Bench Power Supply 3-30V 2A.  - Page 1
Forum section: electronics / projects

author: matty-q
author_rank: Newbie
timestamp: April 24, 2014, 12:11:41 pm
content: Hi guys,I have an assignment which is due in soon. I need to make a Bench power supply that includes:• Channels: 1• Voltage Range: 3V to 30V, adjustable.• Current Capability: 10mA to 2A limit, adjustable• Cooling to be passive or with thermostatically controlled fan• Output voltage / current to be displayed using analogue or digital metersI have chosen a L200 high input and output voltage (Pentawatt) for the chip.I was wondering if you could help me with the procedures to the calculations?My current calculations are...Max unloaded voltage = (Voltage+ 8.5% + 10%) x1.414 – (Volt Diode Drop x 0.4V)= (36 + 8.5% + 10%) x1.414 – (2 x 0.4V)= 36 x 8.5% = 3.06= 36 + 3.06 = 39.06 V= 39.06 x 0.10% = 3.9= 42.96 x 1.414 = 44.374 V= 44.374 – (2 X 0.= 43.5 v= 43.5 AC VThis indicates that the reservoir capacitors and regulator input must be able to withstand at least 44 VoltsMinimum peak voltage = (Voltage - 6%) x1.414 – (Volt Diode Drop x 1.5 V)= (36+ 6%) x1.414 – (2 x 1.5 V)= 36 x 6% = 2.16= 36 – 2.16 = 33.84= 33.84 x 1.41 = 35.254= 35.254 – (2 x 1.5) = 32.254= 32.254 VThe capacitor and regulator should be able to take in 45VCapacitor rating should be able to take in 55VCapacitor-fed rectifierA ‘rule of thumb’ for capacitor-fed rectifier circuits is to rate the power supply to deliver 65% of the stated transformer current.The Current is 2 amps; the result of the equation will be added onto the 2 amps.2/100 x 65%=1.32 + 1.3 = 3.3 AA transformer need a current of 3.3 A for a 2A power supplyChoosing a Reservoir CapacitorA reservoir capacitor is used to smooth the pulsating DC from an AC rectifier. The capacitor releases its stored energy during the part of the AC cycle when the AC source does not supply any power.Capacitor =Cuurent/(100 x 2)= 3.33/(100 x 2)= 16650 ufVoltage = 2 x Nominal= 2 x 40= 80 V……..Peak voltage on load, high mains = (Voltage+ 10%) x1.414 – (Volt Diode Drop x 1.5)= (36+ 10%) x1.414 – (2 x 1.5)= 36 x 0.1 = 3.6= 36 + 3.6 = 39.6= 39.6 x 1.414 = 56 V= 56 – (2 x 1.5) = 53= 53VRipple factor is in fact a measure of the remaining alternating components in a filtered rectifier output. It is the ratio of the effective value of the ac components of voltage (or current) present in the output from the rectifier to the dc component in output voltage (or current).Ripple Voltage = (Current DC)/(18000uF x 2 x 50)= 2A/(18000uF x 2 x 50)= 1.11vAverage voltage = Peak voltage on load, high mains – ( Ripple/2 )= 53– ( 1.11/2 )=52.5vPower Dissipated = Average Voltage x Amps= 52.5 x 2= 105Heat sinkThermal Resistance = (Maximum Temperutre-Ambient Temperture)/(Power Dissipated)Maximum Temperature = 80 oCThe temperature of the air immediately surrounding the power supply, Ambient Temperature = 35 oC= (80-35)/105= 0.476 oC/WI would be brilliant if you could help.RegardsMatty-q

author: mariush
author_rank: Super Contributor
timestamp: April 24, 2014, 02:03:16 pm
content: You have several problems :L200 maximum voltage is 40V DC. This restricts your choice in transformers.L200 voltage drop is 2v typically, 2.5v maximum. This means that your input voltage must be 2.5v above the maximum output voltage.The maximum dropout voltage (see absolute maximum ratings) is 32 volts so if you want 3v out, you can't give it more than 35v input voltage, which again (in theory) restricts your transformer choice.In practice, feeding the regulator with such high voltages to output low voltages is a bad idea - the regulator itself can't dissipate properly more than.. let's say 35-40w. So you wouldn't be able to output 3v @ 2A anyway, that would mean the regulator would dissipate (35v in - 3v out) x 2a = 64w, more than the chip's maximum power.The easy method to work around this is to use a transformer with two secondary windings and use a relay with a voltage comparator or some opamp to switch between using one winding or BOTH secondary windings, depending on output voltage. For example, get a 30Vac