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

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I am going to try to breadboard it with just one MOSFET and make sure it works before soldering anything.
 
Okay, thanks...I do have one (some) already, something that I kind of already asked but still don't really "get", it has been bothering me. The R1 resistor, .13 ohm current sense resistor, it is only rated for 5 watts, it seems like we are running a lot of current through with 5 amps and 24 volts, is that too much for the little resistor? I know it's not because you guys know what you are doing. Secondly, I just want to confirm that I understand the schematic correctly, I hook up the positive terminal to the drain on the MOSFET, the op-amp to the gate, then the negative feedback of the op-amp (not sure if that term is right) to the source on the MOSFET and the .13 ohm resistor to the source as well, then the other side of the .13 ohm resistor to ground?
 
The resistor is working pretty hard at 6 amps. In hindsight we might have made it a little bigger when we went from 5 to 6 amps per FET.
I guess we can give you the secret.....

If you make a little triange

E=Voltage, I = Current , R = Resistance Then there is everyones favorite.. P=IE where P = power

E
------------
I R

So if we know 2 we can calculate the 3rd. Like E=IR. Or R=E/I

Since we know I & R we can substitute E and use P=I2R so 5X5X.13 = 3.25 watts.

The answer to your other questions are "yes".
 
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One thing I want to point out that gets some people. When working with IC's, it is a major PITA getting what pins connect to what parts right in your head. Very convoluted.

For starters, the pin numbers read like a clock... NOT like a text. So if you start at the dot, and count down the chip, you DON'T cross over diagonally back to the top and count down again. You instead jump DIRECTLY over, then count up. So if we say connect pin 7 to part X, you count from the dot, going down 1, 2, 3, 4... (Jump Directly over), then start going up 5, 6, 7. I have burned a few IC's because of this. Secondly, things are mirrored on the bottom and top. This... gets... me... every... single... time... It helps to take the time and write the number on the bottom side next to all the holes the IC goes into. Just make sure you get it right when you do it.
 
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I see, I think where I was getting tripped up was I was thinking the voltage was 24/12, but with ohms law voltage would be calculated as about 0.65v with 0.13 ohms and 5 amps. So power being volts * amps or i*v you get 3.25 watts, or, as you said, i^2*R. I think there is still some part of this I am not quite understanding like I thought the voltage would be constant through the circuit, guess not. At any rate, I will take your word for it that it works

I don't think the .13 ohm resistors will be a problem because the highest the PSU will go is 55 amps, and I probably won't be going that high for any extended period of time. If I do go that high it would just be for a moment to test the over current protection shut off logic.
 
LOL.

So my 60~70 YO grandma is yelling at me for having sent something useless to her and getting her hopes up. She thought some family from AZ was writing her or something. I'm like... What in the hell are you talking about, I didn't send anything to you... Are you getting senile? (she has been) Besides, do you even have family in AZ? Do you need to be checked into a home now? There is NO WAY I had anything to do with this you crazy old bat.

Then... as I was getting repeatedly struck in the head with 20 some P-Fet's taped to a piece of card board... it occurred to me... I don't think I told my dear old grandma that I was having something sent to me in her name /)
 
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It's not working. I know I am probably doing something wrong. I would take a picture but it won't let me upload from my iPhone and I don't have access to a regular computer right now. I will take a picture tomorrow.
 
It happens. Almost every single project I start I have to scratch my head a bit before I get it to work the way it's supposed to. It's almost always a stupid small wiring error, such as crossing two wires that should not be crossed, getting parts backwards, and so on. It comes with the territory. You just gota keep playing with it till it works. And sadly, you are more than likely going to blow a part or two. Also comes with the territory.
 
OK, I attached a bunch of pictures from different angles. It is kind of hard to see how everything is hooked up in the pictures. The logic power supply is the one closest to the front, the DUT is plugged into the back of the bread board. I have gone over it again and again and I can't see anything I have wrong, but it doesn't do anything. When I turn the power on I can turn the pot from one extreme to the other and no change in current, it is always 0 current. One thought I had was if I was connecting to the proper "ground" connection since the schematic doesn't differentiate between the logic supply ground and the DUT ground.

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The variable pot is connected to the white and orange wire at c62 and f49 (the pot itself is not shown in the picture). I know the wire I am using for the load is WAY too small a gauge, I have 10 gauge wire to use when I build the real thing, I was just using this to test it out. I was going to just leave the pot turned way down so there wasn't much current running through it, but I couldn't get any current.
 
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I think it may be that the ground from the big supply is not connected to the ground of the little supply. But before you change that it might be a good idea to make some measurements on pins 1, 2, 3, 4, and 8 on the op amp.
Need to be real careful, because without the big heatsink the FET will get really hot really fast and be toast.
The way it is now pin 1 & 3 will be kind of indeterminate, but 3 should be adjustable between 0 and about .6 volts. Pin 8 should be +12 volts and Pin 4 0 volts with respect to the small supply ground. If all is well leave the pot set to 0 or as low as it will go. If you happen to have say a 10 ohm resistor we could replace the .13 ohm resistor with it just to test. Then we wouldn't burn up the FET.
Here is how it works I=E/R so .13 volts across the .13 ohm is 1 amp. That means there is 11.87 volts at 1 amp across the FET. So the FET would be disappating almost 12 watts. With no heatsink its temperature will go up 65C per watt. Smoke...
 
Doesn't the replacement 10 ohm resistor have to be different than a normal resistor, I thought there is something about that resistor that made it a current sense resistor? I have a bunch I various value resistors I got from radio shack a while back and I am sure I can find a 10 ohm one but it's just a regular small resistor, not the big current sense type like the .13 ohm one, will that work? Or what if I just don't leave it turned on for more than a few seconds to test?
 
Yep a small one will work because the current will now be low.

It will smoke before you can say ahh #$@%%.

Are the 2 grounds the problem?
 
Yep I fried something . Good thing I ordered extra parts! I connected the two grounds. Maybe my problem lies in you said above I should be getting between 0 and .6 v on pin 3 of the op amp, but I am getting a constant 11v and it's alway 11 no matter what th pot is set to.
 
The .13 resistor is what smoked. Is there a quick way to test the op amp an MOSFET to be sure they aren't fried too?
 
Yep, But need to fix pin 3 first. With 11 volts the FET is all the way on so it's probaly just the resistor that is bad.
One side of the pot goes to ground, the center to pin 3 and the other leg to the 68.1k resistor. I'll have another look at the pictures.
 
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