Stepping up 12v to 15v with a lm7815? - Page 1
Forum section: electronics / beginners

author: PotatoBox
author_rank: Regular Contributor
timestamp: November 09, 2016, 02:59:12 am
content: I salvaged a 12v 1200ma transformer from an old wall wart. Now, I want to build a variable power supply at 12v via a lm317. But I know that the lm317 can only go to the max voltage minus 2-3 volts.Now, can I step up the 12v to 15v with an lm7815? This way, once it goes through the lm317, it will be 12v because of the 2-3 volts restriction of the lm317.

author: PotatoBox
author_rank: Regular Contributor
timestamp: November 09, 2016, 03:17:16 am
content: Quote from: blueskull on November 09, 2016, 03:01:29 amAnother thing, when I measuring the transformer with a multimeter, why does it give me 14.8 volts?

author: Bloch
author_rank: Unknown
timestamp: November 09, 2016, 03:23:23 am
content: >12v 1200ma transformer from an old wall wart.With a Full-wave rectification + capacitor you will have 12 x SQR(2) = 17 voltSo it close :-) so it may be ok.

author: PotatoBox
author_rank: Regular Contributor
timestamp: November 09, 2016, 03:26:04 am
content: Quote from: Bloch on November 09, 2016, 03:23:23 amAh ok. The voltage on the label is rated in AC voltage. So the wall wart has no bridge rectifier or a cap. Literally a transformer in a plastic shell.

author: Bloch
author_rank: Unknown
timestamp: November 09, 2016, 03:29:16 am
content: The 12Volt is at 1.2AmpereloadSo you transformer gives 14.8 volts a 0 AmpereIt seems the smaller the transformers is, the higher is the no load voltage

author: PotatoBox
author_rank: Regular Contributor
timestamp: November 09, 2016, 03:33:24 am
content: Quote from: Bloch on November 09, 2016, 03:29:16 amOk so if I used a bridge rectifier and converted it into DC voltage, it will end up at 17 volts?So I wouldn't need a boost converter?

author: Audioguru
author_rank: Super Contributor
timestamp: November 09, 2016, 03:37:47 am
content: The transformer produces 12VAC when it has its rated 1200mA load. It is made to produce 14.8V without a load so that the resistance of its windings drop the voltage to 12V when it has its rated load.The datasheet of the LM7815 says that itsminimuminput voltage is 17V on the datasheet and 17.7V on its sales sheet.The peak of a 12V sinewave is 12V x 1.414= 16.97V but the rectifiers drop it to about 15V and ripple drops it to maybe 14V or less that is much too low for a 7812 or 7815 regulator. The transformer might work fine with an LM7815 if the load current is low because then the input to the regulator might be 17.9V.Why can't your circuit use the 12VDC from an LM7812?

author: PotatoBox
author_rank: Regular Contributor
timestamp: November 09, 2016, 03:49:13 am
content: Quote from: Audioguru on November 09, 2016, 03:37:47 amLets say I fed the unregulated DC voltage into a lm317. What's the highest voltage I would be able to get?

author: Kilrah
author_rank: Unknown
timestamp: November 09, 2016, 05:10:23 am
content: Again it depends on current consumption.

author: Zero999
author_rank: Super Contributor
timestamp: November 09, 2016, 11:38:00 am
content: Quote from: PotatoBox on November 09, 2016, 02:59:12 amA linear regulator works by operating its output transistor in the linear region, i.e. like a variable resistor, to keep the output voltage the same, so the output voltage/current cannot exceed the input voltage/current. Like the LM7812, the LM317 is a linear regulator, which means it can only output a lower output voltage, than the input voltage.To increase the output voltage, a switched mode power supply is required but conservation of energy still applies: if the output voltage is higher than the input, then the input current will need to be higher than the output. P = V*I.In your case, you need a buck-boost or SEPIC converter, as the input voltage could be over 15V, as well as under 15V.This isn't the right way to do this. The correct way is to simply buy a 15V wall wart.

