[SOLVED] [Moved] adding cooling fan in a power supply

[rosemarie321]

good day..

i had my power supply which i built last week and i want to add my 12vdc cooling fan..
if i have this schematic diagram below., how can i put my cooling fan, if my power supply is adjustable from 0-30 v, and my cooling fan must only be receive max of 12v dc.

any idea how to connect a resistor with my cooling fan?

 

[Chullaa]

This is simple! you can use 7812 IC to convert from 30 Volt to 12 Volt. It has three pins (Input, Gnd, Output)
All you need to is
1. connect Input of 7812 with input of LM317.
2. Connect Ground (Gnd) of 7812 to -ve terminal of power supply (internally; also with -ve of DC fan).
3. Connect Output of 7812 to +ve terminal of DC fan.

It will start fan whenever you turn on your power supply.
Hope it helps.
Umair

---------- Post added at 16:17 ---------- Previous post was at 16:16 ----------

P.S. dont need to use any resistor.

 

[rosemarie321]

by d way.. i already connect it to my ckt.. and it works.. but, is it ok that my 7812 become hot? even though it already had its heat sink?

 

[enjunear]

by d way.. i already connect it to my ckt.. and it works.. but, is it ok that my 7812 become hot? even though it already had its heat sink?

That depends on how hot it's getting. If you are making the junction temperature run close to Tj,max (from the datasheet), then you'll impact the lifetime of the part and likely induce early failure.

Remember that a linear regulator will have to dissipate Vdrop*Ipass watts of power. That is, Vdrop = Vinput - Voutput, and Ipass = current passed to load (plus some internal current, but that's typically only a few hundred uA's). So, if you have Vin = 30V and Vout = 12V, then Vdrop = 30-12 = 18 volts. If your fan draws 200 mA (0.2 amps), then the regulator will have to dissipate 18V * 0.2A = 3.6 Watts! If you can put a thermocouple on the flange of the part to get its steady-state temperature, you can compute the junction temperature (Tj).

Tjunction = Tcase + θjc*Pdissipated, where θjc is case-to-junction thermal resistance in degrees C per Watt (from spec sheet).

If θjc = 20 C/W, and Tcase (the metal flange) is measured at 45C, then:
Tjunction = 45 + 20*3.6 = 117 deg C
Most parts are spec'd with Tj,max of 200 C (above that, they burn up and fail).

Use those equations to figure out how hot your part could get (recall that is ambient air temperature increases, like when the power supply is stuck in a corner with inadequate airflow, the case temp rises, and so will the junction temperature).


One way to reduce the strain on your regulator would be to stick a dropping resistor ahead of it (in series with the +30V supply). Dropping a few volts across a small power resistor (wirewound, ceramic, etc) might help keep the regulator running cooler (works longer). Say you can find a 2W power resistor, lets determine the maximum resistance you could use. If you assume that same 200 mA current draw, then let the resistor dissipate 75% of it's rated maximum (same rationale as not running the junction temp of your regulator at 200C... it'll last longer).
P = 75% * 2W = 1.5 W
We know that, for a resistor: P = I^2 * R
So, 1.5 W = 0.2^2 * R ..... R = 37.5 ohms, maximum.

If we use a 37 ohm resistor, then it will drop V=I*R.... V = 0.2*37 = 7.4 volts.
That would make the voltage at the input to the regulator 30-7.4 = 22.6V. Now, the power dissipation of the regulator is reduced to (22.6-12)*0.2 = 2.12 watts, and the junction temperature would drop to approximately 45 + 20*2.12 = 87.4 degrees C.

This has just been a description of the steps you can take to understand the problem, and one way to mitigate the extreme thermal stress your regulator appears to be under. Pull up the vendor's datasheet for your part and look at it's limitations, then use the above method to see where your part is truly operating.

 

[Chullaa]

First you need to know how much of current your fan takes at full speed. then multiple it with 12V. It should not exceed 1A as 7805 can bear up to 1A of current.
If it exceeds you need to replace regulator with higher power OR connect another 7805 in parallel with previous 7805 to achieve 2A current.

Sometimes connecting opposite polarity of 7805 cause heating too. Also check this.
Umair

 

[goldsmith]

Hello my friends ,
You can use LM2576 +12v
It is a switching step down regulator , with high efficiency , and can give you 3 amps .
BTW , you can use your fan to cooling your regulator , and your circuit , at one time !
Best Wishes
Goldsmith