Help with PSU (Temp control fan, load bank, & PWM circuit)

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jocanon said:
So, please forgive me, I am a complete newbie here, but how would I adjust the level of "on"ness of the MOSFETs, with a pot? It seems to simple, but could I just attach a pot to adjust the amount of current running through the MOSFETs?
Click to expand...
Something like that. The pot would control the current, but through another device called an opamp. The circuit is very simple, I can draw it when I get home or someone else might.


jocanon said:
Lastly, by stepped load I take it you mean just having e.g. 15 lines of 100W resistors and then manually plugging in more of them to get less resistance and more current or unplugging some to get less current and more resitance?
Click to expand...
Yes, exactly. If resistors of that rating are expensive, you can use wire or e.g. halogen light bulbs.
 
dougy83 said:
Is there any reason you don't just use a 'linearly' controlled (rather than PWM) load?
Click to expand...

Good idea actually, there is nothing wrong with going full linear now that you mention it. (this is why it's good to have more eyes on the picture jocanon) And it would be FAR cheaper. I'm stuck on PWM for load control as it's the standard for variable load control. This because it's more energy efficient. BUT THAT'S JUST IT!!!, we are deliberately trying to be less efficient here, which hurts my brain I guess. It's hard to break out of what you learn to do normally. It's the whole... "Not a real load" thing that keeps throwing me off.

jocanon said:
dougy83 I am intrigued by cheaper and simpler if it's possible...what do you mean by "linearly" controlled[?]
Click to expand...

When he says linearly, what he is suggesting we do is take out the resistors, heatsink the MOSFET's, turn them half on so they become the resistances. And all the heat is made within them. As long as they can dissipate the Watts it's fine. This would normally be avoided like the plague because you normally never never never want a significant amount of power dissipated in your switching devices, it's very inefficient. It would normally be backwards thinking for such high energy levels, but for this it could work well since we are not concerned about efficiency at all. It should also cut the budget down to nil. Overall I approve of the idea.

There are some details that are worrisome when doing this though.

1) Must protect the device from turning the MOSFET's fully on, it could destroy them.
2) The MOSFET's will need very good heat sinking and forced air cooling.
3) The MOSFET's will need AMAZING interfacing with the heatsinks.
4) The output of an FET is NOT normally linear, it will be logarithmic in nature.
5) Adjusting them will be very very touchy, they are amplifiers after all.

Most of these things will be fixed with a proper Op-Amp feedback design. The heat sinking can be protected with an over temperature shutdown circuit. And the power supply should protect from over current because of it's own over current protection. However, let us not forget that the reason we are making this apparatus is to test if we can induce a power supply to fail. So failure of PSU current safety is a potential problem that should not be overlooked. I suggest looking into possibility having a 50 Amp hard shutdown.

jocanon said:
()blivion I wonder if the schematic is showing up differently for me than it is for you.
Click to expand...

Yeah, that picture is dead wrong, there is a lot more to it than that. I can't explain why it doesn't work for you as the link works 100% for me. You must be using a different browser than Fire fox 13 and/or Java 7 or something?

Anyway, I have attached a text document with all of the simulator information. And HERE is a direct link to the simulator again. You can copy and paste all the text into the import window of the simulator and it should work better. If not don't worry about it as we may go the linear rout after all. And here is a schematic of the PWM version, taken directly out of the simulator window. For those who are still curious.

View attachment 65874
 
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dougy83 said:
If resistors of that rating are expensive, you can use wire or e.g. halogen light bulbs.
Click to expand...

Ni-Cr resistance wire on a ceramic former is the way to go when crafting your own high Wattage resistors.
 
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()blivion said:
Yeah, that picture is dead wrong, there is a lot more to it than that. I can't explain why it doesn't work for you as the link works 100% for me. You must be using a different browser than Fire fox 13 and/or Java 7 or something?View attachment 65874
Click to expand...

I tried opening it in Google Chrome and it looks much better Must be a problem with IE.
 
