Electronic – Washer Machine Fix

Yes; best case the hFe of a BJT will be 100 or so, which means your 0.017 mA will turn into 1.7 mA, which is not enough to power the relay coil.

There's another problem, too: The current out of an Arduino will not be sufficient to drive the coil of a relay, as the typical spec is 25 mA per pin out, and typical relays use 35-100 mA of current for their coils.

I question your assumption, though: What is the "0.2V" that you need for the switch? What do you think that means? Where does this number come from? Specifically, when the switch is open, the voltage gap across the switch will be pretty much VCC, as the switch resistance will be close to infinite. When the switch is closed, the voltage across the switch will be close to zero, as the switch will have close to zero resistance.

There are various solutions to the core problem of "how to turn on a microcontroller with a button, and then keep it on until it's done." You might use a low-side N-channel MOSFET to switch on the relay coil. There would be a pull-down on the MOSFET gate, and the switch would pull it up to VCC. The digital out of the MCU would also be connected to this MOSFET gate, with a current limiting resistor that's lower than the pull-down, but high enough to not interfere with the switch when low. I'd suggest 10 kOhm for the pull-down, and 1 kOhm for the digital pin resistor, and the switch goes straight from the MOSFET gate to VCC. Note that the MCU needs to be able to pull the MOSFET gate both high and low, so a diode wouldn't work in that case.

If you can be more specific about what the "0.2V" requirement actually means, and what it's actually coming from, that would be useful, too. Almost all voltage specifications have to do with insulation ratings, and 0.2V is not within the range of those. Other ratings come from arc gaps, and those are typically 16V or higher as well. Other than that, the main important factor for switches is the amount of current interrupted, and that's typically rated in at least dozens of milliamps.

Thinking about it: Is the reason you need 0.2V differential "between switches" that you want to use an ADC to figure out which of many switches was used to start the MCU? If so, for 5V, a 0.2V differential is achieved with a ratio of resistors, rather than absolute values. A 24 Ohm resistor and 1 Ohm resistor will divide 5V into 4.8V and 0.2V, just like a 24 kOhm resistor and a 1 kOhm resistor will divide 5V into 4.8V and 0.2V, although with different amounts of current running through them!

Your motor that you linked to is a 4.5V 190~250 mA (No Load) motor. At 9v, the current probably increases. You are overdriving it by 200%. And any load/weight will cause it to increase in current requirements as well. Stall current is probably 10x that at least.

You are missing the protection diode across the motor, that can easily kill the transistor.

The Transistor you are using is a 100mA standard, 200mA Absolute Maximum. One of those motors by itself without any load, can easily kill that transistor.

The base resistor is calculated as (Base Voltage - Base-Emmiter Voltage) / Current required. Base Voltage is the Arduino pin, so 5v, Vbe depends on the collector current which is 200 mA here, so typically 1V. Current required is calculated as Collector Current (200mA) / Hfe (From datasheet, 10~30). On the safe side, lets go with 10, so 200 / 10 = 20mA needed at the base.

(5V - 1V) / 20mA or 4V / 0.02A = 200Ω resistor. A 1kΩ resistor would only allow 4mA at the base, which times the Hfe of 10, would only allow 40mA at the collector, probably no where enough to tun on the motor.

TLDR: You need the protection diode, your 9v power source is too high, and your transistor is too weak for the motor you are using. And you need a bigger resistor at the base because the motor requires more current then you are figuring. A common 2n2222 transistor with a 470Ω resistor would do much better.

Edit: Not making the pin an output also puts a damper on things. Answer, Arduino pins default to input.