Fullwave +/- DC power supply design questions - Page 1
Forum section: electronics / projects

author: heatbreak
author_rank: Contributor
timestamp: January 13, 2016, 02:17:31 am
content: Hi there,If I know what DC voltage I need and how much storage capacitor I want to use, anybody know how to calculate the required transformer voltage.Thanks

author: retrolefty
author_rank: Super Contributor
timestamp: January 13, 2016, 03:29:21 am
content: Quote from: heatbreak on January 13, 2016, 02:17:31 amIt is calculable, but differs for half wave and full wave rectification,which do you wish to use?Size of required cap depends on the maximum full rated current of the transformer and maximum desired ripple voltage.

author: mariush
author_rank: Super Contributor
timestamp: January 13, 2016, 04:17:28 am
content: Easy. Instead of going from AC to DC, you do the steps backwards.A bridge rectifier will convert the AC voltage from a transformer and the result will be a peak DC voltage equal to Vdc peak = sqrt(2) x Vac - 2 x Vdiode (Vdiode = voltage drop on one diode in the rectifier)The Vdiode (or better named and written in datasheetsForward Voltageof the diode) varies with the amount of current going through the rectifier and how hot the rectifier is, usually datasheets have graphs or list the estimated Vdiode at a particular current amount. The most average bridge rectifiers will have a Vdiode of 0.7v-1vThe capacitor after the bridge rectifier smooths out the wobbly DC voltage coming from the bridge rectifier, simply put it makes sure the voltage never drops below a minimum value.This minimum voltage can be estimated with the formula : Capacitance (Farads) = Current (Amps) / [ 2 x Mains Frequency (Hz) x ( Vdc peak - Vdc minimum) ]That's pretty much all you need to know.Keep also in mind that if the mains voltage drops below the standard in your country, the voltage on the transformer secondary will also drop and then the peak DC voltage after the rectifier will also be smaller, so if the AC voltage in your country is of poor quality you may want to be conservative and choose a transformer that outputs a bit higher voltage on the secondary.Also keep in mind that at low power consumption, transformers (especially the ones with low VA rating) often output much higher voltage on the secondary side, compared to when the circuit is loaded. This matters when you choose the capacitor used after the bridge rectifier. For example, if you have 10v AC on secondary, you may use the formula and think the Vdc peak will be 1.414 x 10 - 2x0.7 = ~ 12.5-13v and think a capacitor rated for 16v is enough, but if your circuit is idle or uses just a few mA, your transformer may actually output more than 10v, could be as much as 12-14v AC.Let me give you a practical example:Let's say I'm in US where mains is 110v and 60Hz and I want to a stable 12v at 1A. For that, I'm going to use a basic linear regulator, a bridge rectifier, a capacitor and a transformer.Here's the parts:Transformer :http://www.digikey.com/product-detail/en/VPS28-900/237-1283-ND/666169Bridge rectifier :http://www.digikey.com/product-detail/en/KBP206G/KBP206GDI-ND/278624Capacitor :http://www.digikey.com/product-detail/en/UVY1E472MHD1TN/493-12909-3-ND/4329108Linear Regulator :http://www.digikey.com/product-detail/en/L7812CV/497-1452-5-ND/585973The transformer outputs 14v AC or 26v AC depending on how you connect the two secondary windings. Obviously, in our exercise it's set for 14v AC. Since the mains may go down to 95-100v AC, I'm not counting on getting 14v AC all the time, so I'll use 13v AC for the formulas.The transformer is rated for 25VA which will mean a maximum of 25VA / 14v AC = 1.78A AC ... when you convert the AC to DC using a bridge rectifier, you can estimate how much current you'd be able to get with the formula Idc max = 0.62 x Iac max (0.62 is a constant that works well with small transformers) so that means there's going to be up to about 0.62 x 1.78 = ~ 1.1A , which is just about right. Again, 0.62 is just a number, it's for estimation.The bridge rectifier is rated for 600v and 2A. Check datasheet, it says forward voltage per element at 2A of about 1.1v but since our circuit is meant to output maximum 1A, you look at the second page of the datasheet at Figure 2 where there's two curves, one for 25C and one for 150C. With 1A flowing through the rectifier, it will be hot at around 50-60C so the actual forward voltage per diode will be between those curves ... let's go with 0.8v for the 1A we want at 12v.So now if we go with the conservative 13v AC of the transformer and using the 0.8v per diode then we have a peak DC voltage equal to Vdc peak = 1.414 x 13v - 2x0.8v = 16.78v , so let's go with 16.5v.The L7812 linear regulator needs at least 2v above the 12v to regulate properly, so we have our Vdc min of 14v .. therefore now we can use the formula C = I / ( 2 x Freq x (Vdc peak - Vdc min)) to make sure the capacitor is chosen right :C = 1A / 2 x 60Hz x (16.5 - 14) = 1/300 = 0.003333 Farads or 3300uF... that's the minimum. The capacitor chosen is 4700uF so it's a good choice, it's more than enough. We could have used 3900uF, another standard value.So let's say you go backwards and let's say you use the same linear regulator but you need to use a 1500uF capacitor. That's 0.0015 Farads ... so you have0.0015 = 1A / (2 x 60 x ( Vdc peak - 14v) ) ---> 0.0015 x 2 x 60 x (V dc peak - 14) = 1 ---> Vdc peak - 14 = 1 / 0.18 ---> Vdc peak = 14 + 5.555 = 19.5vNow you go backwards and add the voltage drop of your two diodes back and you divide by 1.414 and get the AC voltage you need : Vac = (19.5 + 2x0.8 ) / 1.414 = ~14.9 .... so you'd have to use a transformer that outputs 15v AC on the secondary, but ideally you'd go for 16v or 18v AC

author: m12lrpv
author_rank: Regular Contributor
timestamp: January 13, 2016, 04:30:58 am
content: Use this calculatorhttp://www.trans-tronic.co.uk/rectifier/

author: heatbreak
author_rank: Contributor
timestamp: January 13, 2016, 05:09:03 am
content: Thanks for all the replies. Haven't have time to read them yet. I'm looking to build a simple unregulated Fullwave AC to DC ps for a 2-channel, about 130 wpc, audio amp.

author: rdl
author_rank: Super Contributor
timestamp: January 13, 2016, 06:48:52 am
content: Don't forget that the transformer output voltage will vary between no load and full load current.