add another BJT
-2) that circuit provides really poor line and load regulation, ripple rejection and NO current limit

As a beginner you'd better start with a lower power circuit. A LM317 based variable voltage PSU (e.g. 1.25-15V) would be easier to build and more useful.

[RogerThat]

Thanks for the feedback. I actually live in southern France now but my parents tells me the winter in Sweden is mild this year. Down here we had a proper snow storm and right now it's hovering around 0, which is cold for being here.

Regarding the schematic I was not thinking of using the exact same components but just the layout. I found it nice and simple with just a zener as regulator. Is there a known good regulator design or do most designs use an off-the-shelf LDO(directly or controlling a transistor)?

But before the regulator I need to have a DC which is steadily above the output voltage. This is really the main confusion for me now. Say I want 6vDC (minimum) for feeding my regulator. How do I arrive there?

In my head I can't use a transformer which peaks at 6v beacuse my output will then be too low after rectifying and filtering. How high does it need to be(the peak)? Is there a rule of thumb for this?

What I gather I should use a Toroid transformer which have the closest output voltage(after rectifying and filtering) to what I actually want to use. Let the transformer do the main part of the voltage lowering and then let the regulator just fin-tune it down to the correct voltage.

Btw, 20w power supply can hardly be 'high power'? . I'm aiming for 5v 4A because it will also be useful in my lab when it's finished. I will put two USB ports on it (2A each) and a normal bana jack output. I've built a switched 12v 10w USB charger, so this part I know well.

[andy3055]

Use something like this:

http://www.ti.com/lit/ds/symlink/lm338.pdf

You can size the transformer from the data given for the voltage differential.

[RogerThat]

My idea right now:

Transformer:VPT18-2780
Wire for 9v output

Diod: 2x. D10SBS4-7000
550mv Vf. Rated for 3.4A so putting two in parallel for heat/load spread.

Smoothing Cap: 4x 2200uF electrolyte
Read somewhere that 2500uF per amper was a good rule to follow. I have some 2200uF laying around so will use them. Will probably add a ladder of SMD also, to eat up higher frequency noise.

That's the plan for the transforming, rectifying and smoothing so far. Comments?

[schmitt trigger]

I have also heard the same rule of thumb for bulk capacitors: 2000uF for 60 Hz, and 2400uF for 50 Hz.

The latter is a non-standard value, thus 2200 or 2500 uF should be fine.
Ripple is removed electronically, anyways.

[tunk]

This setup should give you around 11-12Vdc after rectification and smoothing.
It may be better with a 7.5Vac or maybe 6Vac transformer.

[RogerThat]

Tunk, there is a VPT12 which can be wired for 6Vac. do you think this is a better choice? I would be nice to hear how you think/calculate about dimensioning this.

[not1xor1]

My idea right now:

Transformer:VPT18-2780
Wire for 9v output

Diod: 2x. D10SBS4-7000
550mv Vf. Rated for 3.4A so putting two in parallel for heat/load spread.

Smoothing Cap: 4x 2200uF electrolyte
Read somewhere that 2500uF per amper was a good rule to follow. I have some 2200uF laying around so will use them. Will probably add a ladder of SMD also, to eat up higher frequency noise.

That's the plan for the transforming, rectifying and smoothing so far. Comments?

you shouldn't use 2 bridges, just use a heatsink or better buy a 20A bridge to get lower Vforward...

With 8800µF caps the peak ripple voltage should be around 4V.
A 9V transformer should be OK... you should get around 7-7.5V unregulated DC (plus ripple) at max load.

You might use schottky diodes instead of a silicon bridge if the transformer voltage at maximum load is too low.

[not1xor1]

This setup should give you around 11-12Vdc after rectification and smoothing.
It may be better with a 7.5Vac or maybe 6Vac transformer.

it should be:
9V * sqrt(2) - (2 diodes drop at around 20A) - (max ripple)

if we are lucky and the AC line is not much below the nominal voltage, since the transformer is 50VA and should provide a bit more than 9V we get:

13 - 1.8 - 4 = 7.2V

[RogerThat]

not1xor1, thanks for your input.

