Another Dummy Load Project! - Page 1
Forum section: electronics / projects

author: ZachFlem
author_rank: Contributor
timestamp: June 23, 2012, 10:08:20 am
content: Howdy Folks!Now I know there are already a lot of threads here about dummy load projects, but I don't seem to get any definite answers on higher capacity projects.I am looking for a unit to test battery banks over long periods of times with current draw of up to 100a for periods of 3-5 hours.I would like to use something that will maintain a constant current draw (e.g.http://bit.ly/KTbscH) but have come to terms that the heat loading on the componants may be to great.I have in the past used units (http://bit.ly/Nkpa3g) rated at 500a @ 48vdc, but for my current and forseable needs, such capacity is not warranted (I'm not working on battery banks that big any more!)I have a range of fan units (both 240vac and 12vdc) at my disposal, but am limited somewhat by space (approx 200x200x400mm).I have also used units that simply switch a series of 10a coils on and off depending on the required load (i.e. if you need 30a, flick 3 switches, 40a? flick 4)Either solution will be suitable, but having something that maintains a constant load would be preferable.Sorry if this doesn't make any sense, I'm far from experienced in these things, but my current situation dictates I adopt a more DIY approach to these things (read: my wallet somewhat empty)Cheers in advance for your help!Zach

author: Rerouter
author_rank: Super Contributor
timestamp: June 23, 2012, 10:13:55 am
content: hmm, seems i jumped the gun a little, i'll repost my solution to you from the other thread,hmm, 25 of these in parrellelhttp://au.element14.com/international-rectifier/irlb8743pbf/mosfet-n-ch-30v-78a-to220/dp/1740785each taking 4A, would dissipate ~47W (11.6x4) means they would be running 74 degrees above ambient (silicon is 1.11c/w and assume 0.5 for heatsink), or 90-100 degreesyour load resistances would be 10 of these in parrellel per mosfethttp://www.jaycar.com.au/productView.asp?ID=RR1502&form=CAT2&SUBCATID=968#1that will dissapate at half the resistors rating, 1.6W with a 0.4V dropall mosfets must be insulated from each other by a sil-pad or similar, you will want a fan performing active cooling, but without selecting a capable heatsink, cannot calculate air flow rate,the wiring between mosfets should be 2 heavy bus bars such ashttp://australia.rs-online.com/web/p/busbars/4896568/this particular one can be cut to size, so you can cut it into 4 stips of 15 (13 mosfet legs, 1 tie in and a negative sense line for op amps) and have 2 groups of mosfets, 1 of 13, 1 of 12,note i realise it would be quite difficult to solder to these, but insulated spade terminals generally fit onto the tabs and a piece of wire connected to them is very easy to solder to your mosfets and resistors,so far we are up to only$29.50 (mosfets)$11.52 (resistors)$34.70 (busbar)$75.72 + postagethen comes op amps which isnt all that bad, it would just depends on how fancy you want to get with them,the rest of your price will be the heatsink (jaycar do have one that forms and air column of sufficient size but isnt on there site, believe $40-70)the case, the fan and current measurment,an idea for current measurement would be to use an op amp (your already buy a crap load of them) and set one up as a 25 way adder, this will give you effectivly 0-10V (0-5 if in a voltage divider) for 0-100A, (yes your lowest and highest may get funky due to the op amp, but say you drove it at 7V and used the spec sheet to allow a 0V input / output,) to a standard volt meterthat is my thought on how to approach a 0-100A constant current load in a realistic size footprint,

author: ZachFlem
author_rank: Contributor
timestamp: June 23, 2012, 11:17:13 am
content: Are you thinking of the fan Assisted tunnel type heat sinks from jaycar?2x HH-8532http://bit.ly/MD6Tkv- 254mm long, w/ dual fans (push/pull) has a rating of "0.30oC/w approx" according to my jaycar catalogue.I'm no expert, so I'm wondering if that is enough?

author: ZachFlem
author_rank: Contributor
timestamp: June 23, 2012, 12:24:02 pm
content: sorry, I just re-read your reply, and I missed it the first time, but I think I'm reading it correctly now... 10 of the resitors PER MOSFET? as in, 25 mosfets, so 250 resistors?

