EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Flump on October 11, 2013, 10:11:08 pm
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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
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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.
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awesome thanks very much :-+
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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.. :) )
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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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And since the voltage after rectification/smoothing is higher than the AC voltage, the maximum DC current you could draw from the regulator would be less than the AC current, about 0.6 * 500 mA.
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sorry i forgot to include the transformer link, here it is
http://www.ebay.co.uk/itm/271226751676?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649 (http://www.ebay.co.uk/itm/271226751676?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649)
looks like its no good then so back to the drawing board
thanks all.
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If you want 1-2amps 0-40v, you end up having to switch taps (input voltage to the regulator, typically done with relays, but I'm sure you could do it with actual switchs.) Either that or blow a *lot* of effort on cooling. Think big heat sinks, fans and multiple pass transistors to spread out the heat. It's unfortunately non-trivial.
There was a good Radio Electronics 50v/5a power supply article around 1990. I'll try to find it. Had all the bells and whistles, and IIRC the heat sink calculation.
And have a look at TI's LM317 datasheet, pg 19 and 21 to get some ideas.
EDIT: here it is:
"http://archive.org/details/radio_electronics_1990-03"
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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.
Note that the LM317 is a floating regulator which means it's the difference between the input and output voltage which matters, so it will be fine as long as the output voltage remains 40V below the input voltage.
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You can get exactly the same transformer cheaper here:
http://uk.farnell.com/pro-power/ctfcs20-20/transformer-20va-2-x-20v/dp/1780886?Ntt=CTFCS20-20 (http://uk.farnell.com/pro-power/ctfcs20-20/transformer-20va-2-x-20v/dp/1780886?Ntt=CTFCS20-20)
However, they have a minimum order of £20 ex vat.
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Note that the LM317 is a floating regulator which means it's the difference between the input and output voltage which matters, so it will be fine as long as the output voltage remains 40V below the input voltage.
Which happens if the output is shorted or forced in constant current mode by a high load. It may also happen if the overtemperature protection kicks in, something that is not unlikely with 300 mA maximum load if it has to dissipate a decent amount of voltage (eg. 20 V down to 40 V).
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You can get exactly the same transformer cheaper here:
http://uk.farnell.com/pro-power/ctfcs20-20/transformer-20va-2-x-20v/dp/1780886?Ntt=CTFCS20-20 (http://uk.farnell.com/pro-power/ctfcs20-20/transformer-20va-2-x-20v/dp/1780886?Ntt=CTFCS20-20)
Because top-quality-tools scrapes the farnell web site to generate his 234,508 ebay listings, adds a huge mark up, then gets farnell to ship direct to you.
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Note that the LM317 is a floating regulator which means it's the difference between the input and output voltage which matters, so it will be fine as long as the output voltage remains 40V below the input voltage.
Which happens if the output is shorted or forced in constant current mode by a high load. It may also happen if the overtemperature protection kicks in, something that is not unlikely with 300 mA maximum load if it has to dissipate a decent amount of voltage (eg. 20 V down to 40 V).
Yes you need to watch out for that. There are ways round it such as connecting a 39V zener from the output to input but it's probably not suitable for an experimental PSU.
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That better be some beefy zeners if they're going to survive a short on the output. I guess you might be able to put a transistor in series to switch off the power if the voltage across the regulator exceeds 40 V, but any robust workaround is probably going to be more trouble and expensive than an LM317HV.
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hi
beware the input voltage of the 317
good heat sink