Electronic – Why the transistor switched when base was grounded

Your battery is not able to supply enough current to run the motor AND the relay AND everything else. This may be caused by an inadequate battery, or it may be caused by inadequately-heavy-gauge wiring between the battery, the motor, and the relay contacts.

With the relay open, everything is fine. When you key the transistor, current flows through the relay coil, the relay pulls in, closing the circuit to the motor. The motor then draws a huge current, which the battery can't supply. The battery voltage sags, current through the relay coil drops, and the relay drops out, cutting current to the motor. Battery voltage pops back up, the transistor is still keyed, current is restored through the relay coil, and the cycle repeats.

Some years ago, I watched a similar phenomenon on a test rig. We were using a very nice regulated power supply to control some solenoid-operated water jets. With all the jets enabled at once, the total load was more than the supply could handle, and it started pulsing like crazy, as the overcurrent protector kicked in and out. The fix was to energize no more than half of the jets simultaneously.

If you read the section regarding DLY in the datasheet for your sensor you'll see that it suggests using a simple resistive divider to set the delay. Normally, you want to drive your ADC with a low output impedance compared to your ADC's input impedance. Your resistive dividers output impedance would be R1 || R2 (assuming your voltage source is perfect with no output impedance) - so approximately 580 Ω for the circuit below - this should be OK.

But let's suppose you want to set the delay to 15 min - which corresponds to 1.8V from the datasheet. To drop 1.8 V with the divider, R2 would have to be around 5.6 kΩ. The output impedance would then be 3.5 kΩ. This might cause problems if your ADC's input impedance is just 10 kΩ.

But since the datasheet suggests a resistive divider is OK and also states that the input impedance is high, I think you'll be OK with the following (don't quote me on this, though!)

This means something like this:

R2 will drop a voltage of 0.2 V across it. Rest of the voltage (3.3 - 0.2) will be dropped on R1. If your input voltage is 5V instead of 3.3, R2 will have to be around 400 Ohms.

If you do have issues due to the output impedance you can buffer the output of the voltage divider with a unity gain buffer. A unity gain buffer has a very low output impedance and should drive your ADC very well.