author: mariush
author_rank: Super Contributor
timestamp: November 09, 2016, 12:49:43 pm
content: If a transformer says it outputs 12v AC , that means a RMS voltage. The peak voltage will be equal to 12v x sqrt(2) = 12v x 1.414 = ~ 17vIf you convert this AC voltage with a bridge rectifier, you get a DC voltage that has peaks and valleys, the number of these peaks and valleys being twice the mains frequency which is 60 Hz in US and 50 Hz in Europe.Also, you have to remember that a bridge rectifier converts the AC voltage to DC by moving the electricity through the diodes that form the bridge rectifier, so each time the electricity goes through a diode, there's some amount lost. With a bridge rectifier that's basically made up of 4 diodes, there's always two diodes conducting, so after rectification your peak voltage will not be 17v, but in reality it will be 17v - 2 x Forward voltage of each diode in bridge rectifier.The forward voltage of a diode varies with the type of bridge rectifier and the current going through them, but it's common to see values between 0.6v and 1.1v.So in real world, your V dc peak = Vac peak - 2 x Vdiode = ( Vac rms * 1.414 ) - 2 x Vdiode = 17v - 2x0.7v = 17v-1.4v = 15.6vAlso keep in mind that the label of your transformer says 12v at 1200mA... this means if a product uses a reasonable amount of power (let's say anything from around 300mA up to 1200mA) , the transformer should still output 12v AC ... but if your product only uses 10-50mA, the transformer may actually output slightly more than 12v AC ... the smaller the transformer, the bigger this difference could be, which can often trick you into thinking your transformer is more powerful than in reality (when you measure the ac output with a multimeter, the multimeter is actually a load using less than 1mA of the output of the transformer)You also have to remember that at low current, the diodes in the bridge rectifier have a lower forward voltage, but that value goes up with higher current. For example, at 100mA the transformer may output 13v AC and the diodes may have a forward voltage of 0.55v but at 500mA and higher your transformer may output only 12v (the advertised value on label) and the diodes in the bridge rectifier will have a higher forward voltage, let's say 0.7v.So you may have a peak dc voltage of 13v x 1.414 - 2x0.55 = 17.28v at 100mA , but you may have 12v x 1.414 - 2x0.7v = 15.6v at 500mAIt's important to understand that the maximum current with DC voltage will be much less than the maximum current as AC output, as written on the transformer. You can't have a DC voltage up to 15.5v and provide the same amount of current the transformer would provide as 12v AC - some of the current is lost in the bridge rectifier and some will be lost in other places.A simple formula can be used to approximate about how much current you can expect to be usable with DC voltage : Idc max = Iac max * (0.62..0.7) - the value varies with size of transformer.For your 1200mA (1.2A) 12v AC transformer, you can expect that you'll have maybe 0.65 x 1.2A = 0.78 A of DC current.So for example, this tells you that you could use a simple 1A maximum linear regulator, as the most you'll probably be able to provide to devices would be about 0.8A ... no reason to look for a more expensive 1.5A or 3A linear regulator, unless that regulator has other features which would be good for you.Remember, what you determined one or two paragraphs above is apeak voltage, 100-120 times a second the voltage coming out of the bridge rectifier will go down to 0v and back up.If you want the DC voltage to not go below a minimum value, you need to add a capacitor after the bridge rectifier to supply energy when needed.You have to size that capacitor so that it holds enough energy to keep your DC voltage above a minimum threshold all the time.You could say "screw it, i'll just put 100.000 uF capacitor there" but that's not smart.. first of all that's expensive, and second.. a capacitor would act like a short circuit until it fills up, so for a few ms to seconds depending on how big it is, the capacitor can overload the bridge rectifier or the transformer and in extreme cases, with bigger transformers, it's possible to even break fuses.Here's a formula that you can use to approximate how much capacitance your need for some maximum amount of current :Capacitance ( in Farads) = Current (in Amps ) / [ 2 x AC Frequency x ( V dc peak - V dc min) ]So let's go with our maximum 15.6v peak DC voltage which we estimated with 12v AC and 0.7v for the forward voltage in our diodes that make up