Transformer Rating - Page 1
Forum section: electronics / projects

author: irobot
author_rank: Newbie
timestamp: June 15, 2023, 01:57:22 am
content: Confused.How do I interpret the specs of this transformer?How much current available across the 12V winding? My schooling tells me .5A, since the windings are in series.The way it's labeled is confusing.

author: Circlotron
author_rank: Super Contributor
timestamp: June 15, 2023, 02:25:14 am
content: That is pretty dumb. The 6VA rating is the clue. 12V @ 0.5 amps.The only way you could get 1 amp out of that transformer would be to use a current doubler rectifier.https://www.ti.com/lit/an/slua121/slua121.pdf?ts=1686723659638&ref_url=https%253A%252F%252Fwww.google.com%252FEdit -> my statement is not entirely true. A diode from each end of the secondary will get you partway there but transformer utilisation is poor. Likely you will get about 3 watts DC from this 6VA transformer.

author: Ian.M
author_rank: Super Contributor
timestamp: June 15, 2023, 03:42:09 am
content: That's 0.5A RMS @12V. Average DC current from a bridge rectifier followed by a reservoir capacitor would be less - don't expect over 0.31A.

author: mariush
author_rank: Super Contributor
timestamp: June 15, 2023, 04:18:08 am
content: 12v 6VA => Iac = 6/12 = 0.5Arectified using bridge rectifier =>peak dc voltage Vdc peak = sqrt(2) x Vac - 2 x voltage drop of diode in rectifierI dc = ~ 0.62 x Iac = 0.62 x 0.5 = 0.31A - edit : 0.62 is a constant that works well for transformers in the 20VA or higher range, smaller transformers may do a bit better -Capacitor after bridge rectifier defines the minimum dc voltage, value can be estimated with formula C = Current / [ 2 x AC Freq x (Vdc peak - Vdc min desired) ]ex Vdc peak = 1.414 x 12 - 2 x 0.5 (if using1n58171n5819 diodes) = ~16v Idc = 0.3A-edit : 1n5817 have too low voltage at 20v to be safe, go with 1n5819 with 40v max voltage, or other diodes ex 1n400x or a ready made bridge rectifier -AC Freq = 60 Hz in 120v AC / 60 Hz countries, 50 Hz in 2xx v countries , assuming min desired is 14v :C = 0.3A / ( 2 x 60 x (16-14) ) = 0.3 / 240 = 0.00125 Farads or 1250uF ... so a 1500uF capacitor would be sufficientIt would be have to be rated for minimum 35v as at no load transformers often output 10-20%extra voltage and at nights your AC voltage may be much higher than nominal voltage (ex 240v instead of 220v or 125v instead of 110v) so the idle voltage could get very close to 25v instead of a peak of 16-17v

author: Kleinstein
author_rank: Super Contributor
timestamp: June 15, 2023, 09:53:49 am
content: The ratio of DC to AC current is not a fixed number. It tends to be a littel better with small transformers as there is more sereies resistance and thus a slightly better power factor. The size of the capacitors also ave a little effect and if needed one can do some power factor improvements (e.g. get a little extra current via charge pumping).On the other side the voltage goes up more for small transformers when unloaded. With only 6 VA I would expect more than 20%, more like 30 %. It gets really bad below 3 VA.The transformer rating also is not a brick wall. It just the the current that causes a certain increase in the temperature. With good cooling and for short time one may well draw more current.

author: TizianoHV
author_rank: Regular Contributor
timestamp: June 15, 2023, 07:44:39 pm
content: The 1A rating would make sense when you use the "2 diode full rectification" where each winding only gets half the current (positive or negative side). With such wiring you would get half the voltage.Edit: whops, Circlotron just said the same.

author: Gyro
author_rank: Super Contributor
timestamp: June 15, 2023, 07:54:44 pm
content: A useful rule of thumb chart of achievable DC current and voltage for different rectifier / filter combinations...