author: T4P
author_rank: Super Contributor
timestamp: June 23, 2012, 01:58:34 pm
content: Quote from: ZachFlem on June 23, 2012, 12:24:02 pmYes he did, 250

author: codeboy2k
author_rank: Super Contributor
timestamp: June 23, 2012, 03:28:17 pm
content: With this design, or for that matter at anything near a kilowatt or above, I'd seriously start doing some thermodynamic simulations.I just hate guessing at stuff like this. You'll have a lot of heat sources inside an enclosed container... can you really get all that heat out? It will feel good to know for sure.you'll need to create models of your heatsinks, the heat sources, how many sources are on each heat sink, the air flows, the enclosing container, the exit ports, the ambient temperature of the incoming airflow, etc.Once you have the models, feed it all to Elmer.http://www.csc.fi/english/pages/elmer/examplesElmer is an opensource fluid dynamics simulation software. Airflow is considered a compressible fluid flow.Run some simulations. You'll know if your design is thermally sound. (and safe)

author: ZachFlem
author_rank: Contributor
timestamp: June 23, 2012, 11:00:26 pm
content: Someone has suggested I use resistive coils (apparantly found in old DC welding equipment) trimmed to length to get the banks I require? ie, a 120w coil, and run say, 10 of them in paralell to get the 1200w I need... the current will fluctuate a bit with heat and input voltage changes, but it will actually simulate what I'm testing for with more realism.I'm looking at options, because Rerouters solution looks good, but I think it's getting a bit out of reach of my capabilities =)

author: Rerouter
author_rank: Super Contributor
timestamp: June 24, 2012, 07:11:49 am
content: yes zach that was the heatsink i was thinking of, 0.3C/W (twin fan assisted) means its down to 65 above ambient, which is far more realistic, (and yes 250 resistors, but they are only 1/4W so tiny)note this is not something i need, just a way to approach it that seems feasable to me, the mosfets can handle up to 175 degrees max, so effectivly they would be running well under spec, so thermal simulations arent as important to the outcome,also a probably layout for this should only be about 28cm long, with the heatsink, and fans governing the length,12 (or 2 pairs of 6) mosfets mounted onto one side of the column, 13 (or 1 of 6 and one of 7) on the other, and a fan controller and thermal cut out would really not be a bad idea, (fan controller probably another op amp driving a higher power npn running a pair of high flow rate fans, with its input part of a voltage divider with an ntc thermistor,now onto some of the math side of it, to successfully dump 1KW/s of heat, you need 1200L of air per second, for a 1 degree temperature rise (this is put ambient for the heatsink) so now that we have freed up 10 degrees (lets aim for 20), that means 60L/s, still a tad unresonable for an 80mm fan and borderline for 120mm fans,if someone good with airflow modelling would want to work out the flow rate loss from a 120mm fan to 80mm shaft, or if someone knows of a 120mm version of the heatsinks, they would be the best bet to realistically dump all of the heat, (inlet and outlet must be the same) with a realistice temperature rise,alternativly you could use 2 of the columns, so 0.5KW per sink, 30L/s which is well in the scope of 120mm, and borderline on 80mm fans, for a 20 degree increase, bringing us upto 85 above ambient on the silicon junction, (still inside thermal limits)

author: george graves
author_rank: Super Contributor
timestamp: June 24, 2012, 08:42:17 am
content: subscribed!

author: ZachFlem
author_rank: Contributor
timestamp: June 24, 2012, 08:59:27 am
content: Quote from: Rerouter on June 24, 2012, 07:11:49 amPlease excuse my ignorance, but could a higher value resistor be sub'd in to reduce the number required? ie could one use 0.5w resistors and half the number to 125?I'd be pretty keen to see your design come to fruition, as I really like the idea (it's much more elegant than my original heater/nichrome idea) I might need a little help with the technical stuff though =)Cheers again for everyones input!

author: Rerouter
author_rank: Super Contributor
timestamp: June 24, 2012, 09:48:20 am
content: you could even get away with 5W 0.1 ohm resistors per mosfet, this way was just cheaper and a little more accurate,to point out, this design should be able to give a variable load over the range, lower currents may be hard due to your choice of op amps, but using app notes it wouldnt be that unfeasable for it to be able to cope down to 1A or less,

author: ZachFlem
author_rank: Contributor
timestamp: June 24, 2012, 10:53:02 am
content: Quote from: Rerouter on June 24, 2012, 09:48:20 amWould using the 5w 0.1ohm resistors result in a smaller physical size?And as for usable current range, I cant see myself using it for anything under 10a anyway...

author: Rerouter
author_rank: Super Contributor
timestamp: June 24, 2012, 12:00:50 pm
content: yep, you can use them in place and try to save a bit of spacehttp://www.jaycar.com.au/productView.asp?ID=RR3206&form=CAT2&SUBCATID=968#1however when i first approached this, the thought of a few mosfets running hotter than the others was my main concern, such as dave with his original use multiple resistors to get a more accurate value, my though was to get a better average value,also if you wanted to pursue a single resistor, then i would reccomend using JB WELD from jaycar or a thermal epoxy and epoxying the resistors to the heatsink, that or cutting up a bar, and clamping them in place against it, and make sure to secure the wires in something flexible, some silicone gel that can handle decent temperatures, as on the 5 and 10W resistors its pretty easy to have your wiring work harden the legs to a point where they snap off,

author: ZachFlem
author_rank: Contributor
timestamp: June 25, 2012, 02:18:31 am
content: Ok, so lets go with the 250 resistors for better overall performance. Would the resistors be mounted on pcb such as in Dave's blog (#102) or because of the number required, would they go on say a sheet of acrylic or are they mounted on heatsinks?I don't mean to be a PITA, I'm just trying to get an idea of how it would be assembled.Cheers again for you help!