Trying to learn here, how do you arrive at 4v? 'With 8800µF caps the peak ripple voltage should be around 4V.'

At the moment I'm looking at MC7805 for regulation and it states 2v drop...very little head room if the unregulated is around 7-7.5.

[tunk]

I stand corrected - what not1xor1 is saying is right.
You could do a web search for "rectifier ripple calculator", e.g.:
https://www.changpuak.ch/electronics/power_supply_design.php
https://www.fxsolver.com/browse/formulas/Ripple+voltage+%28full-wave+rectifier%29%29

[mariush]

When using a bridge rectifier  :

V dc peak = sqrt(2) x Vac - 2 x Vdiode  (Vdiode = voltage drop in a diode inside the bridge rectifier at the current desired)

Idc = ~ 0.62 x Iac  (0.62 is an approximation that usually works well for 50-200VA range transformers)

ex a 9v AC 50VA transformer will have :
Iac = 50 VA / 9 = 5.5 A 
Idc  = ~ 0.62 x 5.5A = 3.41A 

Vdc peak = 1.414 x  9 - 2 x 0.8v =  ~12.75v - 1.6v = ~ 11.15v

(so 11.15v x 3.41A = 38w .. maybe a bit more ... 0.8v x 2 x 4A = 6.5w lost in the bridge rectifier)

Keep in mind that most transformers (but esp. the low VA ones) will have a higher output voltage at low load... for example at  50-100mA the 50VA transformer may output 10..11v AC instead of 9v AC
Also keep in mind that your AC voltage may fluctuate depending on network load, time of day etc ... in a 230v country, at 2-3 AM I often see 245v AC at my outlet. So, account for these fluctuations on the primary side, which will affect the output on the secondary side.
For example, in the example above the formula says 10.75v, but in all further calculations it would be smart to consider 10v or 10.25v as minimum voltage, and around 12-14v as peak DC voltage (+1-2v at low load, ~+1v if your AC input is high)

Not applicable to you, as it would be too expensive (this would be useful for 20A+ rectification and higher voltages) but it's worth mentioning:
 
Instead of a bridge rectifier, you can use ideal diode controller and four mosfets : https://www.analog.com/media/en/technical-documentation/data-sheets/4320fb.pdf
This way you have less heat produced.. in your example, at 4A of current, the bridge rectifier would produce approx. 2 x 1v x 4A = 8w of heat.
The ideal diode controller would cause a voltage drop of a few tenths of a volt and maybe just 1w of heat (it really depends on the mosfets chosen)

Anyway... the size of capacitor can be estimated with formula :  Capacitance (in Farads) =  Current / [ 2 x AC frequency x (Vdc peak - Vdc min)]

So for example, let's say your "safe" Vdc peak is 10v and Vdc min is 7v (to have 2v above the linear regulator's output voltage) and you're in a country with 60 Hz mains frequency

C = 4A / [2 x 60 x (10-7)] = 4 / 360 = 0.011111 Farads or 11000 uF ... that's the minimum estimated.  So, I'd say 2 x 6800uF 25v caps in parallel would work, or 3 4700uF capacitors.

You can use more capacitance, but too much will cause a very high current when you plug the power supply (capacitors act as black holes sucking loads of energy when they're empty) and that can cause the fuse in front of your transformer to blow up, if you don't calculate the right value for it.

There are better linear regulators than the classic 7805, for example LM1084 is a good example: https://www.digikey.com/products/en?keywords=lm1084


It's also available in packages easy to heatsink, and also adjustable (use 2 resistors - or resistor + potentiometer - to configure output voltage): https://www.digikey.com/product-detail/en/texas-instruments/LM1084IT-ADJ-NOPB/LM1084IT-ADJ-NOPB-ND/363557
See the datasheet on the page for example circuits and detailed description of what components to use.
 