transformer with center tap or two windings ( 18v dc on each after rectification).. Below 15v DC out, use only first winding and then for 3v @ 2A, you only need to dissipate about 25w, reasonably easy with good heatsink and fan. When output voltage goes above 15v, switch to using both windings.I don't know where you got those 8.5% and 10% values for choosing the transformer voltage.30v DC , 2 A is about 60w ... this means you probably need a 100 VA transformer. Such transformers are generally big enough that the unloaded voltage is only about 5% higher. It's worse with smaller transformers. Anyway, you're only interested about this to keep the peak DC voltage below the maximum input voltage of 40v (for L200).I don't know where you found diodes that would have a voltage drop of 0.4v - for 2A currents, you would normally pick 3A+ diodes and at those currents, the voltage drop is usually at least 0.6v. It's safer to go with 0.8-1v.. for example see this 4A bridge rectifier :http://uk.farnell.com/multicomp/vsib410/bridge-rectifier-4a-100v-gsib-5s/dp/1861531- look on second page on forward voltage drop and you'll see the voltage drop is 1V at 3A.So... assuming we pick a 28V AC transformer ( with center tap/2 secondary windings) and a 4A bridge rectifier.Peak DC unloaded = 30v ( 28 + ~ 6%) x 1.414 - 2 x 0.4v (low current, low forward voltage) = 41.6v (problem, more than 40v limit of regulator)Peak DC unloaded (if you use only one winding at low voltages) = 15v x 1.414 - 2x0.4 = 20.41 ( ok)^ you could add a minimum load resistor (BEFORE the voltage regulator) to make the power supply always use 50mA or something like that, which hopefully will bring the transformer down a bit.Peak DC at 2A = 28v x 1.414 - 2 x 1v = 37.5v (ok)Current in the secondary side is about 0.62 x Iac so you need about 3.3A on the secondary ( 3.3 x 0.62 = ~ 2.05A) ... so you'd need a transformer rated for at least 39.5v * 3.3 = ~130VA (i may be wrong here, seems a bit high)So anyway, you have a peak voltage of 37.5v dc. To keep the l200 regulator happy, your minimum input voltage needs to be always 2.5v above the output voltage, but for safety select 3v. So 30v + 3v = 33v. This means your input voltage can hover between 33v and 37.5v and the regulator will be happy at 2A output.A basic formula for Capacitance = Current / ( 2 x AC Frequency x Vripple) where Vripple is that voltage difference between minimum and peak voltage.So for 2A and 50 Hz mains frequency and 37.5v - 33v = 4.5v (but let's say 4v for more safety) then C = 2A / (2 x 50 x 4) = 0.005 F = 5000uF. I would go with a minimum of 6800uF (a popular standard value after 4700uF). As the voltage will be up to 37.5v (or 42v when unloaded and using both windings), the voltage rating of the capacitor should be at least 50v (I would use 63v)The L200 has a maximum power dissipation of about 40w (read datasheet, page 4, typical safe operating area) but that's very hard to achieve even with heatsinks and forced cooling.So again I recommend going with a two secondary windings design, switching to using one or both depending on output voltage to minimize power dissipation.You can't really do 2A without some tricks, with 37.5v input I have if using the 28Vac transformer... Remember at the start of the post... maximum dropout voltage 32v .. can't get 3v out with 37.5v input ... certainly can't dissipate (37.5v - 3v ) x 2A = 69w when the limit is at 40wSeehttp://www.ef-uk.net/data/heatsinking.htmYour L200 has 150c junction temp, Rjcase (junction-case) 4c/w for to-2, 3c/w for pentawatt (i'm going with this one) , Rj-amb 50c/wAssuming you change the design like I recommend to have two secondary windings ( 0-19v dc , 19-38v dc for a 28Vac transformer) let's pick the worst case scenario, when you want 17v out. At 17v out, your regulator needs 17 + 2.5v = 19.5v at input, which is more than what a single secondary winding is capable of, so it means both windings need to be connected, so you have 37.5v input and 17v output.At 2 amps, you're looking at 33v min, 37.5 max ( so let's pick 35v average) , 17v out ... (35-17)x 