How do the two transistors in the picture control the relay?

It should be Q2Q4 is turned on, the relay K1 will be closed, and the main relay IO is to control the on and off of Q2Q4!

The driving of the relay is very simple. The circuit should be a DC-driven relay, so it can be turned on as long as it is energized. At this time, it is determined by the working principle of the device, and it cannot be changed.

There are several questions about the landlord’s circuit. 1. The main relay IO signal in the circuit is an AC signal, and the circuit is not a DC signal. 2. The connection point of R19 and R23 should be connected to the connection point of R21 and EC9, otherwise Q2 should not be turned on.

This output drive circuit is controlled by pulse, C9, EC8, D12, D14, C11, EC9 form a bootstrap boost circuit, which is similar to the principle of an AC voltage doubler circuit. The theoretical voltage value at both ends of EC9=2*VCC-1.4V. The connection point of R19 and R23 should be connected to the connection point of R21 and EC9, but it is redundant to use two transistors in series to drive the relay. The original creator may be that the IO port has another multiplexing function, otherwise there is no need to sacrifice simplicity and complexity.