the bridge rectifier. I also estimated that your transformer could reliably supply up to 0.78A of current, so let's go with a round number of 0.8A.Any linear regulator will need a slightly higher voltage than the output voltage to output a smooth output voltage, some of the cheaper linear regulators may need up to 1-2v above the output voltage to work (like 7815 or LM317 linear regulators), there are some linear regulators that need less voltage to work , for example *1117 linear regulators which may need as little as 0.7-0.9v to output up to 1A, and there are also more expensive linear regulators that can need as little as 0.1-0.3v above the output voltage to output a very stable voltage.Let's say we find a great linear regulator, which needs at most 0.2v above the output voltage at the current we want (0.8A or more) .. in this case we want our DC voltage at least 15.2v all the time :Capacitance = 0.8 A / [ 2 x 60 x (15.6v - 15.2v) ] = 0.8 / 120 x 0.4 = 0.8 / 48 = 0.0166666 Farads or 16.666 uF with a voltage rating of at least 25v (because the peak voltage could in some conditions go over 16v)So you could put maybe 3 x 5600uF capacitors in parallel to get a capacity of 16.800 uF, or maybe go with 2 x 10000uF 25v capacitorsNow we can go at various distributors of electronic components and see if we can find such linear regulators .. here's what I found on farnell.com (newark.com in US and some other countries) :LT3083 :http://uk.farnell.com/linear-technology/lt3083efe-pbf/ldo-0-24vdo-3a-adj-16tssop/dp/2295281Very expensive and seems difficult to use as it's surface mount and has lots of pins but if you look in the datasheet there's only 4 pins, the majority of the pins have the same purpose (input and output). Also, if you look at page 5 at the graph in the corner you can see that this chip can output up to 3 A , but the minimum voltage can be only 100mV above the output for up to 500mA, and only about 150mV are required for 1A of current. So this would work great.SEMTECH SC1592ISTRT LDO VOLTAGE REGULATOR, 3A, 8-SOIC :http://uk.farnell.com/semtech/sc1592istrt/ldo-voltage-regulator-3a-8-soic/dp/9394109If you look at the brief specifications, you may ignore it as it says 260mV voltage which is above the 200mV we wanted, but it's best to check the datasheet to see if that 260mV is the minimum, or if it's the peak voltage at 3A output - since our project will only see up to 0.8A, the voltage may be smaller at only 1A.And what do you know.. go to page 10 and look at the last graph, and you'll see that the dropout voltage is only 100mV at 1A so if you can get over the fact that it's surface mount chip, this would work fine.LINEAR TECHNOLOGY LT1764AET#PBF LDO Voltage Regulator, Adjustable, 2.7V to 20V in, 340mV drop, 1.21V to 20V/3A out, TO-220-5 :http://uk.farnell.com/linear-technology/lt1764aet-pbf/ic-ldo-reg-3a-ln-to220-5/dp/1663484Through hole part, so easy to use but being made by Linear it's expensive. 340mV voltage drop, but that's at maximum 3A it's capable of. At less than 1A, if you look in the datasheet, you'll see that it can work with less than 200mV at 1A of output current.These are just first three randomly pick from the first page of results on farnell, after setting some filters. Plenty of other linear regulators that would fit the bill, you just have to play with the filters at sites that specialize in electronic components (like able to handle at least 18v or so, because you may have that much voltage when the transformer has a small load), adjustable up to at least 15v, at most 500mV dropout voltage (because we may have parts with 500mV dropout voltage at 3A, but only 100-200mV at 1A, and based on the transformer in this example we only need a part capable of 0.8A or more) .. and after you get a list of these, you may filter to have only through hole parts if you're scared of surface mount, or only parts that are 2-3$ if you buy just one.But anyway ... you should look at this in another way... the 12v AC transformer is not really the best to use if you want adjustable voltage up to 15v. Normally, you'd use a transformer that outputs a higher voltage - maybe 15v AC or 18v AC - which means the peak DC voltage will be higher in all situations, which means in order to keep the minimum DC voltage above 15v by some reasonable amount like 0.5v ..1v you'll need less capacitance after the bridge rectifier and you'll be able to use adjustable linear regulators that are not as great (require bigger dropout voltage) but they're much cheaper ... all these will make your adjustable power supply both cheaper and smaller.