author: Rerouter
author_rank: Super Contributor
timestamp: June 25, 2012, 02:33:45 am
content: its no problem to me mate,one thought i looked over last night is that most PCB fab houses give you 10 copies, so if your boards held 3 sets of op amps and load resistors, not only are you within thermal limits, but you end up with 1 board and 2/3rds of another as spares, in case you wanted to use it for 13.8V or 14V instead, or damage one in set up, that should fit on a 10x10cm board easily, maybe even smaller, (5x10 i believe was an option on some)they are dumping a fair but manageable bit of heat, you only would need a tiny bit of airflow across them, (1.6x25 = 40W)(little more than passive)a dual sided board is recommended, as there is less worry of a trace pulling away later, you would have your negative connection come in, the mosfet drain leg connection in, on one end near the resistorsand the mosfet drain leg fed back into some high ohm (say 10K) output for your current, summing amplifier, (or a 10K and a 10K to ground if you wanted 0-5V out) and positive and control signal for your op amps in, on the other sidepick an op amp and what size board you can get and i may even throw together the design for you, but it would be in expresspcb or image format and you would just need to re-copy it out in your desired pcb software,

author: T4P
author_rank: Super Contributor
timestamp: June 25, 2012, 04:16:54 am
content: I smell lots of fans ...

author: Rerouter
author_rank: Super Contributor
timestamp: June 25, 2012, 07:09:34 am
content: so far 4 fans, if a dual heatsink approach is used, if a 120mm varient of the heatsink can be found then 2, + for the control electronics perhaps a low speed silent 80mm,

author: T4P
author_rank: Super Contributor
timestamp: June 25, 2012, 07:54:16 am
content: Quote from: Rerouter on June 25, 2012, 07:09:34 amOr it can be easier to use any bog standard 80mm and drop the voltage down with a resistor

author: Rerouter
author_rank: Super Contributor
timestamp: June 25, 2012, 08:27:32 am
content: ok, i dont normally want to touch op amps, but this would do very nicely if your willing to try hand soldering a ssop packagehttp://au.element14.com/texas-instruments/tlv2774cn/ic-op-amp-quad/dp/1460036your requirements are an input range of 0-4, and an output of 1.8 to 3.2V this covers both, you also end up with an extra op amp per board, so i dont know, could add some functionality on that? or have each module checking its own mosfets temperatures for more reliablility, i dont know :/also its cheaper to buy 10 than 9, and that last one will come in handy as a current summing amp, and probably a voltage follower for your potentiometer, and also a 2 fan regulators,daveXRQ yes true, suppose so, just thats even more heat to dump,

author: T4P
author_rank: Super Contributor
timestamp: June 25, 2012, 09:07:29 am
content: Quote from: Rerouter on June 25, 2012, 08:27:32 amHardly much more heat, let's say it draws 12V at 0.16amps so if you want only 0.06amps you only dump 1.2W via the resistor ... 1.2W is hardly alot

author: ZachFlem
author_rank: Contributor
timestamp: June 25, 2012, 09:18:58 am
content: Quote from: Rerouter on June 25, 2012, 08:27:32 amThis may surprise you *sarcasm*, but that paragraph wen't COMPLETLY over my head!I'm INCREDIBLY gratefull for all your help in throwing this together, but I think I'm out of my depth...

author: Rerouter
author_rank: Super Contributor
timestamp: June 25, 2012, 11:49:20 am
content: this would be your basic circuit, with all of your VIN lines tied together and fed 0-4V (0-4Amp per mosfet) and all of your VOUT lines tied together fed into an op amp in a summing configuration,http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Op-Amp_Summing_Amplifier.svg/200px-Op-Amp_Summing_Amplifier.svg.pngwhere rf = 10K aswell,The power traces are 2mm wide on both layers, and signal wires are 20 mil (yes i am mixing units,)this would be for a 5cm x 5cm PCB, the 5V would be the supply to the op amp, with ground being common to the mosfets, all nice and pretty,D1 is drain 1 on a mosfet, G1 is the gate, and ground goes to your ground bus bar, the large holes are 1.5mm with a 3.3mm pad feel free to change as required,the smaller holes are 1mm with a 2mm pad, for the gates and lower corner inputs, (i consolidated after making the images)there is still the 4th op amp to use, and with some shifting someone could easily break out some pins for whatever they wish, this is just the nicest design that came to mind for 5x5cmred is copper side, and green is component side, though its not that critical, as worst case it just involves flipping the op amp 180 degrees,hope this helps, the last post just covered some of the bacground theory, with 4V in, means 0.4V on the + input of the op amp, and being how op amps love keeping there 2 inputs at the same voltage, it will then regulate the mosfet until it passes enough current to produce 0.4V across the sense resistor, (mosfets linear range from datasheet for 0-4A is ~1.8 to 3.2V)while most of this is to help you i am dumping in bits of info so any luckers or later readers can actually understand how it worksedit: R6 and C1 and there repeated ones are just some optional footprints to try and stabalise the circuit a little if your mosfet is oscillating, though i may have got it wrong and i may need to be added between the inverting and non inverting instead, alsothe pin numbers on the schematic are meaningless, they where just from me throwing together a quick diagram, though the indicator mark on the pcb is correct,

