A design analysis question

[Speedyg8]

The purpose of this circuit is to control a fan using a temperature sensor. When the sensor reaches 50C, the fan turns ON. When the sensor drops below 49C, the fan turns OFF.

A comparator is used to detect when the sensor's output has reached 500mV. The output of the comparator then signals the transistor to enable the fan.

I have limited experience with electronics. I would like some help finding a solution, but I would prefer to learn as I go. So I would like to ask a few questions.

Does the transistor need to create a voltage drop (~12V) across the fan for it to operate?
I'm not sure how to determine R[Load] of the fan for the purpose of configuring the transistor. The specs of the fan are 12VDC, 0.45A.

Thanks in advance.

[Roosvelt Sanchez]

 Speedyg8 ,

Could you please print your schamatic in PDF or copy to MS Word in order I can open it without using multisim 10?

[SimpleEric]

Hello. See the attached MS file.

Question: Does the transistor need to create a voltage drop (~12V) across the fan for it to operate? Answer: Yes if you want it to run at maximum speed (12 V @ 0.45 mA). The motor will work fine even if you don't have a 12 V voltage drop.

Question: I'm not sure how to determine R[Load] of the fan for the purpose of configuring the transistor. The specs of the fan are 12VDC, 0.45A. Answer: Rload can be calculated with Ohms Law, R = U / I, R = 12 V / 0.45 A = 26.66 Ohm. Put a resistor with that value instead of the motor when you are simulating. If you connect Rload between 12 V and ground (0 V) you will get a current that is 0.45 A. If you put a transistor between Rload and ground to switch it on/off there are a couple of things to remember:

1) The collecor on the transistor must be connected to ground in this design or else it will not work. I have rotated it.

2) There is a voltage drop across the collector and emitter of the transistor so the motor can't get exactly 12 V. The drop is about 1.2 V so the motor will run at 10.8 V and 0.405 A. Imotor = U / R = (12 - 1.2) / 26.66.

3) The transistor must be able to handle the voltage (more than 12 V) and the current (more than 0.45 A). Your original transistor only handles 0.2 A. I have changed it to TIP42B (6 Amperes), but many other transistors (1 A or higher) will work fine. The power of the transistor can be calculated like this P = Udrop x Ic = 1.2 V x 0.4 A = 0.5 W. Use 1 W for safety. In a real design som sort of cooling may be needed for the transistor.

4) I have added a diode across the motor (Rload) since the off-switching will generate high voltage spikes because of the inductive nature of the motor. The diode is 1N4007. The diode must be able to handle the same current as the motor.

Other changes:

1) The LM393 pins on your model looked strange so I changed it. If you doubleclick on the model and select "pins" you can see where the pins should go. V+ to 12V, V- to gnd, output to pin 1 and so on. Your model had some other numbers.

2) I got 0.4 A when the gate current for the transistor was about 5 mA so the current gain is about 80 when it is saturated. A 10 Ohm base resistor will work. The 27 kOhm from base to 12 V is just there to make sure that the transistor is shutting off completely. If you

remove this the transistor will draw about 4 mA. (Not enough to run the motor though).

3) Switch (key A) is just there to simulate a voltage over and under the treshold (500 mV).

4) R3 adds some voltage to the reference voltage when the LM393 output is high (motor off). This means that the ON-treshold is higher than the OFF-treshold. This is called hysteres.

5) Use Rtrim to adjust the reference voltage.

W.k.r.

[SimpleEric]

Hello Speedyg8. See the pdf:s Fan_original and Fan_new.

W.k.r.