How increase output current from of a transformer??? - Page 1
Forum section: electronics / beginners

author: Dhanushka
author_rank: Regular Contributor
timestamp: October 01, 2019, 07:29:34 am
content: Dear all,I salvaged a transformer from an old Sony stereo audio system. It has 2 x 12 V/2 A (center tapped) and 2 x 20 V/3 A (center tapped) outputs. Can I connect these center tapped winding in parallel to increase the current (4 A and 6 A)? Please tell how can I do this? I mean that whether should I consider the winding direction?

author: Zero999
author_rank: Super Contributor
timestamp: October 01, 2019, 10:19:31 am
content: Connecting the transformer to a biphase rectifier will increase the current capacity a bit, because each winding conducts half of the time, so allows time to cool down, but not by double, as increasing the current by a factor of two, results in four times the power dissipation.https://www.electronics-tutorials.ws/diode/diode_6.htmlIf you can access the windings, then you could break the centre tap and connect the two separate secondary windings in parallel, but it's not possible to connect two different voltage windings in parallel, because current will flow from the higher to lower voltage one, causing it to overheat.

author: Gyro
author_rank: Super Contributor
timestamp: October 01, 2019, 10:58:33 am
content: I've always found the attached transformer-rectifier circuits guide from Hammond quite a useful quick reference (It used to be in the RS catalogue at one point too)...

author: exe
author_rank: Unknown
timestamp: October 01, 2019, 11:18:04 am
content: Quote from: Zero999 on October 01, 2019, 10:19:31 amMy calculations for tranny I have showed almost no benefits from these, as most losses are on primary, and iron losses. Secondary windings have low resistance. This is easy to verify by measuring resistance of windings.I'd suggest to put windings in parallel (if they have same voltage), that would make it lowest losses, imho.

author: Zero999
author_rank: Super Contributor
timestamp: October 01, 2019, 11:28:21 am
content: Quote from: exe on October 01, 2019, 11:18:04 amI admit I haven't done any calculations, but it does surprise me you found more losses in the primary, than secondary windings. I would have thought the copper losses in both the primary and secondary would be similar and the resistance of the secondary is only lower, because the current rating is higher. If what you say is is correct: the secondary has negligible copper loss, then you should be able to increase the current rating more, not less, with a biphase rectifier, as it only increases the losses in the secondary.I do agree that splitting the centre tap and connecting the secondary windings in parallel, would be the most efficient solution, but they have to both be the same voltage and it might not be practical, if the tap has a wire connected to it, deep in the winding.

author: Kleinstein
author_rank: Super Contributor
timestamp: October 01, 2019, 12:12:52 pm
content: Using the 2 diode rectifier with the split coil does not help much with more current. It only helps a little by splitting the heat to both halves. However the limiting part is more like getting rid of the total heat. If helps to get a little more voltage as there is one diode less.Still the maximum current can be higher than the nominal for the split coil, as only one half is used. At the nominal current, the total power would be only about half. So half the current and 1/4 the power loss in the primary and only half the loss in the secondary, as only one half used.So the current used could be higher by approximately up to a factor of square root of 8/3 = 1.6. This is about the factor one looses when going from AC (RMS) current to DC current after rectifier and filter cap. So the split 3 A winding using the 2 diode rectifier is a bout good for 4.8 A AC and 3 A DC after the rectifier.

author: exe
author_rank: Unknown
timestamp: October 01, 2019, 01:29:20 pm
content: Quote from: Zero999 on October 01, 2019, 11:28:21 amWait, I might did it wrong. So, here is the tranny I have:230V primary (14.5Ohm), 2x15V secondary (0.3Ohm each measured with two-wire method, so not completely reliable data), 120VA. Seems that full power and on each winding the power dissipation is I^2*R and:(120/230)^2*14.5 = 3.9W on primary(120/15/2)^2*0.3 = 6.06W secondary (each).Measuring secondary resistance reliably is hard with tools I have. But even in the best-case scenario losses are comparable. So, I was wrong (at least for the tranny I have).

author: Gyro
author_rank: Super Contributor
timestamp: October 01, 2019, 06:08:54 pm
content: I suspect that, below a certain wattage, primary windings are of larger gauge than strictly necessary, simply for robustness and ease of handling during manufacture. Hence, they may have lower I2R losses than the secondaries.