diff npn motor

Thread Starter

duxbuz

Joined Feb 23, 2014
133
Hi
When I look at this circuit it seems that both motor terminals are connected to 5v and when the transistor opens it then allows one of the motor terminals to ground.
http://gangsterveggies.wordpress.com/2012/01/16/arduino-tutorial-1-way-dc-motors/

But when I look at other examples they seem to have a more common theme of one of the motor pins going only to the collector like this example:
http://quarkstream.wordpress.com/2012/05/31/arduino-12-transistors/

The first example seems weird to me.

Is one more desirable than the other?

Thanks
 

k7elp60

Joined Nov 4, 2008
562
The first example is correct. When the transistor is open the motor is stopped. When the transistor is on (base +) then the motor is on as one lead is to +5V and the other lead is to ground.
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
Thanks.

So the second example is wrong?

I was wondering which is preferable. The second example seems more common.
 

GopherT

Joined Nov 23, 2012
8,009
In the first example, a diode protects the circuit from reverse voltage. Reverse voltage occurs when the motor is turned off (transistor base is switched from 3 volts to 0 volts) and the motor slows down (motor acts like a generator).

The first also includes a capacitor to remove electrical noise caused by the motor and prevent interference with the arduino board.

The second is not wrong, it is just not as good or reliable.

If you use the first circuit, make sure your diode is the correct direction.
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
In regards to the first example here:
http://gangsterveggies.wordpress.com/2012/01/16/arduino-tutorial-1-way-dc-motors/

He mentions using a BC547.

I was looking at the datasheet and it mentions:
http://www.fairchildsemi.com/ds/BC/BC547.pdf
that Ic is max 100mA

He mentions prior to that it requires 340mA

Then I follow the calculations and using the 1k resistor mentioned in article and 5v from arduino pin and using forced beta of 10

I get:
(5V-0.7V)/1000 = 4.3mA
Ic = .0043 * 10(forced Beta) = 43mA

How's that going to work?

I am wondering if I am working this stuff out wrong?
 
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