2 = 36w (risky, but doable).Tdelta = Tmax - Tamb = 150c - 30c (you won't have 25c in a box) = 120c. In practice, you would not want the chip to get to 150c, you would do the math for 140c max but in this case...Power dissipated is 36w in our worst case scenario.thermal coefficient = Tdelta / power = 120 / 36 = 3.33 C/wThis is lower than R j-amb of 50c/w so we definitely need heatsink. For pentawatt Rj-case is 3c/w, so we need a heatsink with 3.33c/w - 3c/w = 0.33c/wSo heatsink should be 0.33c/w or better... keep in mind even the best thermal paste between heatsink and case will kill about 0.1c/w so in reality the heatsink must be even better.

author: matty-q
author_rank: Newbie
timestamp: April 25, 2014, 12:22:57 pm
content: Hi Again,I have read through it and it makes sense but still confuses me slightly, is their an easier way to put this for me?This is the link for the diode I originally used:http://uk.farnell.com/vishay-general-semiconductor/gsib1580-e3-45/diode/dp/1276334I can not seem to find a 28vAC Transformer but could get a 30v AC with two secondarys :http://uk.farnell.com/multicomp/mcta030-15/transformer-toroidal-2-x-15v-30va/dp/9530312and when looking at the L200 Datasheet, which FIGURE would you suggest I use :http://www.circuitstoday.com/wp-content/uploads/2009/03/l200.pdfI am sorry about all the questions, I am new to this module.RegardsMatthew

author: mariush
author_rank: Super Contributor
timestamp: April 25, 2014, 01:45:54 pm
content: Ok, open the datasheet for that bridge rectifier.Go to page 2 and not the thermal characteristics. It says Rja (junction - ambient) of 22C/w , Rjc (junction-case) of 1.5 c/w.As electricity goes through the diodes in the bridge rectifier, there's some energy wasted due to the voltage drop of the diodes. For example, if there's 3A going through the bridge rectifier, two out of four diodes in the bridge rectifier will always conduct electricity, so there's some power wasted on them. If we assume there's a voltage drop of 0.8v at 3A, this means there's going to be 2 (diodes) x 0.8v (v drop) x 3 (amps) = 4.8 watts wasted as heat.If we don't have a heatsink on the bridge rectifier, it means the temperature of the package will raise above the ambient (25c) by Rja x 4.8 = 22 x 4.8 = 105.6C, so the temperature of the bridge rectifier will be ambient temperature 25c + 105.6c = 130.6c .. which is below the 150c maximum temperature for the bridge rectifier. Well, this is a very simplified formula, in reality it may be slightly less warm.But what you have to get from this is that even though the bridge rectifier is 15A rated, it doesn't mean it can actually work at such high currents by itself, without any cooling.If you look down at figure 1, you will actually see exactly what I described above. You have a line called "PCB Mounting" which is more or less the rectifier soldered to a board so the only heatsink is the pcb traces drawing heat from the leads and the air around the case moving over the case (natural convection), and you have a line that says Heatsink mounting.From that graph, you can see that if the air around it stays up to about 50c, the bridge rectifier will continue to work with around 3-4A.. the rectifier itself will be hot and will continue to heat up the air around ... if you have this in a very closed case the air around will eventually heat up and the bridge rectifier may eventually die due to overheating.Normally, most devices have some vent holes that allow the air to move through the case, so the air inside won't heat up that much and with your 2-3A of current the bridge rectifier will be fine without a heatsink.But if you don't want it to be very hot inside, you can of course put it on a heatsink and worry less.Now move on and look at figure 3, instantaneous forward voltage. You have there three curves. From left to right, it's 150c , 125c and 25c , the junction temperature (inside the bridge rectifier). While working, your bridge rectifier will not stay at 25c, unless you really cool the shit out of it (pardon the expression) with a heatsink and fan, so it's best to use the middle curve, for 125c.On the Y axis, you have there the current going through the diode. 