# Toy DC Motor (Size 130)

Fleeting- External reference: https://www.adafruit.com/product/711
- External reference: https://sciencing.com/size-capacitor-electric-motor-7994649.html
- External reference: https://electronics.stackexchange.com/questions/132720/how-do-i-find-the-voltage-range-for-an-unknown-dc-motor
- External reference: https://www.instructables.com/Toy-Motors/
- External reference: https://components101.com/motors/toy-dc-motor

Toy DC Motor

Motor Specifications

- Standard 130 Type DC motor
- Operating Voltage: 4.5V to 9V
- Recommended/Rated Voltage: 6V
- Current at No load: 70mA (max)
- No-load Speed: 9000 rpm
- Loaded current: 250mA (approx)
- Rated Load: 10g*cm
- Motor Size: 27.5mm x 20mm x 15mm
- Weight: 17 grams

Current is like Juice, it depends on our capacity how much can we drink it without getting our stomach burst, similarly its CIRCUIT’S CAPACITY how much current can it bear without getting circuit’s components burst due to heat.

if we are using a 9 V, 1000 mAh battery; it will provide 9 Volts all the way but current again depends upon the circuit. If our circuit. is demanding 1A or 1000mA, Battery will last for an hour, and if it is demanding 500mA the battery will last for 2 hours

high torque to start i.e., due to larger load, thus they will need a higher current to start.

current is required to overcome current generated by back EMF and that is why we need a starter for a DC motor

textbooks don’t provide a simple and practical method.

cap’s store charge when the voltage is turned on and will get charged and thus by connecting it across the terminals or in parallel with the motor, it will increase the voltage slowly to the maximum value

To control speed or RPM of a DC motor we can limit the input power to the motor. But as the current depends on the circuit, we will harshly be able to control it without disturbing the circuit.

only option we are left with is Voltage control and for constant voltage we can only limit the voltage by various voltage regulators like if we want to run a motor at 5 volts we can use IC LM7805 for limiting the input voltage(works up to 35 volts), or we can reduce the RPM by introducing a gear box at the shaft.

simple model for a DC motor is V=R∗i+eV=R∗i+eV= R*i + e, where VVV is the terminal voltage, RRR is the motor resistance, and eee is the back-emf voltage

best way to measure resistance is to take several measurements and average.

lock the rotor and then apply a small current to the terminals.

Measure voltage and current and calculate R = V/I

. Repeat several times and average

rotor.

eee can be determined from e=Kb∗ωe=Kb∗ωe = K_b* ω, where KbKbK_b is the back-emf constant (units of V/(rad/sec) or V/RPM) and ωωω is the speed in the same units as KbKbK_b.

You’ve already found KbKbK_b. It is just 16V570RPM=28.07VkRPM=0.268Vrad/sec16V570RPM=28.07VkRPM=0.268Vrad/sec\frac{16 V}{570 RPM} = 28.07 \frac{V}{kRPM} = 0.268 \frac{V}{rad/sec}. As somebody else mentioned, the torque constant of a motor is equivalent to the back-emf constant, so Kt=0.268NmAKt=0.268NmAK_t = 0.268 \frac{Nm}{A}.

motor needs a little bit of energy to start the rotation of its metal shaft.

— https://sciencing.com/size-capacitor-electric-motor-7994649.html

capacitor is used to supply this initial push to the motor.

— https://sciencing.com/size-capacitor-electric-motor-7994649.html

Multiply 0.5 times the square of the voltage. Call this result “x.”. Continuing the example, you have 0.5 times 11.5 volts times 11.5 volts, or 66.1 square volts for “x”.

— https://sciencing.com/size-capacitor-electric-motor-7994649.html

Divide the start-up energy requirement, in joules, of the motor by “x” to arrive at the capacitor size needed in farads. The start-up energy of the motor is found either in its documentation or written on the motor itself. Assuming a start-up energy of 0.00033 joules, you obtain 0.33 divided by 66.1 which equals 5.0 times 10^-6 farads. The symbol “^-” denotes a negative exponent.

— https://sciencing.com/size-capacitor-electric-motor-7994649.html

Convert the capacitance to microfarads by dividing by 10^6, or one million, since a microfarad equals one-millionth of a single farad. The microfarad is the common unit for capacitors. Completing the exercise leads to 5.0 times 10^-6 farads divided by 10^6 microfarads per farad, or 5.0 microfarads

— https://sciencing.com/size-capacitor-electric-motor-7994649.html

Technical Details

- Operating Temperature: -10°C ~ +60°C
- Rated Voltage: 6.0VDC
- Rated Load: 10 g*cm
- No-load Current: 70 mA max
- No-load Speed: 9100 ±1800 rpm
- Loaded Current: 250 mA max
- Loaded Speed: 4500 ±1500 rpm
- Starting Torque: 20 g*cm
- Starting Voltage: 2.0
- Stall Current: 500mA max
- Body Size: 27.5mm x 20mm x 15mm
- Shaft Size: 8mm x 2mm diameter
- Weight: 17.5 grams