How to Speck out Transformer and Bridge Recifier

I have a project that for past versions I have been using a DC power supply. To take it to the next level, I would like to the AC to DC conversion on my PCB. This is my first time ever specking out parts to do this and I am wondering if I have the right process. I will take you through my thoughts.

I currently use a LM2596-ADJ switching voltage regulator with an input voltage of 12.10 VDC I saw on average a 78 mA current from my circuit. The highest peak I saw was a fairly rare 140 mA. So that brings me to a peak power consumption of 1.69 Watts (Atmega328p, SIM800 and some other small stuff).

With that I went to mouser and filtered for any transformer that accepts a 110-120VAC input and has anything between a 2-3.5 Watt rating. The cheapest one was the F36-065-C2 datasheet http://www.mouser.com/ds/2/410/36-065-C2-263502.pdf.

Now for the bride rectifier. I discovered that bridge rectifiers are cheap and can be small with a high amperage. With that I chose a large 1.5 A in a SMT package and chose the cheapest one the ABS15M RGG. datasheet http://www.mouser.com/ds/2/395/ABS15J%20SERIES_A13-335344.pdf

Okay then I drop a 470uF cap on the input line to my voltage regulator and I should be good to go right????

So my questions: Was my thinking correct with the transformer and rectifier? How big of a filter cap should I put on the incoming line? Is there a rule of thumb or calculation? Why is it that my 1 cubic inch tablet charger can put out 5 watts and I spec out a transformer twice that size and it is much lower power? How do they make it so small? In the end I want a cheap AC to DC conversion and not have it be too massive. (I can make the size of what I specked out work if needed)

Thanks for your help!

Since the capacitor is usually used at the input of a regulator, the current discharging it is constant. If powered by the AC line, assuming full-wave rectification, the capacitor must supply current to the regulator for 1/2cycle (8.3ms or 10ms, depending if it is 60Hz or 50Hz)

The Charge Q (Coulombs) removed from the capacitor is Q=I*t, where I is current and t is time.

Q also = C*ΔV, where C is the capacitance and ΔV is the voltage drop as the current flows out.

So C*ΔV = Q = I*t

Rearranging gives C = I*t/ΔV. | Where t = 1/2 period or (1/60)/2

For a 2A power supply, 60Hz, full-wave, where you can tolerate a 3V sag in the filter capacitor voltage without the regulator dropping out of regulation,

C = 2 * 0.008/3 = 0.0053F = 5300uF

(C = 2 * [(1/60Hz)/2]/3V =2 * 0.008 sec/3V = 0.0053F 0.0053F = 5333 uF )

Its not ripple that is important; it is how low does the voltage sag in order not to violate the dropout spec for the regualtor

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Basic Power Supplies

Why is it that my 1 cubic inch tablet charger can put out 5 watts and I spec out a transformer twice that size and it is much lower power? How do they make it so small?

When I was at the Bremerton Naval Ship Yard in Bremerton WA in 1980, (where the U.S.S. MISSOURI was moored for 25 years)

they showed me a two 6 foot high metal storage cabinets filled with vacuum tubes and told me that was 4 Flip Flops. (that were still working ) Of course I asked why they were still using them and they told me the Navy's policy was : "If it ain't broke, don't fix it !" I think they take that too literally.

Now you can get 32Gb in a tiny keychain fob . WHY ? That's progress . Your tablet charger is a switching p.s. Doesn't that answer your question ? (It's all about efficiency)

I suggest you consider using something like this: http://www.newark.com/myrra/47122/pcb-mount-power-supply-adjustable/dp/90R9579.

Edit: extra caution is needed during development when mains voltages are connected to printed circuit boards.

Hey , That's cool ! I've never seen one of those before. (I guess I'm giving away my age, ha ha)

You are right that is cool I didn't know they existed. Now the hard part is deciding if the added cost is worth it.

Transformer $4.50 Bridge rectifier 0.50 Lm2596 voltage reg 0.50 Inductor 0.50 other components 0.50 Total 6.50

Hmmmm is removing the headache worth an extra 6 bucks

Hmmmm is removing the headache worth an extra 6 bucks

I thought you said your primary concern was size.

Don't forget the heat sink for your regulator, it will have to consume the difference between the peak input voltage and output voltage, times the drawn current. The peak voltage is 1.4*effective voltage of the transformer, and the effective voltage should be at least 2V (+rectifier drop) of the regulated output voltage. For an 8V transformer and 5V output this means that the regulator has to dissipate more power than your 5V devices consume together.

A switching DC power supply provides a more stable voltage, not so much above the effective voltage, will have less weight than the transformer, and will operate at a wide range of AC voltages and frequencies. When I built 5V supplies for my first TTL computers, I soon started to pre-stabilize the input voltage, using thyristors or triacs at the primary or secondary side of my transformer, to get rid of the huge and hot heat sinks for the power transistors. One supply required to press an start button, in order to turn on the primary side triac, with the advantage that the supply shut down automatically on a shorted output :-)

raschemmel: I thought you said your primary concern was size.

Well I actually said

”In the end I want a cheap AC to DC conversion and not have it be too massive. (I can make the size of what I specked out work if needed)”

But you do make a good point that it will save me PCB space. I just have to figure out if the added price is worth the diminished space and pain of soldering on 8 or so extra components per board.