*COMPLETELY OFF TOPIC... BUT*
As a computer professional, and someone who has used most browsers for extended periods of time, I can tell you with great authority that Internet Explorer is mostly garbage. I recommend Fire Fox. Google Chrome is a close second, and is not as "fast" as is claimed (your browser has little to do with how fast you see the web). Only use IE if you absolutely positively have to. That being said, IE is better than the rest of the lesser, older, "no-name" browsers out there.
 
Here, THIS is about what you want for a linear method. It's not exactly perfect as I made it in a rush. Someone can polish it up for you, or I can tomorrow, which ever comes first. And here is a schematic for you in case the simulator decides to act up again.

View attachment 65875

Speaking of simulator acting up, A lot of what your going to see is just me working around it's "quirks". Most notably all the MOSFET's, we don't need that many. Note that this circuit can use as many or as few MOSFET's as you want though, provided you never exceed the Wattage per each unit. The three 10m Ohm resistors are > 6 Watts, and are just for current sense. I have some picked out on Digikey. Here is a link for that...

https://www.digikey.com/product-detail/en/MRS-1298R010FE1401/MRS129801-ND/953741

The slider to adjust the load is on the far right hand side again, (as it always is). And the current is shown by a wire just to the left of the Op-Amp (triangle thingy pointing up with a - and + sign in it). Also note that the OP-Amp's power supply is not shown. All it needs to be is the Vee pin grounded and the Vss needs to be attached to a 5+ volt supply. This is a small thing.

That's all I have to say for now. I'll be back tomorrow though.

Now... SLEEPPPPP TIME......
-()blivion
 
Thanks ()blivion & dougy83. Strange but the simulation looks different again. I was hoping it would work now that I am on a different computer (I was opening it from home last night, now I am at work) but it must be related to IE because it still has the same issue. So FYI to anyone else trying to open the circuit, don't use IE, use some other browser or it will not show up correctly. It worked for me on Google Chrome. I did notice that there is a little triangle with an exlamation point in the top right corner when I open the schematic from within IE so at least it knows there is some wrong and is trying to warn you.
 
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+1 for Fire Fox. I just downloaded it and like it. No problems opening the schematic with it and it imported all my bookmarks and homepage settings from IE, nice.
 
OK. I had an idea for cooling this beastly project.

Originally, we were going with resistors, and I was not too concerned with cooling the MOSFET's because they were not the things dissipating all the power.

NOW however, we are going with the idea that we are intentionally going to loose all our power in the MOSFET's alone. This means we *MUST* cool them very very well. This presents a problem for us because of there small volume, mass, and surface area. Normally what you do for situations such as these is attach them to a heat sink very well to share the heat with, effectively giving them much greater mass, volume, and surface area. Also you would normally use insulating spacers for transferring heat to the heat sink, without making an electrical connection of any kind.

Well... I had already figured earlier that, (A) we don't need to worry about the electrical connection aspect of heatsinking the MOSFET's because all the tabs are going to be shorted anyway, in fact we want them to be electrically connected. And (B) we are going to be pushing so much heat out of these things that cooling them is going to be a full time job, and a good thermal interface is one of the most critical things to good cooling.

So it occurred to me.... Why don't we solder them to our heat sink? Better yet... why don't we solder them to a square or rectangular tube (copper, brass, tin) and run water through it, water cooling them? This would give you max power dissipation and safest operating temperatures possible. It could be as low budget as just sticking a garden hose and barb on both ends and putting one end in a drain, and the other on a faucet. Or as complex as a fully sealed (brazed or soldered) and enclosed system. Complete with reservoir, pump system, radiator, and fans. The works.

