Today my breadboard came through mail and I created my first circuit!

Took a mini fan from an old laptop cooler base, and I made the connection at the breadboard, taking power from a 9V battery. Ok, I might not be Tesla, but I got it working

Now, playing around I also powered the mini fan using a smaller 1.5V battery. It worked again, this time with much lower speed though.

Now my questions are:

1. On the fan's label (photo attached) I see an indication of DC5V. I suppose that this is the required voltage that the fan needs in order to run? If so, why did it run with 1.5V as well?
2. Should I have user a resistor in the first case (9V)? Or a resistor already exists in the fan's circuit board (located under the label sticker)?

Thanks!
John

 

Hello,

To start with, the fan is made for 5 Volts.
According to the datasheet it will work from 3 to 6 Volts.

Bertus

 

According to the fan's label, your fan works at 5V DC. And yes, of course it will run at a lower voltage, but it will run very slowly. It's the equivalent of connecting a 6V flashlight light bulb to two 1.5V batteries in series (3V total)... the bulb will light up, but more dimly. If you run this fan at 9V, in time you're going to burn its coil and it will stop working, you need to place a resistor in series with it to make sure that not too much current is being drawn by the circuit

 

If this is a relatively modern fan it is most likely BLDC, internal circuitry, vary rare to see a DC PC motor fan now.
Max.

 

Sorry, yeah, it says 5 Volts. I'll correct the initial thread.

 

If this is a relatively modern fan it is most likely BLDC, internal circuitry, vary rare to see a DC PC motor fan now.
Actually it shows a 5v with AC/DC symbol?!
Max.

I guess it has, here is what is under the sticker:

 

Yes BLDC. Uses something like the Microchip 625a ic.
Max.

 

What Max is saying, is that it is a Brushless type DC Motor, which is a type of highly efficient motor that requires a special driver circuit.

 

If you run this fan at 9V, in time you're going to burn its coil and it will stop working, you need to place a resistor in series with it to make sure that not too much current is being drawn by the circuit

Thanks, I'd appreciate some help with this. Since it is the first time I am planning to use Ohm's Law (to figure out the required resistor), here are my thoughts:

I=V/R so R=V/I

I know that voltage supplied by the battery is 9V. Do I need a multimeter to find out the current of the circuit? Or there is another way? At the datasheet, I see that the rated current of the fan is 0.09Amp. Is this helpful somehow?

Thanks!
John

 

There are people in this forum that are far more knowledgeable than me. But I doubt that your motor will follow ohm's law to the letter, since it's not a purely resistive load. There are other two factors in there, which are inductance and capacitance.
But, if you look at the label carefully, it says that your fan runs at 5V and consumes 0.4W of power.
What does that tell you?

 

There are people in this forum that are far more knowledgeable than me. But I doubt that your motor will follow ohm's law to the letter, since it's not a purely resistive load. There are other two factors in there, which are inductance and capacitance.
But, if you look at the label carefully, it says that your fan runs at 5V and consumes 0.4W of power.
What does that tell you?

I just googled about Watts. "In an electrical system power (P) is equal to the voltage multiplied by the current."
So, P=VI which means 0.4W = 5V * I so
I= 0.4W/5V = 0.08 amperes

Is this correct?

PS: In the calculation above, I have not used 9V for voltage, instead I have used 5V. Why should I do that since i connect it on a 9V battery?

 

If this is a relatively modern fan it is most likely BLDC, internal circuitry, vary rare to see a DC PC motor fan now.
Max.

As opposed to an ac pc fan?

 

I just googled about Watts. "In an electrical system power (P) is equal to the voltage multiplied by the current."
So, P=VI which means 0.4W = 5V * I so
I= 0.4W/5V = 0.08 amperes

Is this correct?

PS: In the calculation above, I have not used 9V for voltage, instead I have used 5V. Why should I do that since i connect it on a 9V battery?

Yes, that is correct... Now, since you're not feeding voltage to a direct load, but to a driver circuit, I would not advise to just plug a 9V batter directly to it, otherwise you'd fry the circuit.

There are two options here:

1. Use a resistor voltage divider, which is a bad idea, since it will draw a lot of current from the battery and be very inefficient.
2. Use a voltage regulator circuit. You can just google "7805 circuit" and find out for yourself. This is a far better idea.

 

Here is the typical BLDC ic used, rated at 5v supply, so any more could damage the circuit.
http://ww1.microchip.com/downloads/en/DeviceDoc/21452C.pdf
Although the sheet shows a DC fan motor.
Extra needed for BLDC motor.
Max.

 

Use a voltage regulator circuit. You can just google "7805 circuit" and find out for yourself. This is a far better idea.

Actually this looks interesting. I'll give it a try anyways, just to exercise.
However, why shouldn't I use just a simple resistor in the circuit?

Thanks again!

 

Resistor only works as long as the current is a constant, i.e. does not vary.
Max.

 

Max is right, a BLDC motor's driver circuit is a "moving target" regarding how much current it draws. That's because the circuit tries to keep the fan running at a constant speed, and is constantly adjusting the current it delivers to it.
If you place a resistor in series with the fan, it might work for a while (the resistor will heat up, though) but in the long run it will damage the circuit.

 

Actually this looks interesting. I'll give it a try anyways, just to exercise.
However, why shouldn't I use just a simple resistor in the circuit?

Thanks again!

Another thing, if you don't own a multimeter, I suggest you buy yourself one at your local electronics store. You shouldn't spend more than 15 bucks on it, and it will be an extremely valuable tool for you to get started.

 

The problem with using a resistor to limit current is that the fan is not drawing a constant current. It is energizing different coils at different times. The current draw that you calculate from the voltage and power is an average. In reality, it might be drawing twice as much current when a coil is activated and almost no current when no coil is activated. This means that if you calculate a resistor to drop the extra voltage, the voltage seed by the fan at different times will range from less than 5V to nearly the full 9V. And 9V could damage the electronics.

Bop

 

Building yourself a 5V regulated power supply is a very good noob project and will come in very handy down the road. I highly suggest you go for it.

One reason you didn't ruin your fan may be that the battery voltage sagged quite a lot under the load of the fan, exactly as if you had a resistor in series. If the fan was drawing 80mA (or more, if the voltage was higher), this is enough to challenge an aging 9V battery. The voltage while running could well have been 7V or less.