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denny800 · Mar 7, 2010, 03:48 PM

I put together a little computer fan circuit using a PIR 555-28027 sensor through a NPN 2N4401switching transistor and I'm not sure if my connedtions are correct. The fan stays on as though the sensor isn't even connected. I have a drawing, but don't know how to put it in here so I'll try and explain.
Power 4.5v DC negative to the fan, positive & negative to the sensor. Out of the sensor to the collector of the transistor, positive to the base and emitter to the positive side of the fan.
Thanks

cdad · Mar 7, 2010, 04:01 PM

What exactly are you trying to do ? Also are you following directions for warming up the circuit?

denny800 · Mar 7, 2010, 05:09 PM

I'm wiring a fan to come on via sensor. I wasn't aware of warming up a circuit.

KISS · Mar 7, 2010, 05:10 PM

Use Go advanced/Manage attachments to post in various formats with limitations on size.

cdad · Mar 7, 2010, 05:11 PM

Look for the PDf file that outlines the sensor your using. When I looked it up before it spoke about a warm up time for the sensor so it can calibrate itself.

denny800 · Mar 7, 2010, 05:15 PM

Ok, are you saying to wait for the 10 to 60 seconds for the sensor to callibrate and then the sensor should work?

cdad · Mar 7, 2010, 05:17 PM

Yes, they called it warming up the circuit.

denny800 · Mar 7, 2010, 05:36 PM

Yes I have let it go for longer than the 60 seconds. You think the wiring is right. Can I put the diagram on here?

KISS · Mar 7, 2010, 06:04 PM

I posted how in post #4.

Your wiring it as a high side switch which it isn't. It has to be used like the circuit in fig #1 here: http://www-inst.eecs.berkeley.edu/~e...abs/2N4401.pdf

The 200 ohm resistor would be connections to the load or your fan.

The two resistors are needed too.

What's the fan current?

denny800 · Mar 7, 2010, 06:43 PM

In the first diagram the fan runs constant and in the second the sensor works fine but doesn't allow enough power to run the fan.
Thank you for the advanced help now you can see what I have.

denny800 · Mar 7, 2010, 06:50 PM

The fan is DC 12V 0.30a.

KISS · Mar 7, 2010, 07:02 PM

Neither method will work as you found out. I need to hone my schematic drawing skills and I haven't re-installed a package that does it.

The FAN + need to connect to + power
The FAN - needs to go to to the transistor collector
The emitter of the transistor needs to go to ground.

What we don't have is an equlivelent circuit for the PIR detector.
So, the output of the detector would normally go to the base of the transistor through a resistor. A much larger resistor would go from the base to ground. This gives a leakage path.

That's where figure #1 in the data sheet does.

The base needs to be current driven. Saturation (full on) would occur at 600 mA of collector current, so the base current required is 0.6 divided by Hfe(min).

The resistor needed would be sized by the voltage available to turn - a diode drop of the B-E junction.

R < (V - 0.6)/ (0.6 A / Hfe)

I need to look at the data sheet to find those numbers. I'm unhappy there was no equivalent circuit of the PIR detector.

It might work without the resistor.

denny800 · Mar 7, 2010, 07:12 PM

I printed out the 2n4401 switching transistor information. Thanks again for leading me to the right one. I'm still not clear on how I should make the connections to make the fan work along with the sensor. I have no problem paying for the information to get the right answers. Radio shack was not sure about a lot of this stuff and as you probably can see I'm not the very up on it either.
Thanks again.

KISS · Mar 7, 2010, 07:48 PM

Does this make sense now?

Modified fig #1 from data sheet.

denny800 · Mar 7, 2010, 08:08 PM

What I was told from Parallax this Pir was 555-28027.

KISS · Mar 7, 2010, 08:17 PM

Yep, http://www.jameco.com/Jameco/Product...DS/2082927.pdf But the data in the datasheet is useless. It doesn't give me much to go on for the characteristics of the output like an equlivelent circuit.

denny800 · Mar 7, 2010, 09:31 PM

If I'm reading what your saying and I have the connections correct, the fan comes on and stays on. The sensor must be working somewhat because the fan slows down 5% when it kicks on.
Denny

KISS · Mar 7, 2010, 09:43 PM

The drawing shows C and E shorted. E has to go to power minus.

denny800 · Mar 7, 2010, 11:03 PM

OK, then where does c go?

DanielF · Mar 7, 2010, 11:47 PM

Your second circuit is the correct one, but 2N4401 is not a power transistor, and is not really suitable for switching the fan.

Also, I'm guessing that the fan is probably intended for 5V operation, but your second circuit will deliver only about 3.5V (4.5V supply minus ~1V transistor B-E drop), which is probably why it's running slowly.

You'd be better off using the 2N4401 as an inverter stage and then driving a PNP power transistor with it. A 5V or 6V supply would also help compensate for transistor losses.

You should also put a 'freewheeling' diode across the fan motor to kill spikes that might otherwise kill your transistor.

Daniel