Let me know what you think.
-()blivion
 
Water cooling sounds like a great idea! Seems quieter than fans too, or then again, maybe not quieter, but I think it would work better since water has such great cooling properties. The works sounds cool, but perhaps expensive and complicated...I am leaning toward a simple garden hose...()blivion, if you don't mind, could you give me a quick dumbed down explanation of what each of the components of the circuit are for? For example, I am having a hard time finding anything on google that makes much sense to me regarding the current sense resistors and the purpose they serve. Also, it seems like they are way under powered at only 6.3watts (I am sure they are not as you know what you are doing, I think I just don't get it yet). Also, what is the point of all the other various smaller resistors in the circuit? Do I need to step down the voltage to get 5 volts to power the op amp? Will this work on both 12.5v up to 55 amps and 25v up to 55 amps as well (I think 55 amps is where the over current protection kicks in and the PSU shuts off according to some tests a guy did on YouTube). Lol, maybe it would be easier to tell you what I DO understand versus what I don't .
 
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HERE is a new and improved version of the linear load circuit.

First, some answers to your questions.
The previous circuit could draw ~75 Amps at 24 volts with the knob in the full load position. I believe the circuit would only go up to about 37 Amps on 12v though. I have and will continue work on this. Note that the simulator stops at 75 Amps on 24v, and I am getting 59.1 Amps from 12v. But the real circuit would go much much farther than this.

The current sense resistors are not a mistake, they don't need to be very high Watts because they are very low resistance. Our Watts are turned in to heat by what is called Joule heating. These Watts are dissipated in the total resistance of the circuit. before we were using resistors that took up a larger fraction of the total resistance of the circuit, so they were where all the Watts went. The MOSFET's for that circuit were either fully OFF, so no current was going through them, thus no power. Or they were fully ON, so all the current was going through them, but they had almost no resistance for heat to be generated in.

With this new circuit, we are making the MOSFET's have a significant resistance compared to the whole resistance, so almost all the heat is going to be made inside them. The sense resistors will continually have much lower resistance compared to the MOSFET's, and thus generate much less heat. In fact, the resistors and MOSFET's have sort of switched places as far as resistances in concerned.

The current sense resistors purpose is a bit tricky to explain in lay persons terms. But it sorta acts like a "speed bump" or "backs up" the flow of electric current just a bit. We are not doing this to slow it down or anything like that. This "speed bump" is to create a voltage that gets bigger with more current (Amps) going through it, because more current "spills over" so to speak. In the end it is nothing more than a way to measure current. It's output goes to the (-) side of the OP-Amp and is only millivolts.

Here is a schematic of the new and improved circuit. And the following is a bit of explanation on it.

View attachment 65904

How It Works
Basically, all the things in line across the top of the simulation are your MOSFET's. We are using them as if they were our resistors now. As stated above, heat is lost in a circuits resistance. We can change the resistance of these MOSFET's by changing the voltage applied to their gate, more voltage means less resistance and vice-versa. The problem we have is that MOSFET's are WILDLY sensitive and do not change flatly by them selves (non linear). So we need something to stabilize them out a bit if we are to use them reliably. Enter the Op-Amp.

The triangle thing in the center is an Operational Amplifier (Op-Amp for short). Op-Amps are very flat (linear) amplifying devices. They have one output, and two inputs. The best part about them is that their inputs are "differential". Without going into too many cloudy hazy details, we can take advantage of this differential nature to make the circuit keep the two inputs at exactly the same levels. You will see this level out behavior if you move the slider in the simulator, then hover the mouse pointer over the (+) and (-) wires on the Op-Amp. We achieve this effect with what is known as "feedback".

As stated before, the current sense resistors give us a small voltage, based on how much current is going through the circuit. This voltage gets sent into the inverting (-) input of the Op-Amp. The inverting input will cause the output to do the opposite of whatever it is seeing. So, if the current rises, the voltage of the inverting input will also rise. This makes the output drop. The output is connected to the MOSFET's gates, and as stated, they will increase in resistance when the voltage on the gate goes down.

Since the current going through the whole circuit is controlled by the MOSFET's, and the MOSFET's have just increased in resistance because of the drop in gate voltage, the current would of course drop too, in accordance with Ohms law (Volts/Ohms=Amps). If the total current drops, so too must the current on the sense resistors. Finally, if the sense resistors are what are feeding the inverting input it's voltage, and that voltage just dropped, the output would increase again. The opposite of what had started this.

