Post by thread:[EE]: Transformer power rating...

This started out a PIC "ASM weirdness" thread... >> using a 6V secondary to get 5.7V out of an 7805.... Djeez.... > > If it's 6VAC then you might be OK... OK. So exactly how do transformer ratings work, anyway? Like many people, my first 5V power supply had a 6.3V filament transformer and an LM309k, since the 6.3V is RMS voltage and yields 8.8V when rectified and filtered, which is plenty for the regulator dropout. And yet, there's not necessarily a lot of power in the peaks of those waveforms. If a transformer is rated 6.3V at 100mA, just how much current can I draw from the filtered output without the voltage drooping significantly? (I wouldn't think that the size of the filter cap would matter that much for a continuous load, right?) BillW

> -----Original Message----- > From: spam_OUTpiclist-bouncesTakeThisOuTmit.edu On Behalf Of William Chops Westfield > Sent: Thursday, March 24, 2005 12:58 AM > > OK. So exactly how do transformer ratings work, anyway? Like many > people, my first 5V power supply had a 6.3V filament transformer and > an LM309k, since the 6.3V is RMS voltage and yields 8.8V when > rectified and filtered, which is plenty for the regulator dropout. > > And yet, there's not necessarily a lot of power in the peaks of those > waveforms. If a transformer is rated 6.3V at 100mA, just how much > current can I draw from the filtered output without the voltage drooping > significantly? The voltage will only be 6.3 if the current is 100mA. At lower current the xformer will give a higher voltage. This is specified as xformer load regulation, it is expressed as a percentage rise in voltage at no load. Load regulation of excellent xformers can be 10% or less and with very cheap wall warts it can be 30% or worse. A cheap 6.3V @ 100mA xformer may deliver 6.3V + 30% = 8.2V with no load. The load line is usually very linear so @ 50mA the voltage will likely be 6.3V + 15%. Above when you calculated the output as 8.8V you forgot to subtract the diode losses. On average you can estimate 0.7V drop per diode for a total of 1.4V. So that 8.8V would really be closer to 7.4V. If you check the data sheets for the rectifiers you use you will find that the Vf will vary depending on the current. A typical 1N400X 1A diode has a Vf of 0.5V @ 1mA and 1.1V @ 1A. These values are for a steady state DC current, in an AC/DC power supply the average Vf will be lower because of the alternating current, the 1N400X has an average Vf of 0.8V @ 1A. Next you need to consider line regulation, some high end xformers will actually reduce the effect of line voltage variations. However most xformers currently available have zero line regulation so the entire line fluctuation is reflected at the output. For US 110/120VAC power it's usually safest to design for line voltage varying 90VAC to 140VAC. The xformer should specify the line voltage at which the xV @ xA spec is measured. An example: Xformer spec - primary 120VAC secondary 12VAC @ 100mA. This shows a 10:1 reduction @ the rated current. So, with a line voltage of 90VAC you will get 9VAC @100mA and a line of 140VAC will give 14VAC @ 100mA. I hope you can see that it is easy to have a supply drop out of regulation or burn up do to the variations in the line voltage and/or load current, if they are not considered during the design. > (I wouldn't think that the size of the filter cap would > matter that much for a continuous load, right?) The magnitude of the 120Hz ripple will increase with current so you need to size the capacitor such that the ripple is acceptable at maximum current. If the capacitor is too small the magnitude of the ripple may cause the regulator to drop out in the ripple valley. If you can find a copy of the 1980? National Semiconductor "Voltage Regulator Handbook" it has excellent information on xformers, rectifiers and filter caps for designing linear supplies. I searched national.com but, it is no longer available. IIRC the 1980 "Audio/Radio Handbook" had some of the same good power supply design information and you can buy a reprint of it at http://www.audioxpress.com/bksprods/books/bkaa59.htm. Paul

> regulation of excellent xformers can be 10% or less and with very > cheap wall warts it can be 30% or worse. A cheap 6.3V @ 100mA > xformer may deliver 6.3V + 30% = 8.2V with no load That's something you need to be aware of when using wallwarts/ plug-packs/battery eliminators. A neighbour of mine was caught out badly when he used a higher amp 12V in an alarm system. The circuitry was designed to load the original p-p down to 12V. When he swapped that p-p for another, the circuit loading had virtually no effect and the system saw 18V rather than 12V. Not a forgiving design, and it was his fault, but still ......... You have to assume that most people do not have a DMM and wouldn't know or care how a transformer works. They see the "12V" and that's what they were looking for I recently picked up a load of plug-packs (thank you Russell) that are unregulated and nominally 12V @ 150mA. Open circuit they measure 19.2V or +60% !!!!! That's OK, I knew that beforehand and have the circuit designed for them. There's a label on the blister pack (but not on the p-p itself unfortunately) stating "WARNING - UNREGULATED. Output voltage will be higher if your appliance draws less than the rated current" Dangerous if you don't know the implications of that - especially as the warning is not on the item but rather on the packaging, which would typically be in the bin pretty smartly - and I'll be including more specific information in the product's manual as well as moving the warning sticker to the p-p