author: Rerouter
author_rank: Super Contributor
timestamp: June 25, 2012, 08:55:00 pm
content: my copper layer has an error with R31, i will post the revised board this afternoon,

author: codeboy2k
author_rank: Super Contributor
timestamp: June 26, 2012, 12:52:03 am
content: Your schematic.... it has the old style pen plotter lettering. I miss that in today's toolswhat schematic capture are you using?

author: Rerouter
author_rank: Super Contributor
timestamp: June 26, 2012, 04:22:01 am
content: express pcb's schematic editor, (i don't use their services, just the fasted thing i can bash together designs in)

author: Rerouter
author_rank: Super Contributor
timestamp: June 26, 2012, 07:32:10 am
content: ok, the corrected pcb's are up now, and i even added screw holes for convineince (3.1mm hole pad not important) and upon reading up, moved the capacitor footprint across the inverting and non inverting input of the op amp, the capacitor and gate resistor is if it oscillates, i cant actually tell you what values might work, but the footprints are there to experiment with,

author: ZachFlem
author_rank: Contributor
timestamp: June 26, 2012, 08:37:29 am
content: Rerouter, are you happy for me to put a call out to any freelance EE/Hobbyist's in Australia that might be able to help me get this together? I don't want you to go un-thanked in all of this =)Is there anyone else out there that might be reading this, that might be interested in making this more than a one off?

author: Rerouter
author_rank: Super Contributor
timestamp: June 26, 2012, 09:02:03 am
content: well the board design was far from a one off,and sure do what you like with it, as far as i am concerned its open source hardware, give my name a referance somewhere if you want to mass produce or whatever, but other than that your free to do as you like,

author: ZachFlem
author_rank: Contributor
timestamp: June 26, 2012, 09:12:09 am
content: I like to ask the question, even if it was implied... credit where credit is due after all =)

author: codeboy2k
author_rank: Super Contributor
timestamp: June 26, 2012, 09:13:02 am
content: Quote from: Rerouter on June 26, 2012, 04:22:01 amoh, I have never used their editor. But I like the retro look:)

author: Rerouter
author_rank: Super Contributor
timestamp: June 28, 2012, 10:54:54 am
content: i figure i have designed it this far, why stop so close to making a product,so here is my thoughts on the mechanical layout,good luck

author: Bloch
author_rank: Unknown
timestamp: June 28, 2012, 02:03:36 pm
content: How low did you need the Thermal Resistance ?Fischer Elektronikhttp://uk.farnell.com/jsp/search/browse.jsp?N=100852+2009+204385Whey have a Thermal Resistance down to 0.06°C/W and can be stacked as needed.

author: Rerouter
author_rank: Super Contributor
timestamp: June 28, 2012, 02:05:56 pm
content: take a good look at the price difference there mate, i was aiming for bang for buck,