 

[not1xor1]

[...]
Anyway... the size of capacitor can be estimated with formula :  Capacitance (in Farads) =  Current / [ 2 x AC frequency x (Vdc peak - Vdc min)]

2 x AC frequency is for full wave rectification otherwise (half wave) it is just x AC frequency
that comes from Vpeak=f(C) => Vripple = I / (f * C) where f is the frequency of the rectified wave (double for full wave rectification)
for various reasons the real ripple is a bit less than the theoretical one so even with 8800µF you would get less than 4.5V...

You can use more capacitance, but too much will cause a very high current when you plug the power supply (capacitors act as black holes sucking loads of energy when they're empty) and that can cause the fuse in front of your transformer to blow up, if you don't calculate the right value for it.

AFAIK the problem with larger capacitance is a decrease in the power factor, with larger and shorter current peaks

BTW Rod Elliot wrote lots of useful information on his site... (about this and other subjects). I think that RogerThat may find those quite useful.

[RogerThat]

Really good information 

I played with the calculator and was struck by the large amount of ripple. For efficiency, would it be more beneficial to use many small capacitors to lower the ESR or will this give some issues? I'm thinking something like 11x 1000uF alu polymer would be better than my electrolyts.

The ideal diod solution might be interesting as the improvement is quite large, I will look into it.

[not1xor1]

I noticed Rod Elliot disagrees with what I've written and confirms what mariush wrote

I resurrected an old LT spice simulation and according to that (but the transformer model was quite rough) there is only about 10% increase in RMS current for an ten-fold increase in capacity...
so I was probably wrong 

[mariush]

Really good information 

I played with the calculator and was struck by the large amount of ripple. For efficiency, would it be more beneficial to use many small capacitors to lower the ESR or will this give some issues? I'm thinking something like 11x 1000uF alu polymer would be better than my electrolyts.

Modern electrolytic capacitors already have very low ESR, and very high lifetime.
For example, look at Panasonic FR datasheet: https://industrial.panasonic.com/cdbs/www-data/pdf/RDF0000/ABA0000C1259.pdf

A 3300uF 25v Panasonic FR capacitor has a 14 mOhm ESR value, 3.8A current ripple and it's rated for 10k hours

You're working with classic transformers and very low frequencies (60/120Hz) so polymer capacitors are overkill.
Paralleling so many capacitors is waste of money, you're better off with 2-3 electrolytic capacitors, or a single big one (but generally one goes for 2 or 3 capacitors with smaller volume.

The ideal diode solution might be interesting as the improvement is quite large, I will look into it.

Yeah, but look at it like this ... pay 2-4$ for the ideal diode controller and 4 mosfets (in volume), or you could pay 50 cents more for a bigger transformer (ex instead of 9v 50VA, go with 12v AC 50VA or 75VA) and that may also allow you to use smaller bulk capacitors (and save money there as well)

You get a cheaper product but you're losing efficiency (and more heat produced maybe) and weight (because transformer would have more copper) and the transformer may be a bit bigger (which may affect the case you use for the psu box)

[David Hess]

I would not recommend any power supply circuit which does not include short circuit current limiting.

As a practical matter, low noise designs will include a differential pair in the error amplifier and remote sense.  Something like the design shown below but with an LM338 replacing the LM109 is very high performance with low complexity.  The JFET can be replaced with a PNP transistor.  I made a tuned version with a 7805 and LT1007 that had a load regulation of better than 0.0001% which is beyond what common multimeters can measure.

[RogerThat]

Thanks again for interesting reads.

I checked price for bigger transformer and price for copper is high: double the current capability and price is up roughly 30% (10€).

Correct me if I am wrong but isn't ESR close to 1 ohm in the 50hz frequency? The stated ESR is usually at 100khz and as such much lower, in the mohm range.