Continued from the previous article "Circuit Analysis and Maintenance of Bread Maker (1)".
Principle:When the base of Q2 comes from the signal of the control board to start heating (non-zero volt high level), Q2 is turned on, the voltage on R12 is pulled down, the potential difference makes the relay close, and the heating tube gets 220V power. Problem Process:Q2 opens, the relay closes, this action causes the 12V voltage to be pulled down to 94V, while the 6V supplying the control chip is pulled down to 36V, the main control chip is at 3At 6V voltage, it is equivalent to being powered off, then the high-level control signal sent to the base of Q2 is interrupted, the level of the control signal received by the base of Q2 becomes zero, Q2 is cut-off, and the relay is disconnected again. Try to solve:When trying to reduce the current flow of the relay closure circuit and increasing the R12 from 100 ohms to 300 ohms, it was found that the relay could be continuously closed. However, the brightness of the control panel is significantly reduced, indicating that Q2 conducts on to close the relay, which consumes most of the current capacity. If the Q1 conduction drive motor is running at this time, the light on the display board will flash again and again, and it will be at the boundary of resetting at any time. Continue to increase the R12 resistance from 100 ohms to 400 ohms, the circuit is very stable, and the voltage will not fall below 10V, but the relay will not close. It is explained that after the Q2 base receives the conduction signal, the relay cannot be closed due to the significant reduction of the current at 400 ohms. Therefore, adjusting the R12 resistance value does not solve the problem. Using the common 9014 tube instead of S8050, the result is the same, indicating that the Q2 transistor function is normal. Here D8 is a relay protection diode, I suspect that it is leakageThe measured forward and reverse resistance values are normal, and the fault is still the same if you change one. The 12V power supply was connected to the driver board separately from the outside, and it was found that the fault disappeared completely, proving that the DC working voltage generated by the C part of the circuit could not drag the load. Clams can't add an extra 12V power supply, which is too troublesome. Therefore, the problem is focused on the C part of the circuit. However, when the circuit C part is not energized, the resistance value of each point is not abnormal. After powering on, the standby working voltage is also normal, and the control motor is also running normally. It's just that after the switch Q2 is turned on and the relay is engaged, the tens of milliamperes of this branch of Q2 will immediately pull down the DC working voltage of the entire circuit. Replace EC1, Z1 and Z2, and find that the fault is still there, and it is not a problem caused by the voltage regulator filter (the voltage regulator tube is IN4735A). In the end, there was only C4, the yellow shell that I never doubted (it was really you). Usually this kind of low-cost resistor-capacitance step-down circuit is designed to be much higher than the working environment when considering safety, so it is the most durable and almost undamaged. 
Try to connect the 1UF capacitor in parallel on C4, boot up, everything is fine, the working voltage of 12V is maintained very well, even if the two switch tubes are turned on at the same time and the relay is engaged, 12V only falls down to 104v。
The nominal value on the C4 shell is 15UF, then in the circuit environment of 50Hz220V, its capacitive reactance is about 2123 ohms, and the obtained current capacity is about 103 mA. By connecting 1UF in parallel on C4, the obtained current capacity is increased to 172 mA. (Reactance = 1 (2*3.)14*50*0.0000015), current capacity = 220 capacitive reactance).
Of course, to increase the load capacity of the C part of the circuit, you can also consider changing the half-wave to full-wave rectification, but that is not as convenient and simple as the C4 expansion. Note: Most of these small household appliances use low-cost resistor-capacitance step-down circuits to get low-voltage DC, which is a non-high-voltage and high-voltage isolation circuit, and it is necessary to pay attention to safety in disassembly and maintenance, because 220V is not completely isolated. Okay, let's try to make a big steamed bun now. After 3 hours, it is automatically done. 