author: codeboy2k
author_rank: Super Contributor
timestamp: June 29, 2012, 12:24:32 am
content: Is it possible? Maybe I've missed something, or I'm doing something wrong, but by my numbers you can't do what you are trying to do here.Those heat sinks from Jaycar are 0.3°C/WONLYif you put two of them together in a tunnel like configuration and keep the length at 254mm.That's the rating of the system as a whole.If you cut that heatsink in half you now have only half the total surface area available and thus I would expect that the heatsink is now rated 0.6°C/WNow you want to put 12 MOSFETS on a heatsink rated at 0.6°C/W ( in your drawing, the bottom sink will have 13 MOSFETS).You want 4A @ 11.6 V on each MOSFET. Let's say 12V, and that's 48W. Lets derate a little and say 50W.50W x 12 FETS = 600W of power going into that top heatsink, and 650W into the bottom heatsink.600W of power into the top heatsink at 0.6°C/W = a temperature rise of 360°C above ambient for the heatsink. (390 °C above ambient for the bottom heatsink)By my calculations, it's just not going to work.So what will work with this design?For the chosen MOSFET, The Rthjcis 1.11°C/W and the max junction temperature is 170°C. Let's stay safe at Tj=150°CYou specified a Sil-Pad to mount these to the heatsink.When the junction reaches 150°C and Rthjcis 1.11°C/W, and Rth(Sil-Pad)= 4°C/W (average), what's the temperature of the heatsink going to be?Let's find the watts that raises the junction to that temperature:Tsink = 150 - (Rthjc+ Rth(Sil-Pad))*WandTsink = 0.6*W*12 (total for all mosfets)therefore 150 - (Rthjc+ Rth(Sil-Pad))*W = 0.6*W*12solving for W yields W = 12.18 Watts (per MOSFET)so this system would be good for 25 * 12.18W = about304 Watts maximum total powerbefore the junction got too hot and MOSFETS start popping like corn.The heatsink temperature would be around 0.6*W*12 = 87.7°C (this is consistent with your original calculation of a heatsink temp of 90°C)If you put a thermister on the heatsink for thermal shutdown and set it for around 87-90°C, it would shutdown at this max wattage.(by the way: I totally recommend a thermal shutdown be added to this design)Someone check my numbers please. Did I miss something obvious, or am I correct here?When I get such wildly different numbers from another engineer I have to question my methods.Great thread though! I love a good thread that makes you really think.But I hate to see you build this and it doesn't work as you expected.Cheers!

author: Rerouter
author_rank: Super Contributor
timestamp: June 29, 2012, 02:44:03 am
content: Quote from: codeboy2k on June 29, 2012, 12:24:32 amok you got me there, must have been off my game there, ok leave full lenght,Quotelooking more into it, a thin mica pad with grease is the better option, 1.3C/W, which leaves it too hot.. hmm good thing this design is easy to break out into 28 instead,QuoteI hadn't touched more on that than saying to use one in a voltage divider fed into the spare op amp (probably should have catered for this on the PCB) where if any modules area is overheating it pulls the input signal to ground and brings it down to minimal current, (probably using a nice cheap signal mosfet, as the signal is through large resistances,)thanks for pointing it out, my head must have been funny last night,

author: codeboy2k
author_rank: Super Contributor
timestamp: June 29, 2012, 12:20:26 pm
content: yes, it's not an easy design. Even if you break it out into 28, I'm still not sure that's enough.Assuming you want to hit 1 kW, then 1000 / 28 = 35 W per device. Let's do some more math:First, let's talk about the heatsink.. The manufacturer says they characterized it at 0.3° C/W. You absolutely won't hit this number, and here's why:The manufacturer will characterize this heatsink with a full on heat source across the entire planar surface of the milled sides.This design, however, will have several point sources of heat on each side; the area of the heat source is much smaller compared to the area of the surface they are mounted to. The heat has to spread out, and this spreading encounters resistance, and takes time, and causes a localized heat zone at the point where the FET mounts onto the heatsink. This actually causes the FET to be hotter than a simple calculation shows.Here's the heat spreading calculationsLet's assume just one FET dissipating 35W on the entire heatsinkThe average temperature of the heatsink is simply the total power x the thermal resistance = 35W * 0.3° C/W = 10.5° above ambient.The local point source heating will cause a localized rise in temperature above the average. The thermal resistance that governs that riseis Rc, the constricting thermal resistance.Asis the contact area of the point sourceApis the contact area of the baseplatet is the thickness of the baseplatek is the thermal conductivity (for aluminum, k = 200-250. lets pick 210)R?is the average thermal resistance of the heatsink ( the manufacturer's number)whereFor this design:AsTO-220 = 7 x 13 mm (varies) = 91 mm2= 9.1*10-5m2Ap= 80x254 mm (assumption) = 20320 mm2= 0.0203 m2t = 2 mm (assumption) = .002 mk = 210 W/m•KR?= 0.3° C/Wplugging it all into the equations:thus the thermal resistance at the point heat source of a single TO-220 MOSFET all alone on the large heatsink isTo do it for more than 1 TO-220 device on the heatsink, just divide the area Apby the number of devices. I did further calculations, with 4 devices, 8 devices, 10, 15 devices. The value for Rcgets closer and closer to about 0.200 °C/W. It gets even lower with more devices on the heatsink because more devices spreads the heat more evenly, and any one point heat source is not above the average temperature by very much any more. But at some point, you can't add more devices than the heatsink can actually dissipate, as seen in my previous posting.So you'd be best to design with the assumption that the heatsink is actually going to yeild 0.3°C/W(avg)+ 0.2 °C/W(local)= 0.5 °C/W(total)per MOSFET, then calculate the maximum power you can dissipate from the heat sink to keep the heatsink average temperature below about 85CFinally to wrap it up, going back to my previous postTsink= 150 - (Rthjc+ Rth(mica+grease))*WattsandTsink= (Rth(avg)+ Rth(local))*Watts*Number_Of_FETStherefore 150 - (Rthjc+ Rth(mica+grease))*Watts = (Rth(avg)+ Rth(local))*Watts*Number_of_FETS150 - (1.11 + 1.3)*W = (0.3+0.2)*W*Nsolving for Watts gives a formula to use to keep the Junction temperature below 150 °C : Watts(per_fet)= 300/(4.82 + Number_of_FETS)run it for 1-25 FETS and ask google to plot it(remember this is all still on your original choice of heatsink, which is made up of 2 heatsinks from Farnel)google.com/search?q=plot%20300%2F(x%2B4.82)%2C%20x*(300%2F(x%2B4.82))%2C%20from%200%20to%2025You get a nice graph. The blue line is the number of devices vs watts per device. If you multiply the number of devices * watts per device (the blue line) you get the red line, and the goal is to maximize the red line.Conclusion :What you see is that this heatsink is only good up to about 300W. beyond that it gets too hot, and the junctions get too hot.You can have any combination of number of devices (blue line, X-axis) * Power per device (blue line, Y axis). More devices, less power per device. Less devices, more power per device. You can find a spot around 25-30 devices and 10-8W per device that yeilds about 250-260W total on the heatsink. Then you need 4 channel heatsinks to get 1000W. you can put 25 devices (side 1 has 12, side 2 has 13, spaced 20mm apart or so). So you need 100 MOSFETS.Here is where the more expensive heatsink from Farnell starts to make more sense. You only need 1. it can dissipate about 1000W. (0.06 °C/W)finally, no one really needs to do all this math. Just always derate your heat sinks. about 20%, that might be a good rule of thumb.--------------------The blue line , X axis is number of devices. The Y axis is the Power per Device. The red line is the Devices*Power product.

