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transformer rating

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Flump:
hi All

I want to build a variable LM317 based psu and seen this transformer

if i join the secondaries together can i get 40v ?
and im a bit unsure of the va rating
as the advert says 500mA
and I thought 20vA was higher than that  ?

I wanted to be able to draw between 1 to 2 amps from it

Jebnor:
40V * 500mA = 20VA which is correct.
You can loosely think of VA as "AC watts".  It's not really that but it works for back of the envelope calculations.

Flump:
awesome thanks very much  :-+

Paul Moir:
Just a quick caveat:  I don't see the transformer, but if the output is 40v AC, then when you rectify it, it will be higher.  The RMS for a sine wave is sqrt(2) or 1.41...  So your peak voltage is 40*1.41 = 56.4v, which is what you'll have over your rectifer's filter capacitor if there's no load on the power supply.  That's too much for a regular LM317.  There is a high voltage variant though called the LM317HV or LM317HVK which is good to 60v. 

Also, dissipating those 20w is going to take some serious heatsinking and maybe a fan.  (EDIT:  too slow with the edit.. :) )

mariush:
I hope you know you can't get 3.3v-5v out of a LM317 with a 40v input  without making it super hot.

Linear regulators dissipate the difference in voltage as heat, so  if you want 5v @ 100mA (0.1A) , you'll dissipate (40v - 5v )  x 0.1A = 3.5w... with a basic heatsink, lm317 can dissipate about 1-2 watts  ... no way you're going to get 1 or 2A with some a voltage difference.

ps, since I just saw the message above mine... yeah a 40v AC rms 20VA  will give you a maximum of 40x1.414 - (2x voltage drop on rectifier diodes of about 0.8-1v) = 56.5-2 = ~55v . The maximum current would be about 0.62 x (ac current ) = 0.62 x 0.5a = 0.31A or about 310mA.  Better put those secondary windings in parallel to get a about 26 v DC and 600mA

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