Transformer problem? - Page 1
Forum section: electronics / beginners

author: AxleD
author_rank: Regular Contributor
timestamp: April 07, 2015, 08:09:50 pm
content: Hi All,I was having an issue with getting the rated current out of a 1 Amp plug in wall transformer so i decided to open it up. It is one of those plugs where you can select the output voltage.Pics attached.I was surprised to find that the taps have no insulation and that a few of them are touching each other. I haven't checked with a multimeter but i believe the non-braided wire on the far right coming out of the transformer (the lower picture) is the ground and it is touching the 2nd and third taps.I assume that this is the problem as those taps are being shorted to ground? or is this arrangement normal?Many thanksThanksMuch appreciated.

author: picandmix
author_rank: Frequent Contributor
timestamp: April 07, 2015, 08:17:13 pm
content: Don't panic, if you read up on transformer winding, the copper wire used is treated with a form of enamel which insulates it.

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 07, 2015, 08:18:44 pm
content: There is a lacquer finish on the wires so they are insulated. If you have a multimeter you can test it using the ohms setting.You probably can't get the rated current because... well, it's overrated

author: AxleD
author_rank: Regular Contributor
timestamp: April 07, 2015, 08:24:12 pm
content: Great, thanks for the quick responses.Cheers

author: SeanB
author_rank: Super Contributor
timestamp: April 07, 2015, 08:46:14 pm
content: 1A rating, but it realistically will only do 300mA, and even then it will vary in voltage. The 12V setting will be around 16V unloaded, and at 300ma it will drop to around 11V, and at 1A it will be close to 1V. Those are very optimistic, and with the cheap capacitor in there ( probably 470uF 16V or 1000uF 16V) the ripple will be a little high as well at higher currents.

author: AxleD
author_rank: Regular Contributor
timestamp: April 08, 2015, 03:19:26 pm
content: Out of interest a test with a multimeter confirmed (as suggested) that the copper wire is in fact insulated! I would have never guessed - certainly doesn't look insulated!SeanB, that's pretty much what I'm getting out of the transformer. 16v unloaded, 12 volts loaded, 400ma and large ripple.Is this is an issue with most transformers (given all sources will have some "internal resistance") or is it that these wall plug ones are particularly bad/poor quality?If I want 1A at 18v, what is the minimum specs i should be looking for in a transformer (15V AC, 2.2A perhaps)?Thanks

author: SeanB
author_rank: Super Contributor
timestamp: April 08, 2015, 04:25:14 pm
content: It is worse with the wall wart ones as they cut down on everything inside, the transformer is very much undersize, the capacitor is also undersized and thus you get poor performance. Depending n what you need, and what ripple is acceptable ( all simple rectifier circuits will have ripple, you need a linear regulator to reduce it afterwards) you can get away with the transformer you suggested, using 3A diodes and a 4700uF or larger 35V capacitor. You will get just around 20v unloaded, and with 18V and 1A it will be around the 18-19V mark. Ripple will be around 1V though. If you want lower ripple you will need a different transformer, 18VAC at 2A again, and then use a linear regulator to give 18V.

author: mariush
author_rank: Super Contributor
timestamp: April 08, 2015, 05:18:37 pm
content: Here's some simple formulas to work with.You get a transformer that's rated in VA , for example 50 VA, and has a secondary winding of let's say 18v AC.When rectified to DC using a bridge rectifier (or four diodes), the voltage and current become these :V AC = 18v I ac = 50 VA / 18V = 2.77V DC peak = 1.414 x V ac = 1.414 x 18 = ~ 25.4v I dc = ~ 0.62 x Iac = 0.62 x 2.77 = 1.72AThis is assuming the four diodes are ideal, but in practice there's always 2 diodes out of the 4 diodes conducting, so there's some voltage drop on each of those 2 diodes. Therefore, the peak DC voltage you will see isV DC peak real = V DC peak - 2 x V diode dropThe voltage drop on each diode in the bridge rectifier depends with how much current goes through the diodes and the diodes themselves (the type and construction).If you design for 100mA for example, you may use four Schottky diodes which each have a voltage drop of 0.3-0.4v or even less. If you design for outputs of 1A or more, the voltage drop with normal diodes could be 0.7-1v on each diode.Going with average example of 0.7v , that meansV DC peak real = 25.4 - 2 x 0.7 = 23vSo you have a peak DC voltage of about 23 volts, but every 100-120 times a second (depending on your mains frequency) the voltage goes down very close to nothing, so you need some capacitance to provide the energy when the transformer doesn't.The size of the capacitor you can choose depending on how much current you expect to provide to devices, and how much you're willing to let the voltage go down, from the peak dc voltage.So for example, let's say I want a very smooth 18v output and up to 1.5 A from this "power supply". Since I'm going to use a linear regulator to be extra sure the output will be smooth, that means the linear regulator will need to always have at least 1-1.5v more than 18v to work properly, so my input voltage needs to stay between 19.5 and 23v. Therefore, the maximum "voltage ripple" is 23-19.5 = 3.5vNow I can use this formula to get a basic approximation of how much capacitance I need:C = Current / ( 2 x AC Frequency x V ripple) ... in our case assuming we're in US with 60 Hz mains frequency then C = 1.5A / 2 x 60 x 3.5 = 1.5 / 420 = 0.003571 Farads or 3571 uFSo 3571 uF would be the minimum to guarantee the linear regulator always has at least 19.5v at the input. So the next standard value would do. Of course you can use higher values like 4700 or 5600 uF, but there would be minimal benefit, and would only make the linear regulator dissipate more energy as heat (the difference between input and output in a linear regulator is wasted as heat, so it's in your best interest to keep the input voltage as close as possible to the minimum needed).You could also connect 2 or 3 smaller capacitors in parallel, if the power supply will look better like that ( for example, use 2 x 2200 uF capacitors instead of one 4700uF capacitor)Now as a note, notice that with this example I figured out that the peak dc voltage the input capacitor will see is 23v. It's NOT smart however to assume this will be the case and use a 25v rated capacitor for example, because there's lots of reasons why in practice this won't happen, the capacitor should be rated for 35v at least (for my example).For one thing, the AC voltage at the mains isn't always fixed at 110v or 230v, there can be fluctuations which can make the output ac voltage of your transformer different than the expected voltage - oh and by the way, this must be also taken into account when figuring out the capacitance, as the "voltage ripple" value could be much smaller if your mains voltage is often much lower than what transformer is designed for.Another thing, most transformers that are small (let's say under 25VA) are by design going to output up to 5-15% more than their advertised AC voltage at low currents. For example, if a device connected to your "power supply" uses only 50-100mA, it could happen that the transformer will output 20-21V AC instead of 18V AC, which when rectified will be much more than 23v.So hope these help... you can change my numbers with yours and figure out what transformer is needed.

author: AxleD
author_rank: Regular Contributor
timestamp: April 08, 2015, 07:26:03 pm
content: Thank you all!Clear explanation and this will definitely help me out!Cheers