Author Topic: Another Dummy Load Project!  (Read 13325 times)

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Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #25 on: June 26, 2012, 07:32:10 am »
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,
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #26 on: June 26, 2012, 08:37:29 am »
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?
 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #27 on: June 26, 2012, 09:02:03 am »
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,
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #28 on: June 26, 2012, 09:12:09 am »
I like to ask the question, even if it was implied... credit where credit is due after all =)
 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #29 on: June 26, 2012, 09:13:02 am »
express pcb's schematic editor, (i don't use their services, just the fasted thing i can bash together designs in)
oh, I have never used their editor. But I like the retro look:)
 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #30 on: June 28, 2012, 10:54:54 am »
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 :)
 

Offline Bloch

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Re: Another Dummy Load Project!
« Reply #31 on: June 28, 2012, 02:03:36 pm »
How low did you need the Thermal Resistance ?

Fischer Elektronik http://uk.farnell.com/jsp/search/browse.jsp?N=100852+2009+204385

Whey have a Thermal Resistance down to 0.06°C/W and can be stacked as needed.
 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #32 on: June 28, 2012, 02:05:56 pm »
take a good look at the price difference there mate, i was aiming for bang for buck,
 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #33 on: June 29, 2012, 12:24:32 am »
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/W ONLY if 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/W
Now 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 Rthjc is 1.11°C/W and the max junction temperature is 170°C. Let's stay safe at Tj=150°C

You specified a Sil-Pad to mount these to the heatsink.
When the junction reaches 150°C and Rthjc is 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))*W
and
Tsink = 0.6*W*12 (total for all mosfets)

therefore 150 - (Rthjc + Rth(Sil-Pad))*W =  0.6*W*12

solving for W  yields W = 12.18 Watts (per MOSFET)

so this system would be good for 25 * 12.18W = about 304 Watts maximum total power before 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!


 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #34 on: June 29, 2012, 02:44:03 am »
Those heat sinks from Jaycar are 0.3°C/W ONLY if 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.

ok you got me there, must have been off my game there, ok leave full lenght,

Quote
You specified a Sil-Pad to mount these to the heatsink.
When the junction reaches 150°C and Rthjc is 1.11°C/W, and Rth(Sil-Pad) = 4°C/W (average), what's the temperature of the heatsink going to be?


looking 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,

Quote
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)

I 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,

 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #35 on: June 29, 2012, 12:20:26 pm »
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 calculations

Let's assume just one FET dissipating 35W on the entire heatsink

The 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 rise
is Rc, the constricting thermal resistance.

As is the contact area of the point source
Ap is the contact area of the baseplate
t is the thickness of the baseplate
k 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)


where


For this design:

As TO-220 = 7 x 13 mm (varies) = 91 mm2 = 9.1*10-5 m2
Ap = 80x254 mm (assumption) = 20320 mm2 = 0.0203 m2
t = 2 mm (assumption) = .002 m
k = 210 W/m•K
R? = 0.3° C/W

plugging 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 is



To do it for more than 1 TO-220 device on the heatsink, just divide the area Ap by the number of devices.  I did further calculations, with 4 devices, 8 devices, 10, 15 devices. The value for Rc gets 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 85C

Finally to wrap it up, going back to my previous post

Tsink = 150 - (Rthjc + Rth(mica+grease))*Watts
and
Tsink = (Rth(avg) + Rth (local))*Watts*Number_Of_FETS

therefore 150 - (Rthjc + Rth(mica+grease))*Watts = (Rth(avg) + Rth (local))*Watts*Number_of_FETS

150 - (1.11 + 1.3)*W = (0.3+0.2)*W*N

solving 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%2025

You 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.




 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #36 on: June 29, 2012, 07:29:05 pm »
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?
 

Offline SeanB

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Re: Another Dummy Load Project!
« Reply #37 on: June 29, 2012, 07:39:06 pm »
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.
 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #38 on: June 30, 2012, 03:01:56 am »
would that be why the commercially available units don't use mofsets, but rather  coils

Yes.  The nichrome resistance wire is (a) cheaper than using multiple MOSFETs and (b) easier to cool. 

You can reach 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

« Last Edit: June 30, 2012, 03:04:56 am by codeboy2k »
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #39 on: June 30, 2012, 07:44:41 am »
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...
 

Offline SeanB

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Re: Another Dummy Load Project!
« Reply #40 on: June 30, 2012, 08:16:17 am »
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.
 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #41 on: July 01, 2012, 03:03:31 am »
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?
R=V / I = 12V / 10A = 1.2 ohms

From 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 feet

So 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 ohms
NiCr C   1.2 ohms @ 20°C  becomes 1.2 + 5.2% @ 315°C = 1.2624 ohms

and to answer the question...

Quote
how much does the current flowing through nichrome wire fluctuate with temperature?

since I = V / R

NiCr A   I = 12/1.2396 = 9.68 A
NiCr C   I = 12/1.2624 = 9.505 A

It'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^2

You can similarily curve fit the Temperature Vs. Current data table to get temperature as a function of current for any given wire AWG

http://bit.ly/KLdVjg  <-- short link to the wolfram alpha page below


« Last Edit: July 01, 2012, 03:28:23 am by codeboy2k »
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #42 on: July 03, 2012, 10:58:33 am »
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!
 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #43 on: July 03, 2012, 11:13:02 am »
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,
 

Offline codeboy2k

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Re: Another Dummy Load Project!
« Reply #44 on: July 03, 2012, 04:47:21 pm »
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.


 

Offline T4P

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Re: Another Dummy Load Project!
« Reply #45 on: July 03, 2012, 07:36:15 pm »
IGBT's have relatively crazy ratings and they don't need a complement but heck.
More heat concentration on one spot is another problem
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #46 on: July 04, 2012, 07:57:51 am »
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,

The 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?
 

Offline Rerouter

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Re: Another Dummy Load Project!
« Reply #47 on: July 04, 2012, 08:25:23 am »
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,
 

Offline ZachFlem

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Re: Another Dummy Load Project!
« Reply #48 on: July 04, 2012, 09:25:51 am »
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?
 

Offline SeanB

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Re: Another Dummy Load Project!
« Reply #49 on: July 04, 2012, 03:25:52 pm »
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.
 


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