build a simple 5v power supply for digital circuit

"robb"

Hundreds of watts, not 5 watts.

........ Phil

The characteristic of a capacitor is I=C*dV/dt.

From this, you can deduce that the voltage ripple is roughly T*I/C, where T is the period of the AC waveform. For a 60Hz supply which has been full-wave rectified, T=1/120s = ~8.3ms. So, if you are drawing 1A and can tolerate at most 3V ripple, you would need ~2800uF.

Using a higher secondary voltage will allow you to tolerate more ripple, but will increase the power dissipation of the regulator.

Also, a larger capacitor will result in a smaller conduction angle and increased RMS current (and thus power dissipation) in the transformer and rectifier.

"Nobody"

The value of the filter cap has almost no effect on the rms current igure - allowing only that the ripple voltage is under 25% of the peak value.

This is one of the silliest and most oft repeated myths in all PSU design.

Go try it out if you doubt this.

...... Phil

Thanks Phil, for all the answers robb

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Thanks again Phil, (in case you missed top post) for all the answers robb

It has a marked effect on the *peak* current, however.

You can't go wrong by making 2piFCR >= 25 or so. R is an estimate of equivalent load resistance which in your case should be about 8V/Imax, gives a conservative C, Imax is maximum load on the 5V output, F is

Yes- a ceramic 0.1u would be a standard for high frequency bypass...and this is in parallel with your electrolytic.

A conservative estimate is to select an electrolytic that prevents the output from going out of regulation when stressed with a no load to full load step increase in current. 5% of 5V is 0.25V so to be conservative you would want the e-cap to hold up a full load step increase for 500us without drooping more than that. Example, if full load is 1A then you have C,min>= (0.5u/0.25V)*1A=2uF, use a 4.7uF, e-cap values change a lot with temp and initial tolerance. This is in addition to your 0.1u high frequency bypass...

Yeah- that would not hurt- you mean a diode from the regulator output pin to its input pin with anode on the output. A 4001 would be good for this.

The back emf diode is the same as the short circuit protection diode- prevents blowing the regulator if the bridge filter cap sputters or goes short for some other reason.

Bleeders allow the high energy capacitors to discharge when the circuit is powered down- it would not hurt to put these at least on the bridge filter capacitor- usually a 1 MegaOhm or so...

Yep...

Not sure- that is just another form of back emf diode, and it would make sense if a second circuit was loading the bridge filter circuit in addition to the 7805...drawback is it reduces headroom, which is the differential voltage across the regulator circuit input and output...

You don't need it for a 7805 circuit...

Sure- there are tons of complicated mods you could do. Are you compiling a compendium, or is it you're not confused enough?

Real smart question, now you're starting to sound like a lamebrain..

The 7805 or LM340 series datasheets would be a good start. Most of the manufacturers now link to applicable application notes for the parts. It was so hard for people to figure things out for themselves...

And don't forget about the stepdown transformer rating, this could be important. To make a long story short, you need an 8VAC with RMS current rating at 2x the numerical DC maximum loading current. Then there's all that rigmarole about heat sinking the IC, max working voltages and ripple currents for the caps, and other stuff computed for worst case conditions which typically range over +/-20% nominal and some other things like fusing. Anything less gets you a pos that flakes out. Come to think of it, just buy a surplus open frame...or use a plug-in...

would be (0.5m/0.25V)*1A=2000u, make that 4700u...

Fred, You sound like some one I may have run into over the years, your last name is ringing a bell and I just can't seem to bring it to the top of my head.

If you wouldn't mind, could you tell me a brief location and history? I'm in CT , USA btw.

...... Phil

Jamie wrote:

would make

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output...

i think that is where i saw an example, a +/- 5v supply using

could be

RMS current

there's all

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plug-in...

currently just taking simple approach of using a 18/24 Vac 1 Amp wall-wart while experimenting with 5v psu

thanks for time and answers to the questions robb

I came across this universal psu design while searching for 5V psu info ... is there a problem with the universal psu schematic (page 2) shown in the following document ? more specifically with the selectable voltage part of the schematic

also i am trying to understand how that PSU can produce higher voltage than 5V while using a 7805 5V regulator and no path for the bridge rectifier (+) voltage to bypass the 5V regulator ?

thanks for any help robb

Hint: The 7805 forces 5V across R1. Where does the current through R1 go?

