Hello fellows,
What do you think about this circuit? https://ibb.co/8gvxgjS
I intend create a circuit where I can activate this motor (https://datasheet.octopart.com/MOT1N-Velleman-datasheet-22075344.pdf) and regulate your rotation through this potentiometer (https://www.tme.eu/fr/details/ow20bu-10k-lin/potentiom-plastique-a-rotation-unique/omeg/ow20bu-10ka-bush-cp7mm-spindle-f2-l-16mm/).
Firstly I think use four this batteries (https://data.energizer.com/pdfs/l91.pdf), theoretically will last 18 hours (6000mAh/320mA = 18Hours and 45 minutes). But to don't waste battery and harm de environment I am thinking use this power bank (https://www.mi.com/global/10000mAh-mi-18w-fast-charge-power-bank-3/specs). Where will she last around 17 hours (I am using parameter Rated Capacity: 5500mAh/320mA = 17 Hours).
What do you think about this circuit? The goal is turn on the led, while the circuit is turn on and regulate the velocity this DC motor.
Thanks
A few points.
Here's a link to a better data sheet, which I've also attached to this post in case that site stops working.
https://www.jprelec.co.uk/pdffiles/450-010.pdfWhat load is the motor driving?
A motor behaves like a DC voltage source in series with a small resistance. The DC voltage (back EMF) opposes the supply voltage and is proportional to the shaft's speed. When stalled the current is determined by the resistance of the motor's winding. As the shaft spins faster, the voltage generated by the motor approaches the supply voltage and the current drops. When unloaded, the back EMF is near the supply voltage and the current is low. As a mechanical load slows the shaft down, the current increases. If the shaft is spun fasted by applying a rotational force in the same direction as the shaft is spinning, thus speeding it up, the current direction will reverse and the motor will act as a generator, charging the battery.
The 300mA specified on the data sheet is when nothing is connected to the shaft. When stalled the motor current will be close to 4A, at 3V, which indicates its internal resistance is V/I = 3V/4A = 0.75R. The stall current is briefly drawn when the motor is started, but can't be sustained for long without damage. Try powering it up and measuring the current with your finger pushing against the shaft to slow it down. Obviously don't do this for too long, otherwise you can burn your finger, as well as the motor. According to the data sheet, the rated current is just over 1A, with a load of 14g·cm, at which point the speed will be 11k5 rpm. The motor might overheat if this load is exceeded for too long.
A linear controller which dissipates the voltage difference between the supply and motor, gives off V
difference*I Watts of heat. This applies regardless of whether the element dissipating the heat is a transistor, or plain old resistor. A PWM controller i.e. one where the transistor is either on, or off, does not have this issue, which is why this has been recommended, even though it's more complicated.
If the motor voltage is reduced, the motor can't draw as much current, due to Ohm's law.
If you're using linear method to drop the voltage, the maximum possible power dissipation in the resistor, or transistor will be a quarter of the motor's power consumption when stalled. Half the voltage will be dropped across the motor and half across the resistor or transistor and the current will be half of the rated stall current. P = VI = 1.5*/2A = 3W. Your resistor/transistor will need to be able to dissipate at least 3W for short periods of time. The continuous power rating of the resistor/transistor should be about half the motor's full power rating, so 3*1/2 = 1.5W.
If you don't have time to read the above, make sure your resistor/transistor can safely dissipate 3W peak and 1.5W continuously.