Thermostat question for fermentation chiller

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yusupov

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I am building a son of fermentation chiller. The thermostat I purchased is a ritetemp 6020 which I want to connect to the 3inch cooling fan I got from radio shack. I want to run the fan entirely from the thermostat which has batteries, without using an adapter.

Where do the red and blue wires from the cooling fan connect to on the thermostat e.g. RH, RC, Y, G???

Thanks, for any help!
 
I'm assuming Green is your ground. RH is hot, RC is cold. Connect red to RH, blue to RC, and ground your G wire. Not sure what the yellow could be without looking at it. I'm no electrician though, so double check before you trust me on that. It's just what I would assume is right.
 
Thanks for the answer. I didn't word my question very well (sorry), I have one red wire and one blue wire, where do I put each wire? there are no other wires.
 
At the moment I have it going off a 9v battery just to test that the circuit works (RH and G terminals (works)). But when it comes time to put it in the fridge I might use a big battery like the ones you put in a big flashlight. What do you think?

You don't think a 9v battery will last long enough for the two+ weeks of fermentation?
 
Why not use a Sears Garden Tractor or motorcycle battery if you must have a remote battery power supply system vs using a wart cube power supply?
I'm surprised your stat will still operate on a 9 vilt battery operating the relay in the stats coil not alone any fan as all the house states I have used operate off a 24 volt AC transformer Any reason why you can not use a 12 VDC wart cube then run the wires to your fermenters location? That would power up the stat as well the cooling fan, plus a cheap wire run. Heck even run it underground unless your locations are between solid concrete. JMO.
 
At the moment I have it going off a 9v battery just to test that the circuit works (RH and G terminals (works)). But when it comes time to put it in the fridge I might use a big battery like the ones you put in a big flashlight. What do you think?

You don't think a 9v battery will last long enough for the two+ weeks of fermentation?

A typical 9v alkaline battery has at best 600 mAh. Assuming 100 mA draw from your fan, (which is very low), at 9 volts, you get 6 hours of run time. With a more reasonable draw of 500 mA, you drop down to a little over an hour. This isn't including the draw from the thermostat itself, (which granted will be very low).

I suppose it depends on the cycle time of your system.

A 6V lantern battery has 26000 mAh. Wire two in series to get you to the 12v that the thermostat would want, and two lantern batteries should last 260 hours with 100 mA draw, (although your fan will probably draw more like 500-1000 mA, dropping you down to 52-26 hrs, respectively).

I'd definitely go with a lantern battery setup if you are really gung ho on that....or like suggested above, just get a 12v "wall wart", (a transformer, like is used to charge a cellphone), and use that. You can get 12v wall warts for less than 5 bucks at surplus stores.
 
Yeah, good idea, I got an old charger and used that, which is working great!

Thanks.

Is it better to use the ice packs with the coolant/gel stuff than to use ice trays or frozen soda bottles? Does anybody use these?
 
If you have those packs laying around, go ahead and use them...but ice bottles work just as well and for a fraction of the price.

I always used old 1/2 gallon apple juice bottles (the round ones) because they stack nicely and won't rupture when frozen. I kept 8 of them. 3-4 in my SOFC at a time. Only had to change them about once every 2-3 days when the house was at 75F to keep beer at 65F.
 
Yeah, good idea, I got an old charger and used that, which is working great!

just be careful as many chargers when not under a load the voltage is increased by a rather large amount. A 12 volt DC cube will produce 15-16 volts without a load just a caution for your equipment that it can handle this higher voltage.
 
Can you dumb that down a little? My fan is 12VDC, 1.9 W, 160mA and my charger is 5.5VDC, 280mA.

Thanks.
 
That power supply will kick out more than 5.5 VDC but not near the limits of your fan so your cool.

Are you sure about this brewbeemer?

Power transformers work off of AC current, and step down power based on winding ratios. It was my understanding that the voltage was locked for a given power supply based on the geometry of the windings...

I agree that, either way, it doesn't matter....but I'm curious as to why having a load or no load would change the voltage....voltage is just potential difference, which shouldn't depend on load.

EDIT! My bad, you seem to be correct!

So I went to a wall wort I have plugged into a computer fan, (was using it as a stir plate until I got a real one). It's a 12v wart. Fan switched off, 21 volts across the wires. Fan switched on and left to run, 10.2 volts. Fan switched on but held stationary with a finger on the fan, 9.8 volts.

You are definitely right....but do you have an :)off:) reason why? I'm very curious now....only guess I have is that the inductance of the circuit is changing due to the change in load?
 
Are you sure about this brewbeemer?

Power transformers work off of AC current, and step down power based on winding ratios. It was my understanding that the voltage was locked for a given power supply based on the geometry of the windings...

I agree that, either way, it doesn't matter....but I'm curious as to why having a load or no load would change the voltage....voltage is just potential difference, which shouldn't depend on load.

EDIT! My bad, you seem to be correct!

So I went to a wall wort I have plugged into a computer fan, (was using it as a stir plate until I got a real one). It's a 12v wart. Fan switched off, 21 volts across the wires. Fan switched on and left to run, 10.2 volts. Fan switched on but held stationary with a finger on the fan, 9.8 volts.

You are definitely right....but do you have an :)off:) reason why? I'm very curious now....only guess I have is that the inductance of the circuit is changing due to the change in load?

I'm a dumb azz retired electrician (early by SSD) that liked to tinker around with equipment. Maybe an EE will chime in why this higher idle cube voltage voltage vs the cubes rated output voltage. On the locked rotor of your fan you might of pulled more Ma than the cube could provide hence the reduced to 9.8 voltage readings. My guess?
 
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