author: ZachFlem
author_rank: Contributor
timestamp: June 29, 2012, 07:29:05 pm
content: would that be why the commercially available units don't use mofsets, but rather coils, wound at approx 20mm spanning around 400mm mark, i'm not sure what they were made of, but the wire dia would have been around the 2mm mark.these coils had a bank of 8, 80mm fans blowing over them.Could the fets be substituted for another form of resistor?

author: SeanB
author_rank: Super Contributor
timestamp: June 29, 2012, 07:39:06 pm
content: Wire would be nichrome, as it has the good properties of a lowish TC and good high temperature corrosion resistance. for a simple quick and dirty version you can use steel springs. A genset load I used was a big steel bed frame cut in half, with a thick Bakelite spacer in the gap, and with a lot of long springs along the length. Ran at close to red heat, as it had to handle a 8kA load. Setting was by adding or removing springs from the frame when cold.

author: codeboy2k
author_rank: Super Contributor
timestamp: June 30, 2012, 03:01:56 am
content: Quote from: ZachFlem on June 29, 2012, 07:29:05 pmYes. The nichrome resistance wire is (a) cheaper than using multiple MOSFETs and (b) easier to cool.Youcanreach 1 kW with MOSFETS, but you will need two or more good heatsinks with a Rth of <0.1°C/W or lower and lots of FETS. Better heatsinks will reduce your FET count because you can run more wattage through each one. Then you can modularize those to get more total watts if you want infinitely adjustable power settings over your range.The difficulty with nichrome is that it's a fixed resistance. You have to switch the resistance wire into the circuit to achieve the desired load, so you won't be infinitely adjustable like you could with a bunch of MOSFETs. You'll have discrete steps. So back to the original design goal, 12V battery bank @ 100A for 1200W, so you need 120 mOhm of resistance wire that can handle the heating of 1200W, with a fan cooling it, etc. ( I won't do the math now).You could choose 127 mOhm and have a series string of 7 nichrome coils, switched in binary, 1,2,4,8,16,32,64 mOhm. You willl need more resistance if you want to go to lower amps or voltages, ie. 128,256, 512, etc. At some point nichrome becomes unfeasable and you have to use a wirewound power resistor. You will need 100A switches. You could use 12V automotive relays rated at 100A or MOSFETS , but these need to be low RDSon ( <1mOhm ) to minimize heating in the MOSFETS and make it easy to heatsink them without needing a massive heatsink again.You could make a hybrid, with nichrome wire as the course setting and use a few MOSFETS controlled by the feedback op-amps to get the fine setting. This might give infinite adjustability again, but it won't respond fast to load changes. For example, you could limit your MOSFET to 50W, then whatever load is selected by the user the controller will switch the nichrome wire elements to get to within 50W of that desired load and use the opamp feedback to control the MOSFET to get the remaining 50W. I'll have to give this idea more thought.There's some info on nichrome wire here:http://www.wiretron.com/nicrdat.html

author: ZachFlem
author_rank: Contributor
timestamp: June 30, 2012, 07:44:41 am
content: how much does the current flowing through nichrome wire fluctuate with temperature?and I know I'm being lazy (so feel free to tell me to sod off) but, how long a piece of nichrome would I need to make, say, a 10amp coil?I can't quite make heads or tails of the tables I've googled so far.Cheers...

