EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Mighty Burger on March 19, 2016, 01:15:13 am
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Hello! I'm kind of new here and I'm new to electronics. After learning a little bit about electronics, I want to start my first project. It will be a speed-controlled fan (from a desktop computer fan) that can be powered either by a wall adapter or battery. The fan runs on 12 volts 0.55 amps. The wall adapter provides 12 volts 600 milliamps so that shouldn't be a problem. When I want to use a battery, I will use 2 9v batteries in series to get 18 volts, then use a buck converter to step it down to 12 volts. (I know it'll work with 9v but I want it to run at max speed)
Here's the link for the buck converter I might use www.amazon.com/RioRand-LM2596-Converter-1-23V-30V-1Pcs-LM2596/dp/B008BHAOQO/ (http://www.amazon.com/RioRand-LM2596-Converter-1-23V-30V-1Pcs-LM2596/dp/B008BHAOQO/)
And here's the schematic.
(http://i.imgur.com/t41ViNQ.png) You might notice I have two switches, one SPST one SPDT, instead of one SPDT center-off. This is because of what I plan to do irl. I will use the S1 to switch between batteries and wall power which will be located in the back of the unit, and I will use S2 to turn it off and on which will be conveniently located in the front. Also, I'm not using PWM, because even though it would be the best option, it's a little beyond me at this point.
My first question is: will it work? If not, what can I do to fix it?
My second question is: if I have the switch set to draw from the wall adapter, thus leaving the buck converter outputs in an open circuit, will the buck converter still draw current from the batteries? It seems like something that could be answered with something as simple as using google, but I could not find my answer after a lot of searching.
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If you reduce the voltage to a DC motor then the motor gets reduced torque then it cannot start running. That is why we use Pulse-Width-Modulation with full power pulses for lots of torque and the width of the pulses determines the speed of the motor. Since the switching transistor or Mosfet turns fully on and turns fully off then it stays cool, instead of wasting power making heat like in your circuit.
Also you show a P-channel Mosfet connected backwards so it will not work, the motor will always be at almost full speed
A buck converter always uses battery current even when it has no load. You do not need a buck converter, simply design the circuit for whatever maximum speed you want from the motor.
Two ordinary 9V alkaline batteries produce 18V at 0.55A for only 15 minutes when their voltage has run down to 12V.
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@audioguru is correct. PWM is the best method to control a motor or fan. Use a small compact microcontroller like a pic and it should be a good excersise in programming to get it running well. Good luck and check back with the forum if you need some help. Plenty of bit banging know-it-alls here. :-+
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My second question is: if I have the switch set to draw from the wall adapter, thus leaving the buck converter outputs in an open circuit, will the buck converter still draw current from the batteries? It seems like something that could be answered with something as simple as using google, but I could not find my answer after a lot of searching.
The buck converter will still draw current even when there's no load across it. This is known as 'quiescent current'. The module here uses the LM2596 buck converter chip from TI. Looking at the datasheet (http://www.ti.com/lit/ds/symlink/lm2596.pdf), page 4 states a typical quiescent current of 5 mA to 10 mA.
If you still like to consider PWM, another way of implementing it would be to use a 555 timer. It would be a great way of playing with the 555 timer as well.