...Jim Thompson

(images on a.b.s.e)

i built and experimenting with the 5v DC psu i came across a strange (to me) oscope waveform.

using most basic psu {18vac in, bridge, 7805, 5v out} with a (470 Ohm + standard LED) power indicator on the 5v side

When i zoom in on the ?fuzz? it is a series of about 1 uSec wide spikes that are space about 10 uSec apart and the magnitude follows shape of the sine wave ? (see pic in a.b.s.e) thats the best this amateur can dexcribe

- add a .1uF cap between the -/+ bridge output

- remove the LED

- add smoothing cap to either side

any ideas to help understand what this fuzz is ?

thanks for any help , robb

If you'd read the whole article, it would have answered your question.

You really think so?

Try this:

Version 4 SHEET 1 880 680 WIRE 128 16 -304 16 WIRE 128 112 128 16 WIRE 128 112 80 112 WIRE 176 112 128 112 WIRE 288 112 240 112 WIRE 416 112 288 112 WIRE 80 144 80 112 WIRE 176 144 176 112 WIRE -304 208 -304 16 WIRE 80 224 80 208 WIRE 240 224 240 112 WIRE 240 224 80 224 WIRE 288 224 288 112 WIRE 416 224 416 112 WIRE 176 256 176 208 WIRE 240 256 176 256 WIRE 80 272 80 224 WIRE 176 272 176 256 WIRE 80 368 80 336 WIRE 128 368 80 368 WIRE 176 368 176 336 WIRE 176 368 128 368 WIRE 240 384 240 256 WIRE 288 384 288 288 WIRE 288 384 240 384 WIRE 416 384 416 304 WIRE 416 384 288 384 WIRE -304 400 -304 288 WIRE 128 400 128 368 WIRE 128 400 -304 400 FLAG 288 384 0 SYMBOL cap 272 224 R0 SYMATTR InstName C1 SYMATTR Value {C} SYMBOL res 400 208 R0 SYMATTR InstName R1 SYMATTR Value 10 SYMBOL voltage -304 192 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 10 60) SYMBOL diode 64 144 R0 SYMATTR InstName D1 SYMBOL diode 160 272 R0 SYMATTR InstName D2 SYMBOL diode 160 208 M180 WINDOW 0 24 72 Left 0 WINDOW 3 24 0 Left 0 SYMATTR InstName D3 SYMBOL diode 64 336 M180 WINDOW 0 24 72 Left 0 WINDOW 3 24 0 Left 0 SYMATTR InstName D4 TEXT -32 440 Left 0 !.step param C 1000e-6 10000e-6 1000e-6 TEXT 56 560 Left 0 !.tran 0 25e-3 15e-3 uic TEXT 80 496 Left 0 !.opt plotwinsize=0

shown in

schematic

voltage than

rectifier

question.

thanks for reply stephen,

your correct, i originally only read the circuit description so i went back and read entire article and really the only part relevant to my questions was the circuit description.

if there was an error in the schematic i think they would fix it before printing ? and i am sure the circuit description answers the question of how the circuit works for those familiar with divider junctions and regulator potential driving ? but my request is to understand how this variable voltage works.

AFAIK the (Pin 2) is 0V ground and the (Pin 3) is 5V out and the job of the 7805 is to maintain that relationship however in this application it talks about changing the voltage of (Pin 2) ground. i did not see where that was possible in the datasheet for a LM7805.

but more imporatant is the schematic (? error ?) that confuses me

i thought electricity takes path of least resistance ? that is if i take a 0 Ohm wire wire and i put a 1k Ohm resistor in parrallel with thta wire then the resistance would be 0 Ohms for tha t parralell circuit ?

in the schematic there is a piece (a segment) of wire in the bottom right corner of the schematic that ultimately connects the (-) output of the bridge rectifier to the (Pin 2) wire of the 7805 and connected along that wire is the ? resistor divider stuff ? it is the wire that goes by and around word "VARIABLE" to the right side?

Now i do not see how it is that any resistance value selected in the divider stuff is going to change the resistance between BR1 (-) out and the IC1 (G - Pin 2) ? because all the divide stuff appears to be ? shunted ? shorted by the lower right wire segment ?

any ways even if it was an error i do not see how any differential in voltage at (Pin 2) is generated using the BR1 (-) line out as that represents (0 V) to me as well ?

thanks agian for reply, robb

When in doubt, return to my hint ;-)

...Jim Thompson

Some 78xx regulators oscillate a bit especially with out the proper load on them. even then, you may need to place some by pass caps on it as you did.