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Help with Water Pump

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Well....oops. No big deal-I have a few extra mosfets.
I did identify source, gate and drain from the datasheet, so all is good there.
It looked to me that gate, and if I have it right gate is where the arrow head symbol is, and source which I think is the bottom connection on the mosfet, are connected to a common wire that goes to ground. No doubt I am misreading the schematic. Any chance you can straighten me out?
 
[QUOTE Any chance you can straighten me out?[/QUOTE]

Hey! There it is. I read your response way too quickly, sorry.

Oh dear. Only the source should be connected to ground. The FET gate (check the datasheet for the FET you have, to identify the gate terminal and indeed all three terminals) should connect to the pots and appropriate IC pin. The 'bolt' symbol is in fact an arrow-head symbol and represents an internal diode inherent in the FET. Ignore it. The FET drain terminal connects to the pump. My schematic shows g,d,s to help you.

That part threw me for a loop. I'm gonna run back down to the basement and get another mosfet and hook the gate to the soft start, then on to the pots and IC. Thanks for keeping me moving. Other than this goof up, I'm pretty sure it's wired as shown. We'll see, I guess.
 
Is it OK to lay the wire coming from the gate across the middle tab and one end tab of a pot and solder?
 
Is it OK to lay the wire coming from the gate across the middle tab and one end tab of a pot and solder?
No. I suggested way back to use 10 Ohm resistors on the FET gates (they suppress oscillation). So connect the gate wire to one end of a 10 Ohm resistor, then connect the other end of the resistor to the pot middle and end tabs if you're not using the soft-start circuit. If you are using soft-start then the gate connects via the 10 Ohm resistor to the output of one opamp of the LM324 IC.
Note: the FET has two drain connections. One is the middle 'leg' and the other is the tag with the hole. So be sure the tag isn't touching anything it shouldn't.
 
If you are using soft-start then the gate connects via the 10 Ohm resistor to the output of one opamp of the LM324 IC.
QUOTE]

Thanks for reminding me about the 10 Ohm resistor, I do have that marked on my schematic. I also appreciate the warning about the tab on the FET. My first attempt was such a mess, I started over and am using ()blivion's soldering method. Now it's methodical and neat and I can tell where I've been. Thank you ()blivion.
Anyway, I have the FET installed except for the gate. I was going to install the 10 Ohm resistor, wire in the four components of the soft start, then send a wire to pin 10 of CD40106 IC and the pots. On the pots, I was going to solder to the middle and an end tab by laying the wire across the two tabs. Do I have this right?
 
On the pots, I was going to solder to the middle and an end tab by laying the wire across the two tabs. Do I have this right?
Yes. Pick the end tab which provides decreasing resistance between it and the wiper as the pot is rotated clockwise. If you pick the 'wrong' end tab the result will be that turning the pot clockwise will decrease, rather than increase, the pump on/off time. No big deal but a tad confusing.
 
I fired it up, but the light bulb stayed on no matter how I adjusted the pots. No smoke or pops or anything like that. Any idea what the problem is likely to be? I did adjust the power supply to 24v.
 
It just occured to me that I failed to ground any pins in the CD40106 IC. I connected the soft start to pin 4 on the IC and two tabs on the pots, then connected pin 3 between the diodes and capacitor on the other side of the pots.

Did I probably fry some components? If I only run this one module, which pins from the IC should go to ground?
 
the light bulb stayed on no matter how I adjusted the pots. No smoke or pops or anything like that.
It's quite usual for a project to have a teething problem or two. At least we can assume the FET is working and nothing is disastrously wrong.
Questions:
1) Are you getting 12V from the 7812 regulator output?
2) Are you driving the FET gate from the 40106 or from the LM324?
3) Have you built just one pump driver module or all four?
3a) If just one pump driver module, which pin numbers are you using on the 40106 and LM324?
4) Have you grounded all unused input pins of the 40106?
5) Have you double-checked the ICs are inserted the right way round in their sockets and no legs have been bent underneath?
6) What voltages do you have at pin 14 of the 40106 and pin 4 of the LM324?
7) Are pin 7 of the 40106 and pin 11 of the LM324 grounded?

Your answers should guide the next stage in getting things working.
 
