Arduino > Arduino
Electric motor with a power supply
macrojd:
Hi guys. This is not completely related with Arduino but it's for an Arduino project. I have an electric motor that runs with 1.5 volts. It was running with a battery, so I bought a small power supply of 4.5 volts DC and created a voltage divider with 2 resistors. The 4.5volts are transformed to 1.5 volts. It should be fine, but when I connect the output to the electric motor, the motor doesn't work. It works with an AA battery but not with the power supply.
Is there something I'm missing? It could be the frequency of the power supply? I'm new and completely lost. I thought that DC current was DC current and a power supply should provide the same type of current than a small battery, even better, but I can't make it work.
Any help would be appreciated.
Thanks
JD
mariush:
You will need to use a voltage regulator for the motor, it's not possible to power the motor with voltage obtained using a voltage divider (2 resistors). As you add a load on one of the branches, that load basically is like a resistor put in parallel with the resistor from the voltage divider, hence the voltage will drop.
Use an adjustable linear regulator like LM317 and configure it for 1.5v, or if you need better efficiency, use a switching regulator like MC34063 (more complex, requires inductors and more parts but there are online calculators out there which give you the values required for the parts)
You can turn the motor on and off by using a npn transistor
macrojd:
Hi mariush. Thank you so much. Now I have a better idea. Do you think a module like this could work?
http://www.ebay.ca/itm/1PC-New-2-5V-5V-to-1-8V-AMS1117-1-8V-Power-Supply-Module-AMS1117-/130978017819
It looks like a good voltage regulator and provides 1.8 volts, that should be fine for a motor that works with a AA battery.
JD
dr_p:
That's basically a AMS1117-1.8V linear voltage regulator with filtering caps.
It will be fine for your needs.
mariush:
macrodj: as dr_p says, that board basically consists of a linear regulator and the input and output capacitors required by the linear regulator to work. The -18 in the AMS1117 tells you the chip is "locked" to 1.8v output, it has the resistors normally required for adjustment of output voltage integrated inside the chip.
It would work for you, but it is not a perfect solution.
For example, if your motor locks, gets blocked, whatever, it's possible that the energy stored inside the motor to flow back into the board and the linear regulator and damage it. Same with solenoids or relays... when you trigger a relay it would work fine, but when you disconnect the relay all that energy in the coil has to go somewhere and if you're not careful, it can go back into the linear regulator and damage it.
That small board is missing a diode which has the role of protecting the regulator and giving the current going back towards the regulator a path around the regulator, for example back into the input voltage.
You gain more, you learn more by making your own such board. Buy a LM317 chip, they're cheap and easy to find. Then have a look at the datasheet and read the application hints which explain everything you need to know.
For example, here's the datasheet for LM117 and LM317 (information is valid for both) : http://www.ti.com/lit/ds/symlink/lm117.pdf
If you go down to page 10-11, it goes through and explains why each component is needed, so you can easily figure out what you need for 1.5v. On page 11, it explains why the diodes are required for some applications, like running your motor.
Just note that pretty much all examples are made using LM117 and works the same as LM317 with just one minor difference .. while LM117 needs a R1 equal to 240 ohm, with LM317 you need 100-120 ohm. So pretty much everywhere you see R1 or 240 ohm, you just change that in your mind to 100-120 ohm (pick whatever you can find around).
The solution for this is to use a diode or a couple of diodes to make sure there's a direct path back to
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