0.1A , 1A , 10A .. each line is a step between those values (so between 1 and 10, each line represents 1A of current)On the X axis, you have the forward voltage per diode.Note that when you have current going through the bridge rectifier, there's always TWO diodes conducting, so if you want to figure out the voltage drop per diode, you have to divide the current by twoand as the two diodes are in series, each will have the same current , but they'll separate have a voltage drop (so at the end you multiply the voltage drop by two).When you're not going to have anything connected to the voltage regulator, you're only going to have a few mA of current flowing through the bridge rectifier but to keep it simple, let's assume there's 0.2A through the bridge rectifier, so about 0.1A per diode. You look at the center curve and go up to where the curve cuts the horizontal line that represents 0.1A and you see they meet somewhere around 0.52v.. let's make it simpler and say 0.5vThis means that at 0.2A through the bridge rectifier, there's going to be 2 diodes x 0.5v per diode = 1v drop in the bridge rectifier.When you have 2A going through the bridge rectifier, you're going to have 1A going through each of the two diodes conducting, so you look again at the curve in the center and follow it up until it cuts the 1A horizontal line.. and then you look down to the x axis and you get about 0.7v there. So at 2A through the bridge rectifier, you're going to have a voltage drop of 2 (diodes) x 0.7v per diode = 1.4v.These are just "typical" values though, they vary from rectifier to another in the series depending on how lucky you are, how well it was manufactured, how hot the bridge rectifier will be and so on. So even though you get from the figure that you should expect 1.4v at 2A, you wouldn't design with this figure in mind, you would add some margin in your design.. for example I would simply expect 1.6-1.8v.That transformer is fine, but pay attention to its VA rating. The VA rating is how much power it can provide. A 30Vac , 30VA rating transformer means (in very simple terms) that if you connect a 30Vac lightbulb rated for 30W, the bulb will light up perfectly. If you put a 30Vac 100w ligtbulb, the bulb will only be 1/3 of the brightness.So the transformer above can do 30VA at 30Vac ... this means it can provide 1A of current at 30Vac (if you put the two secondary windings in series), or 2A of current at 15Vac if you put the two secondary windings in parallel.However, this 30Vac (or 2x 15Vac) is a RMS value, when you're rectifying this using the bridge rectifier you get a DC voltage with a peak voltage of Vac x 1.414, but you lose that voltage on the diodes, so the peak dc voltage will actually be Vac x 1.414 - 2 x Vdiode.So at two amps, using that theoretic 0.7v drop per diode, you would have a peak DC voltage of 30Vac x 1.414 - 2 x 0.7v = 41.02vBUT, the transformer is still only rated for 30VA, it can't magically continue to provide the same current when the voltage increased after rectification ... so now the maximum current can be calculated : 30VA / 41.02 v = 0.73AIn practice, a more conservative formula is to just multiply the current by 0.62 : the transformer was 30Vac 30VA so 1A ... after rectification we should expect a maximum of 0.62 x 1A = 0.62 A of current.If you look at the above, it's obvious that the 30VA transformer is much too small for your needs. If your power supply needs to output up to 2A, then your transformer must have a current big enough that after multiplying by 0.62 you're still going to have more than 2A.2A / 0.62 = 3.22A , so whatever you pick must be rated for about 3.22A. If you go with the 30Vac , then it means at least 30 x 3.22 = 96.6 VAAnyway, again, if you do choose the 30Vac transformer, you saw that we may have a peak dc voltage of 41v. This is at 2A ... all transformers at low currents will generally have a higher peak voltage, so your 100VA transformer may actually output