author: SeanB
author_rank: Super Contributor
timestamp: June 30, 2012, 08:16:17 am
content: Work out the resistance required, then look up the table to get how many feet of the wire is needed to make this ( i know feet, but almost all of this is old machinery turning it out, you might get a translation to metric, but check it is correct against the original table) then cut it off the reel, wind around a suitable ceramic former ( or an air spaced coil for the thicker wires) and crimp on the wire leads.

author: codeboy2k
author_rank: Super Contributor
timestamp: July 01, 2012, 03:03:31 am
content: Quote from: ZachFlem on June 30, 2012, 07:44:41 amR=V / I = 12V / 10A = 1.2 ohmsFrom the tables I linked to, choose a wire gauge. The resistance changes with heating, so if you want more stable, less resistance change choose a large wire diameter. Lets say 15 AWG. The tables show two different alloys, NiCr A and NiCr C. The last table shows the chemical composition of NiCr A is 80/20 Ni/Cr , and NiCr C is 61/15/25 Ni/Cr/Fe.15AWG NiCrA is 0.2001 ohms/foot and NiCr C is 0.2078 ohms/foot (@ 20°C)So you need 1.2 ohms / 0.2001 = 5.999 feet , NiCr C is 1.2/0.2078 = 5.775 feetSo assuming 15 AWG wire, look in the the tables again, there's a Current Vs Temperature table. Look at 15 AWG on the left, there is a 10A value on the right, and above that in the headings it says 600°F/315°C. So 10A raises the temperature 315°C (probably above ambient, but it doesn't say)(if you don't want 6 feet of wire, you need to choose a smaller diameter wire (AWG) but then it will heat up more, for example 1.2 ohms of 24 AWG is just 8.94 inches, but it will heat up to over 1000°C, and 30 AWG is just 2.21 inches of wire, but it will overheat and burn up)Further down there are two more tables, that are titled "Increase in resistance with temperature". They have two tables, actually for NiCr A and NiCr C . NiCr A increases less with heating, and shows 315°C causes a 3.3% increase in resistance. For NiCr C, 315°C causes a 5.2% increase in resistance.NiCr A 1.2 ohms @ 20°C becomes 1.2 + 3.3% @ 315°C = 1.2396 ohmsNiCr C 1.2 ohms @ 20°C becomes 1.2 + 5.2% @ 315°C = 1.2624 ohmsand to answer the question...Quotesince I = V / RNiCr A I = 12/1.2396 = 9.68 ANiCr C I = 12/1.2624 = 9.505 AIt's not 10A any more. And since it starts to heat up immediately as soon as the current starts flowing, it probably never quite reaches 315C, instead maxing out somewhere below that, at a slightly lower resistance than calculated here (but still higher than 1.2 ohms). That may or may not be important in your specific application.Finally. here's a curve fit of the NiCr C temperature data. You get a quadratic equation that represents the percent increase in resistance for any temperature. The X axis is temperature, the Y axis is the percent change in resistance. Y = -0.37277 + 0.0216365 x - 0.0000106609 x^2You can similarily curve fit the Temperature Vs. Current data table to get temperature as a function of current for any given wire AWGhttp://bit.ly/KLdVjg<-- short link to the wolfram alpha page below

author: ZachFlem
author_rank: Contributor
timestamp: July 03, 2012, 10:58:33 am
content: OK,So I went to the local dump and picked up some resistance wire (read: oven elements) and after cutting and shutting one until i achieved the desired resistance, in this case, 65mm of element == 10.2a, which flattens out at 10.1a once at operating temp.Now I may be going all low-tech on this one, compared to the solutions provided earlier in this thread, but I have been thinking, 6x 2mm sheets (approx 250x350mm) of aluminimum, spaced 4mm apart, with 10 * 65mm lengths of element, evenly spaced across the sheets.While at the dump, I also picked up an assortment of fans from PC power supplies, 3 of which are 120mm fans, and rated at 3.0a each, one of these in the case of an old welder (also found at the dump) provides an absolute gale force air flow.I'm happy with 10a stepping, but can't seem to find a 10-step rotary switch, so am looking at using 10 micro flip switches and 20a automotive relays as a control.Please feel free to shoot this idea down in flames, or disect it and tell me where I'm going VERY wrong!Cheers!

