Electrical – How to test phone vibrator motor

You probably need a motor-run capacitor to make the AC induction motor work properly. It does say 8uF on the label.

Also, I really, really doubt the first motor is a stepper motor. It states 100V, 50/60 hz. 100V is the Japanese line voltage standard, and I have never seen a stepper motor with a specification in hz.

Edit Ok, the first motor is a 4IK25RGN-A. This is a 1/30th horsepower AC induction motor.

Realistically, neither of these motors are appropriate at all for a servo system. They both look to be fractional HP induction motors.

You can control speed of an AC motor with a variable-frequency drive. However, VFDs (and induction motors in general) do not have much 0 RPM torque, and don't provide control of the actual motor position, just the motor RPM.

Really, you need a brushed servo-motor, or a brushless AC servo-motor (or even stepper motors). All of these use a different mechanism to generate motion then an induction motor, which is actually a rotating transformer.

Are the reflective sensors the only components missing from this schematic?

What are the reflective sensors? Do they provide a digital output, or are they ordinary analogue parts? If they are analogue, and there are no other parts, then there are some issues.

I assume this is for window blinds, in which case, it needs to deal with the possibility of the sensors being blinded or confused by sun-light.

I'd expect to see something controlling the emitter current, to modulate the reflective sensors light source to avoid 'sun' problems. Some sensors might do that internally, but it would help us t see the part number for that sensor.

How is the reflective sensor signal derived? For an analogue sensor, I'd expect to see a resistor, or something so that the sensor can provide a voltage dependent on the intensity of light.

I'd concerned about switching logic gates with an analogue signal. I'd have expected Schmitt triggers to convert the analogue voltage to a logic level.

Edit: I would expect to see pull-up or pull-down resistors connected to the switches.

When switches are open-circuit, they have no defined voltage on one of the pins. That voltage could be set by static, or radio reception. The inputs for many modern CMOS chips requires a very, very small current. So they could be switched by that 'stray' current.

Normally, the connection to the switch which is going to change (when the switch is pressed) has a pull-up resistor connecting it to V+, or pull-down resistor connecting it to ground. That resistor pulls the state of that connection to a known 'default' value, and hence prevents 'stray' electricity accidentally triggering the circuit. When the switch is pressed, the short circuit 'beats' the resistor and defines the voltage on that connection.

Edit: This issue is not about debouncing. The circuit might work okay even if the switches did 'bounce'. The issue is having connections from open switches with no defined voltage value. Every single throw switch must have a pull-up or pull-down resistor, to reliably define the voltage on the open-circuit wire. I might debounce the switch too.

I'd expect to see some decoupling capacitors across all of the ICs. When an IC switches, it will require a 'pulse' of current. That pulse could cause a glitch in other chips using the same power supply. To ensure stability across the whole circuit, that pulse is supplied by a small capacitor connected very close to each chip.

Edit: What is a decoupling capacitor and how do I know if I need one? looks like it covers all the bases.

I'd like to see some capacitance across the power supply input.

I'd be tempted to use one power source (e.g. 12V) for the entire circuit, and use a voltage regulator to step down to 5V. The chips only under a few 10's of mA on the logic 5V supply, so using an appropriate linear regulator should be okay.