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Topic: 24VAC - 5VDC Power Supply Design (Read 5473 times) previous topic - next topic

DavidOConnor

I need a 24VAC to 5VDC power supply for my project. Load is 100-200 mA, 24VAC source is a 30VA xfmr. Attached is my design, I'm a newbie to circuit design so any suggestions would be greatly appreciated. Thanks.

CrossRoads

Sooo old school.  7805 is gonna be really hot dropping 19V and more.
Get one of these instead:
http://www.pololu.com/category/131/step-down-voltage-regulators
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

DavidOConnor

Thanks CrossRoads. I have radio in my project, do I need to do anything special for filtering if I use a switching regulator? The voltage divider is suppose to give 9V to the 7805, do you think it will be higher? Can I lose the voltage divider if I use a switching regulator?

CrossRoads

Yes, ditch the voltage divider. That would only provide 9V at one current anyway, and would change as the current changed - effectively putting another resistor in parallel with 1K2.
Filtering - I don't know. Read the data sheet for the board you are interested in, see what switching frequency it uses.  Might need to put it in a grounded enclosure, and run shielded power cable to whereever the 5V goes.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

jackrae

#4
Apr 24, 2014, 08:14 pm Last Edit: Apr 24, 2014, 08:16 pm by jackrae Reason: 1
You MUST lose the resistor divider, whatever type of regulator you use.
Assuming your load is 200mA and you are using a linear regulator.
Current into regulator equals at least 200mA.  This current must come via the 1k7 resistor and at 200mA the voltage drop across the resistor will be I x R = 0.2 x 1k7  = 340 volts !!!   I think not.
Even without the resistor the regulator will be dissipating a fair amount of heat.  Say output of your bridge is around 30v, the regulator voltage drop will be 25v and heat dissipation will be V x I = 25 x 0.2 = 5 watts
Using a switch-mode regulator will reduce the input current and heat dissipation.

DavidOConnor

Thanks jackrae. I didn't know that the current in and out of the regulator was the same. I was thinking that since Vin > Vout then Iin < Iout.

DavidOConnor

Here is version 2. Using http://www.tracopower.com/products/tsr1.pdf
Does this seem reasonable?

CrossRoads

Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

jonwhite

have a look at the lm 2575 pretty good value from china 10 for $3.74

DavidOConnor

CrossRoads: Thanks for the thumbs up.

jonwhite: I looked at the 2575, but it requires external components and I did not feel comfortable that I could size them correctly.

Paul__B


Here is version 2. Using http://www.tracopower.com/products/tsr1.pdf
Does this seem reasonable?


Good switchmode module.

What's the idea of the extra 1N4004 and the two filter capacitors?

jackrae

The peak voltage from your 24vAC is 34 volts which is just about the limit for your intended regulator unit.  I suggest you measure the off-load 24vAC so ensure it doesn't actually exceed 24 otherwise you might seriously shorten the life of the regulator module.

The transformer is rated to 30VA and I suspect the off-load voltage may well be around 30 volts which will take the rectified output to 42 volts - well in excess of the module rating.

That being the case, might I suggest going for half-wave rectification and a suitable smoothing capacitor, say 200 microfarad  or there abouts.  That will drastically reduce the rectified voltage and your regulator module will be much happier.

jonwhite

#12
Apr 25, 2014, 01:49 pm Last Edit: Apr 25, 2014, 02:07 pm by jonwhite Reason: 1
hi david, i was in the same boat as you i needed a 400ma 5v supply so lots of heat to dissipate i stumbled on the 2575 there are only 4 components, i learnt loads making this circuit up especially getting to grips with data sheets give it a go,

i have the circuit already layed out in design spark if you want it, then go make  the pcb yourself      

tsr-1 blimy they are very expensive, there are much cheaper alternatives  there you go its a very simple circuit.

DavidOConnor

Paul__B:  The extra diode is there for reverse polarity protection. I was thinking of a modular design. The large cap is to smooth the rectifier output and the smaller cap is recommended in the Traco data sheet.

jackrae: The measured voltage is 28VAC. Won't the smoothing capacitor trim the peak voltage?

jonwhite: I'll take another look at the 2575. Can you recommend any other regulators?

jackrae

The purpose of the smoothing capacitors is to "fill in" the missing portions and so eliminate ripple.  With a "perfect" capacitor and no load on the circuit the smoothing capacitor would completely eliminate the ripple and the output voltage would sit at the peak value.  In the case of your 28vac system this would be (28x1.414) - 1.4 = 39.6 - 1.4 = 38.2v  (the 1.4 being the voltage loss across the bridge diode series pair)

You do not require the 4004 for your stated purpose since there is no way the completed circuit can "see" a reverse polarity on the input to the regulator

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