MCU drive relay mode

Circuit 1:

 
Many designs use this circuit to drive the relay. Although it can work, it is unreasonable to do so. After careful analysis, it is found that Q3 cannot be fully saturated.
Let's not count the resistance of R1. Assume that we make the base current of Q1 the largest and take R1=0. When the control signal voltage is 0, the voltage of Q1 eb is 0.7V, and the voltage of ec is also 0.7V. When the 9012 tube is fully saturated, the voltage of ec should be 0.2V. Obviously, the tube is not fully saturated; the maximum voltage on the relay can only be 11.3V.
In order to fully saturate the tube, the base current must be large enough, so the base voltage needs to be below -0.7V.
 
Circuit 2:
 
Let's take a look at this circuit again.

When the control terminal voltage is 0, the base voltage of Q1 is (12-0.7=11.3V). Changing the size of R1 can change the base current. When the base current is large enough, the triode is saturated.
In order to verify the above analysis, we built a circuit, R1 is 4.7K, the base current is 2.4ma, the Q1 ec voltage is 0.2V, and the voltage across the relay is 11.8V.
Note: The value of R1 cannot be too small to ensure that the base current is within the safe range, and it cannot be too large to ensure that the transistor can be fully saturated. This can be calculated through voltage and resistance.
The first circuit can work because the relay has a wide voltage range. Sometimes it can barely work even if it is undervoltage, but the situation is unstable. Therefore, we do not recommend this method when designing. The
correct circuit should be circuit 2. The correct connection method and the appropriate base resistance can ensure the rationality and stability of the design.
Finally, it should be noted that the 12V relay used in this experiment cannot be directly driven by the IO port of the microcontroller, otherwise it will not be turned off. If a 5V relay is selected, it can be used, and the principle is the same as above.