Stop delay on cooling fan

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Jimbo45

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I am using a 0.17 amp, 12 volt cooling fan on a Peltier plate. The fan is controlled by a thermostat. On shut down, I want a delayed stop on the cooling fan. I assume that I can accomplish this with a capacitor wired in parallel with the fan. I would like it to power down for at least one minute. Am I on the right track? How do I size this capacitor? Any help would be appreciated.
 
You would need a huge capacitor. 0.17 A for 60 seconds would be 10 Coulombs of charge. If the fan is OK down to 9 V, then you would need a 3.3 Farad capacitor, which is big and expensive.

You would be far better to have a transistor, maybe a MOSFET, to turn on the fan, and a capacitor on the gate of the MOSFET, so that the MOSFET says on for a minute or so.
 
The power to run the fan must come from somewhere. Battery or Capacitor or power line.
How often will the power down sequence happen? How long will the device be on? (long enough to charge a battery or cap?)
 
Diver300 said:
If the fan is OK down to 9 V, then you would need a 3.3 Farad capacitor, which is big and expensive.
Click to expand...
Smaller and less expensive, but way more complex for a relative newbie: charger > 3.6 V LiPO battery > boost converter.
You would be far better to have a transistor, maybe a MOSFET, to turn on the fan, and a capacitor on the gate of the MOSFET, so that the MOSFET says on for a minute or so.
Click to expand...
I think all power goes away, so a time-delayed switch won't work.

ak
 
Thanks for your interest and help. Project is Peltier cooler. Power supply is 12 volt battery. Thermostat calls for cool and turns on Peltier plate and fan. Both run minutes until thermostat is satisfied. Plate and fan de-energize from battery power, but I want fan continue to run on stored capacitor potential for stop delay to continue to cool the plate. If it can't be a minute, then as many seconds as practical. Voltage can decay from 12 to 6 to zero. Capacitor would recharge on next cycle. Goal is to minimize the transfer of residual heat in the Peltier plate back to the cold side for better thermal efficiency.
 
A 1 F cap is $76 at Digi-Key in ones. This will get you 30 seconds down to 6 V. However ...

A typical low-cost tube-axial fan does not always behave as a resistor. As the applied voltage decreases, so does the drawn current - for a while. At lower voltages the current becomes almost constant as the voltage keeps decreasing, decreasing the running time below the calculated estimate.

If fan power can be connected to the +12 V *before* the thermostat switch, then this becomes a monostable-and-MOSFET circuit, 1-2 dollars.

ak
 
Assuming the thermostat is doing it's job properly, won't it just turn everything back on a couple of seconds earlier without this addition?

Mike.
 
As previously suggested, I second using a MOSFET to drive the fan from the battery, with a capacitor to keep the fan on for your desired time.
The thermostat 12V signal is the In signal to the circuit.

Example circuit below:

 
You could use a supercapacitor module such as this, from ebay.
**broken link removed**

Connect negative to the fan negative.

Connect positive to the fan positive via a resistor to limit the charging current. That is a balance between how much load the power source can manage vs. how long it will take for the capacitor bank to reach near full charge.
If the normal "on" time is in minutes, 100 ohms should be OK. With that value, it need to be rated at 2W.

Add a rectifier diode across the resistor (any 1N4000 series type will do.) Connect cathode [banded end] to the motor.
That will allow the motor to draw the full current it needs while the capacitors are supplying the power.
 
I have a feeling the OP has some misunderstanding of (at best) efficiency, or (at worst) over unity devices.
 
Is there battery power available for the 1 minute that you want the fan to run?
 
gophert said:
I have a feeling the OP has some misunderstanding of (at best) efficiency, or (at worst) over unity devices.
Click to expand...
I'm not sure. He said
Goal is to minimize the transfer of residual heat in the Peltier plate back to the cold side for better thermal efficiency.
Click to expand...
and I can see a valid point there. If the hot side of the Peltier cell and it's heatsink are still warm, it's better to use the fan to get some of that heat to ambient than back to what is being cooled.
It could well increase the efficiency over a complete thermostat cycle. It's a bit like a pump continuing to run after a boiler shuts down. The residual heat in the boiler is better transferred to the water than lost to ambient as the boiler cools down.

It does depend on a lot of things, like thermal capacities, how much heat is transferred through an un-powered Peltier cell, and how much power the fan takes compared to the Peltier cell, so it could work against him.
 
Last edited:
crutschow said:
As previously suggested, I second using a MOSFET to drive the fan from the battery,
Click to expand...
Note that in Wally's circuit, "In" is connected to the 12 V *after* the thermostat, and "+12V" is connected to the 12 V *before* the thermostat.

ak
 
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