slightly more than 41v, it could be 43v, it could be 45v.If you go again in the L200 datasheet and look at the electrical characteristics, or absolute maximum ratings, you would see that the maxim voltage is 40v ... now having peak voltages of 41-45v can't be good...So the conclusion is that using 30Vac transformers is not a smart idea, because you may damage the voltage regulator.As fpr what figure would be better, well the best one would be the one that makes better contact with a heatsink. Look at Thermal Resistance Junction-case on the first page .. The smaller, the better.. so obviously the Pentawatt case wins at 3c /w.It has worse "Thermal Resistance Junction-ambient" but that only means that the Pentawatt package will be much hotter without a heatsink, which won't be the case here. You MUST put this voltage regulator on a big heatsink.PS. Watch this video, explains RMS voltage (the ac voltage on your transformer) :

author: mmilejski
author_rank: Contributor
timestamp: April 25, 2014, 02:38:26 pm
content: Quote from: mariush on April 25, 2014, 01:45:54 pmI think that you mean that the voltage has to be multiplied by two. Diodes are in series, current is the same through both of the diodes.

author: mariush
author_rank: Super Contributor
timestamp: April 25, 2014, 02:40:09 pm
content: Right, yes, thank you for the correction.The diodes are in series, so the current will remain the same. If they were in parallel, then the current would be half (more or less). I've corrected in the post above.

author: matty-q
author_rank: Newbie
timestamp: April 26, 2014, 10:47:43 am
content: Hi Guys,Its starting to make sense now, I really appreciate your help.However, would using a 30va, 2 x 12v secondary as I have here:http://uk.farnell.com/multicomp/mcta030-12/transformer-toroidal-2-x-12v-30va/dp/9530304Be a good solution? like you explained using 2 x 15 would be slightly higher than the l200 max voltage which is 41.5.?Matthew

author: mariush
author_rank: Super Contributor
timestamp: April 26, 2014, 02:40:02 pm
content: As I explained already in the post above, the voltage is not the big issue with that transformer. The issue is the VA rating.. (think of it like the wattage on a computer power supply).30VA is too small, it's impossible to make an adjustable power supply that would work up to 1.5A with so little. At best you're going to go up to about 0.5A with such a small transformer.Please read my long posts again.And don't say you don't have transformers to choose from, there's plenty, you're just looking in the wrong section.. that one's only for toroidal transformers which are generally slightly expensive and you don't really need one.Here's classic transformers :http://uk.farnell.com/isolationFrom Primary, pick 2 x 110v , 2x115v, 2x120v (you can put the primary windings in series to work on 220-240v mains) or 1x220, 1x230v, 1x240vFrom secondary, pick whatever combination would give you less than 40V or around 40v .. for example 2x 14v ... or 2x15v ... if you're willing to make your power supply up to about 28v, 24vac transformers will work so 2x12v will work. You could also get 2 transformers with smaller voltage and put the windings of those separate transformers in series.Then pick the Power rating, which like I said, for your 1.5A 30v requirements, would have to be at least about 75VA but I'd be more comfortable saying you need 100VA or more.There are voltage regulators that have lower voltage drop, like 1.3v for example, which would allow you to use 24Vac transformers and get very close to 30v if not reach it... 32v DC is possible with a good bridge rectifier and 24Vac transformer.See for example LT3081 :http://uk.farnell.com/linear-technology/lt30817-pbf/ic-ldo-linear-reg-rugged-1-5a-7to/dp/2356006works with up to 36v, has a voltage drop of 1.2v and a pin that makes it easy for you to adjust the current limit with a simple resistor. It's just more expensive. The lower voltage drop means you could use 24Vac transformer and you'll just have to be making sure you have a 31.5v or more when you want to output 30v.