How to selecting the correct transformer for a voltage regulator - Page 1
Forum section: electronics / projects

author: bob_60
author_rank: Contributor
timestamp: November 21, 2014, 08:19:27 pm
content: I like to know if there is a simple rule of thump that I can follow when selecting a transformer.Lets say I have a voltage regulator that needs 20V, 2A as input. The mains is 230V and the input is connected to a standard bridge rectifier configured with a smoothing capacitor.The question is, what transformer do I select.I have done a test with a small transformer (15VA, 2X 6V, Farnell product#:1675043) a bridge rectifier (Farnell product#:1861540) and a 3300uF capasitor and I get the following output values using one of the secondary outputs of the transformer:With a 225mA load, the circuit makes the Vmin voltage drop down to 6.56V (VPP: 0.96V)A 905mA load makes it drop down to 3.72V (VPP: 2.28).All this is fine but I'm not sure about how to translate these numbers to a 20V, 2A circuit.A prober capacitor has to be selected of course, but for now I'm just looking for a rough rule of thump that can help me get an indication of the size of transformers I need when feeding regulators./Bo

author: dannyf
author_rank: Super Contributor
timestamp: November 21, 2014, 08:30:05 pm
content: QuoteThe calculation is simple and not so simple at the same time.The 20v Vin is a minimum DC figure. The rectifier's output is a pulsed DC - its high is just shy of the transformer output voltage * 1.4 (=square root of 2) - 2 * Vfwd. So you are talking about at least 14-15v Vac 2ndary.However, what matters is how low that recified ac will go? That, unfortunately, depends on the load as well as the capacitance. Assuming a persistant 2a load current, I would imagine that you have at least 2200uf capacitance on the rail. Typically, you get 1.1 - 1.2x of the 2ndary voltage. Ie., the transformer needs to output 16 - 18v Vac at a minimum.Higher output gives you more headroom / room for error, at the expensive of higher dissipation.

author: bob_60
author_rank: Contributor
timestamp: November 22, 2014, 12:51:00 pm
content: QuoteExactly. That's what I need a rule of thumb to estimate.QuoteAnd again, here is where a rule of thumb is needed.

author: mariush
author_rank: Super Contributor
timestamp: November 22, 2014, 01:35:04 pm
content: The output of a transformer is AC.With low power transformers (let's say 10-50VA), the output of the transformer can often be a bit higher than the value listed on the transformer if whatever is connected to it doesn't use all those 10-50VA. It could be up to 10-15% extra. When the device uses all those VA, the transformer's output will be closer to the label.You need to convert the AC to DC and this gets done using a bridge rectifier. The output of the bridge rectifier will be a DC voltage that varies between 0v and a peak voltage of Vac x 1.414 but since there are always two diodes conducting in the bridge rectifier at any time, you have to take out two times the voltage drop on a particular diode in the bridge rectifier.The voltage drop on each diode in the rectifier varies with the amount of current and the temperature of the bridge rectifier. For example, a bridge rectifier rated for a maximum of 2A may have diodes with a voltage drop of 0.6v at 100mA but these will go up to let's say 0.9v at 1A. So when you select the parts, you should beVdc peak = 1.414 x Vac - 2 x VdiodeNow remember, since we still have fluctuations in this DC voltage, we need to smooth out this DC output and for that you use capacitors. There's a very simple formula that approximates how much capacitance you would need to keep the DC voltage always above a particular level:Capacitance = Current / ( 2 x AC Frequency x Vdrop)where V drop is how much you're willing to let the DC voltage go down from the peak DC voltage.For example, let's say the peak DC voltage out of the bridge rectifier is 17v and you want to produce a 12v at maximum 1A using a linear regulator, let's say a 7812 which requires about 1.5v above the output to work properly. So that means the DC voltage it receives at the input should always be between 13.5v and 17v.So now we can use the formula to get the minimum capacitance required to always have 13.5v even when the regulator does 1A of current:Capacitace = 1 Amp / ( 2 x 50 Hz x (17v peak dc voltage - 13.5v minimum voltage) = 1 / ( 100 x 3.5) = 1 / 350 = 0.002857142 Farads or about 2857uF.I'm assuming 50Hz because that's the AC frequency in Europe and some other countries. In US it's 60 Hz.So 2857uF is not a standard value, but we can choose the next value up, 3300uF.Of course, you can use much larger capacitance but that will only keep the minimum dc voltage closer to 17v which is pointless in this situation because the linear regulator will dissipate the difference between 12v and 17v as heat anyway.Note though... you have to keep in mind that your mains AC voltage is not always super stable... You don't always have 230v or 110v, and you have to also consider who's going to use your device. Depending on the input voltage fluctuations, the AC voltage on the transformer could also vary, so you have to be careful when designing such a classic power supply and don't assume the transformer will always output a minimum voltage written on the label.