So to summarize, if current goes up, sense output goes up, then inverting input goes up, output thus goes down, so gates voltage is lowered, resistance is raised, current goes back down because of Ohms law. This is simple negative feedback. With out this feed back, a slight breeze could change the circuit enough to cause massive damage to the power parts. And we would like to avoid this.

This is all well and good, but we are missing one critical detail. "At what current does this feedback prefer to hover around?" Well, that is what the other input is for. It happens that by virtue of how Op-Amps work when in a feedback configuration, it will tend to try to make the the two inputs the exact same voltage. This means that what ever voltage we apply on the (+) side, the Op-Amp will try to create on it's (-).

The voltage that the current sense resistors output is very small, but that doesn't matter to the OP-Amp, as they are very sensitive devices. However, we need to create an equally small voltage for the control signal, or the Op-Amp will try to match 24V and destroy things. We make this small voltage with a voltage divider. On the right side of the Op-Amp is the voltage divider. The top resistor is 47k Ohms, and the bottom is 1K. This alone makes ~500mV at 24v (At 12v it's about half this). That ~500mV is farther divided by the fact that the second resistor is actually a variable potentiometer. By adjusting this pot, you can make any voltage between ~500mV and 0V. And the higher this voltage, the more current the system must pass to match it with the current sense voltage. Thus giving you control over current passed through the system.


My explanation wasn't as small as I had hoped it would be, but that is close to how the circuit works without going into more details.
 
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That's a beautiful circuit, I love the simplicity of it. I can't wait to start ordering the parts and build it. I guess the next step is to figure out what to get to keep the mosfets cool.

Thanks for the explanation too, that makes sense now for the most part. I think where my initial confusion was is that I was thinking the watt rating on a component like the current sense resistors had to be equal to or greater than the current that was being passed through them, i.e. 58.1i / 3resistors, or 19.37i x 12v = 232 watts, but it sounds like my thinking was incorrect. You said that since the resistance is so low on these, the watt rating does not need to be as high since they will mostly pass the current through and not dissapate it in the form of heat. So, is the watt rating more related to the amount of current the component has to be able to absorb, if you will, in the form of heat, than it is to the amount of current it can conduct through it?
 
So, is the watt rating more related to the amount of current the component has to be able to absorb, if you will, in the form of heat, than it is to the amount of current it can conduct through it?
Click to expand...

Pretty much exactly correct.
 
()blivion,

It is hard to hook up FETs in parallel when they are running in a linear region. The difference in the threshold voltage og the FETs (often 2 volts) means a few will hog all the current. From that standpoint old fashon transistors are better. They still need some resistance in the emitter to make up for the difference in base emitter drop. but not so big. 1200 watts is huge. Trying to think of some kind of load box.
 

Yeah, I had thought about BJT's. But I wasn't sure which was going to be better. So I had figured just stick with parts the OP already knows about and has worked with before. But we can go with BJT's if you really think that's going to be better? (And you ARE prolly right in reality.) It would more than likely be better for me and the falstad simulator.

Thing about MOSFET's, they have positive temperature coefficients (if I am remembering correctly). So, at least in theory, they can balance them selves out to a certain degree. I have no idea how much that is going to be applicable to this project. But I was kinda banking on it.

Your saying it's no good?
 
 
I am currently working on a test of the MOSFET method using 7 FDB7030BL's (60 Watts each). It is aimed at a total power load of 350 Watts, and will be low budget water cooled. Here are some pictures of the MOSFET's and the copper pipe they are soldered to...

View attachment 65928View attachment 65929

I plan on using pretty much the exact circuit that I suggested in post #32. We will see what happens when I'm finished. Chances are that ronv is correct and I will end up destroying one or more of the FET's. I can live with this as they were salvaged parts, so they are worth nothing to me.
 
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