author: Rerouter
author_rank: Super Contributor
timestamp: July 03, 2012, 11:13:02 am
content: actually automotive relays and 10 position rotary switch is a fairly commonly used method for sub 30V, (for DIY non electronics based people) its not particuarly cheap, but it works,

author: codeboy2k
author_rank: Super Contributor
timestamp: July 03, 2012, 04:47:21 pm
content: You're solution is perfectly fine and suits your needs. I like it.I was thinking back to the original design, and I think this design should not use MOSFETS at all now.I think the original high-tech solution with MOSFETS and heatsinks actually should be IGBTs not MOSFETs. Paralleling IGBT's is harder than paralleling MOSFETs but it can be done. I've never worked with IGBT's so I don't know all the pitfalls here, but I think that they are the right solution, as they can handle much more power than even power MOSFETs can.More power handling capability means fewer devices on fewer heatsinks = less cost.

author: T4P
author_rank: Super Contributor
timestamp: July 03, 2012, 07:36:15 pm
content: IGBT's have relatively crazy ratings and they don't need a complement but heck.More heat concentration on one spot is another problem

author: ZachFlem
author_rank: Contributor
timestamp: July 04, 2012, 07:57:51 am
content: Quote from: Rerouter on July 03, 2012, 11:13:02 amThe types of industries I work in, means that I have ready access to automotive relays, and should anything shit itself in the field, this should mean I can get myself up and running faster.Now I just need to find a switch and the smallest automotive relays available!Does anyone have any experience with 10+ position rotary switches, or have another simple way to switch 10+ loads?

author: Rerouter
author_rank: Super Contributor
timestamp: July 04, 2012, 08:25:23 am
content: smallest? just use generic 30A horn relays,as for what to use for the switches, i think you would need a proper industrial switch, as i cannot find anything with a common pole that as you rotates switches on each position plus all the previous,

author: ZachFlem
author_rank: Contributor
timestamp: July 04, 2012, 09:25:51 am
content: as it's only switching a relay, could I not use a diode (http://bit.ly/NkQLm7) between the contacts of the relays so that when the "20a" relay is on, it feeds back to the "10a" relay as well? and just series the relays up to the "100a" mark?

author: SeanB
author_rank: Super Contributor
timestamp: July 04, 2012, 03:25:52 pm
content: Smallest I have seen were OEM for Toyota and Ford. most were used for headlight switches, and they are half the width of the standard 30A automotive relay. Most automobile relays either have a built in diode ( actually 2 so that it will only work one way round) or have a resistor across the coil to attenuate the pulse. Nothing else is needed for use with a switch.

author: ZachFlem
author_rank: Contributor
timestamp: July 05, 2012, 06:48:46 am
content: Quote from: SeanB on July 04, 2012, 03:25:52 pmI think you misunderstand where I want to put the diode.I was thinking of using these (http://bit.ly/MW9t5Vor similar)with the DC input going into the relay on pin 30.the load output on pin 87pin 86 goes to earthand pin 85 recieves 12vdc from my switch.assuming the switch goes up in 10a steps, so relay 1 = 10a, relay 2 = 20a etc, could i not put a 6a diode between the load output (pin 87) and the switch input (pin 85) so that when the 20a relay is activated by the switch, the diode back feeds power to any previous relays?I'm assuming that the switch contacts of the relay aren't going to draw any more power then they need and that the 6a diode will be enough.Unless someone can point me in the direction of a rotary switch that step and hold the previous step... if you know what I mean!the other option, is to simply use 20a 12vdc rocker switches and forget the relays all together...

author: codeboy2k
author_rank: Super Contributor
timestamp: July 05, 2012, 09:46:00 am
content: Zach, your idea can work, with the load side switch back-feeding the previous relay coil. I don't like it because my preferencewould always be to keep the control and the load separate. Your idea would be implemented better if your relay had a secondpole and contact, basically a double-pole, double-throw switch or a double-pole, single-throw switch. Then you can feed theprevious relay from a contact on the next relay, and you don't even need a diode then.Here is an example circuit with diodes on the rotary switch. As you switch up to the next level, the diode feeds power tothe previous level. The downside is that each new level adds a diode drop, so you should use Schottky diodes to minimizethe drop, and secondly, each higher level has to supply all the current to itself and all the relays of the previous levels. Sothe diodes have to have enough DC current and power handling capability for all the relays before it. The worst case is the lastdiode, which has to power all the relays.

author: ZachFlem
author_rank: Contributor
timestamp: July 05, 2012, 10:47:20 pm
content: assuming i use the Narva relays mentioned earlier, with their current draw of 0.15a @ 12vdc, I can't see why a 6a diode wouldn't be more than enough, my reasoning says that at 6a, it's 4 times the capacity of my 10 relays (1.5a in total) and should be more then enough.I'm also considering going down the individual switch route, as it will reduce the overall cost of the unit considerably...

author: codeboy2k
author_rank: Super Contributor
timestamp: July 06, 2012, 01:33:58 am
content: Quote from: ZachFlem on July 05, 2012, 10:47:20 pmexactly. Using 6A diodes is more than enough. 3A diodes would suffice with your 150mA relay coils.Quotesimple too.