Ah, our posts crossed. Nevertheless your answers to confirm all the above points would be useful.
Did I probably fry some components?
Probably not.
If I only run this one module, which pins from the IC should go to ground?
For the 40106, you should have pin 7 permanently grounded and if you're using pins 3 and 4 then temporarily ground 1,9,11,13 (I assume you are using pins 5 and 6 for the soft-start triangle generator). For the LM324, connect the input pins of the three unused opamps to the +12V supply line temporarily.
 
I grounded pin 7 and this thing came to life!!!!! It works-it's alive! I was having all kinds of trouble trying to temporarily groung the odd pins so I decided to give it a go and what a pleasant surprise. Pretty exciting stuff. Anyway, I'm getting 11.9v at the regulator and pin 14. I'm going to salvage what I can from my first mauling of components and wire & put three more modules on this board.

Thank you Alec for the superb schematic and top notch advice. Thank you very much for dragging me across the finish line on this project.
 
Glad you've got there. Always pleasing to know that a circuit design works in the real world. :). The reason for properly connecting unused pins is to prevent erratic behaviour of the ICs. Good luck with the rest of the build. When you come to drive pumps rather than lamps it's advisable to connect an interference suppression cap (0.1μF) between the two terminals of each pump (if it doesn't already have one).
 
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Got the other three modules built and grounded pins 13 & 5 on the 40106 IC. Everything is working as planned. Amazing schematic. I ended up trimming over an inch from the board-at first I thought there's no way all this will fit.

"The reason for properly connecting unused pins is to prevent erratic behaviour of the ICs."
You must have read my mind, I was going to ask about that.

"When you come to drive pumps rather than lamps it's advisable to connect an interference suppression cap (0.1μF) between the two terminals of each pump (if it doesn't already have one)."

I don't understand. I do remember you said a cap needs to have a voltage at least twice that of the circuit. I have no idea what's inside the pump. Is that something I need to determine?
Or, are you saying that I should add a cap to the positive lead to each pump and another cap to the negative lead to each pump?

The controllers that came with the pumps have little LEDs to indicate power to the pump. I plan to T off the pump leads and hook up those LEDs. As long as the LEDs can take 24v, that shouldn't be a problem, is that right?
 
Everything is working as planned.
Congratulations!
are you saying that I should add a cap to the positive lead to each pump and another cap to the negative lead to each pump?
Even if the pump has a capacitor already it won't hurt to add another. One end of cap to pump pos, the other end to pump neg. The cap will be a plastic or ceramic type, rated about 50V or more.
As long as the LEDs can take 24v, that shouldn't be a problem, is that right?
LEDs are essentially current driven, not voltage. The average red/yellow/green LED drops about 2V across it (Vf in the datasheet) when passing current. The spec for the LED gives its max current (usually about 30mA). Allowing a good safety margin I'd suggest run the LED at half its max, say 15mA (0.015A). So you will need a resistor in series with it to limit the current. R = (24 - 2) / 0.015 = ~ 1k5 for that example. (If R = 1k then the LED current would be ~ 22mA).
 
Thanks. As luck would have it, I have four caps leftover that will do the trick.

Negative pump lead is soldered together with the FET drain and a diode. Positive pump lead is soldered to a 24v line I ran down the middle of the board. Would it be OK to solder one cap lead to the pump lead, FET drain and diode, and the other cap lead anywhere on the 24v line?
 
That hopefully will be ok. Ideally, though, the cap should be on the pump motor itself, as close to the pump motor terminals as possible so as to suppress electrical noise at source.
 
I got the caps and LEDs installed and have a tin to be my Faraday cage.

Would a coat of polyurathane floor finish be good as a conformal coating? Being oil base it's nonpolar and would leave a fairly thick film. I could dip the whole circuit and swish it around. Then let it drip dry. The pots and jacks are suface mount, so I would only dip the circuit board. That would not hurt the 40106 IC, would it? It most definitely would make it hard to replace the 40106. Maybe I could encase the 40106 in vaseline first.
 
There are latex-based conformal coating liquids available, which have the advantage that you can peel the coating off if necessary. However, if you don't want to buy any then available polyurethane varnish should be ok. You can solder through that if you have to (WARNING from the Health & Safety Police: burning polyurethane gives off toxic fumes).
I wouldn't use vaseline on the 40106. Not sure about its conductivity (after all, it's recommended for use on car battery terminals). Perhaps masking/insulating tape?
 
Thanks. I'd think if vasiline is reccomended for car battery terminals, it'd be because it does not conduct. Generally speaking, anything oil base or grease is non polar.
 
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