Simple PSU question

I am building another generic PSU to pass power to 24VDC relays and 12V illumination lamps

I have doen some reading around and would like to make sure I am not frying myself

I have a spare 30VA 24v+24v transformer around...
Wire the transformer so it gives 24v and 0v

This is my idea...

+voltage Out of the bridge rectifier - split into two paths
- voltage to common ground

first path
1000uf (50v and upward) electrolytic to common ground
7824 with heatsink - common to ground
100uf (50v and upward) electrolytic to common ground
24v out

second path
1000uf (50v and upward) electrolytic to common ground
7812 with heatsink - common to ground
100uf (50v and upward) electrolytic to common ground
12 v out

common ground out

I was planning to wire the secondaries to give more current as I don't know what to do with the spare one...
Similar to Mark Burnleys explanation in this...

http://mysite.wanadoo-members.co.uk/deltaharmonic/Technical/Transformer_Connect.gif
The paralled secondary picture

Problem is this doubles the current - will the 78xx handle the larger amount (I am using some 6CW heatsinks)

The bridge rectifiers - no problem - I sussed that out a few months ago - I seem to be only building PSUs at the moment

Could you just expand on why I would need a 100ohm or so resistor of the 24v rail before the 7812's 1000uf capacitor? Is it to get the voltage down so the 7812 doesn't throw away all the extra volts in heat?
If so is it an ohms law thing to find the resistor?

Use the two 24V secondaries in parallel for more current.

Depending on how much current you're planning to use on the 12V line, you may not need a feed resistor at all - just some heatsink.

Also, use a single bridge rectifier and a common reservoir cap for the two voltages, and simply regulate that to 12 and 24V.

Calculate power loss, and mount an adequate heatsink.

Jakob E.

Jakob
Many thanks
it's a 30VA transformer
so the current is 0.625amps - does this mean that if I attach both secondaries it goes to 1.3amps
Reading the datasheets I see that the cuurent rating of the 78xx is 1A..

The rating of the 78XX is more to do with how much current you`re drawing through it with the load, than how much current the source is capable of supplying to it.

Yes.

And 78S12 (or 78S24) are 1.5A types. If you need that much current.

yes, it will give 1.3 A in parallel, the voltage regs will also supply around 1.5A max depending on brand/quality.

:guinness:

also it might be overkill for two caps before the Vregs, try 1000uf and 10uf tant and scope it to see if that smooths it out enough for you. the Vregs are not very picky but you also want to put 1000uf lytic and 10uf tant/film on the outputs, maybe sub the 10uf for .1uf if you have HF noise? best to scope it and tailor for lowest noise in the circuit.

it may be best to design with lm317/337 since these are more precise and offer less noise. It'll add a few resistors to the design but may be well worth the 5 minutes.

:guinness:

many thanks guys...
it's all been printed out and noted - and some of it actually sticks...

If your powering relays, don`t forget to use a diode reverse biased across the relay coils (particularly if you go the LM317/337 route) to prevent the back EMF destroying things when the coil is de energised.

[quote author="Rob Flinn"]If your powering relays, don`t forget to use a diode reverse biased across the relay coils (particularly if you go the LM317/337 route) to prevent the back EMF destroying things when the coil is de energised.[/quote]
I am using somebodys (buttachunks or NY Daves) general purpose relay driver circuit which has a 1n4148 in parallel to the relay and a 2n4403

i would use a MOSFET switch for the relay. you can get one with an integral body diode to freewheel the current/voltage generated by the dropping field. pretty much the same as adding the transistor/diode but one part instead of a few. I use them all the time and they work well. just be sure to pick on rated at least twice what the coil voltage is rated at and at least 5-10 times(guessing) what the current is rated. that should cover it for such a small inductive load as a relay.

:thumb: