Working principle of shutdown bright spot elimination circuit

In the circuit diagram of the tail board of the OM8373 super chip color TV, there is often a shutdown bright spot elimination circuit composed of CY07 (100μF), VDY09 (1N4148), VY10 (2SC1815), etc., as shown in the attached figure. Among them, CY07 is used to store energy. RY33 provides a charge and discharge circuit for CY07, and YDY09 mainly acts as a switch to control the current added to the VY10 pole. In the normal state of the circuit, when the power is turned on and +8v voltage is generated. The +8v voltage forms a loop through CY07 and RY33 and charges CY07. At the beginning of charging, its charging current is large, the voltage drop across RY33 is high, and VDY09 is in the cut-off state. The conduction current of CY07 passes through the emitter junction of RY31 and VY10, providing a forward bias current to VY10, so VY10 is turned on when the power is just turned on. YDYOIA/02A/03A is turned on, the RCB signal current is bypassed, the cathode of the electron gun is cut off, and the picture tube is black. However, as CY07 is charged, the left positive and right negative charging voltages are gradually established at both ends. When the negative voltage reaches 0.7V, since the positive end of VDY09 is grounded, VDY09 is turned on, the current flowing into RY31 is bypassed, VY1O is cut off, the RGB circuit is not affected, and the fluorescent screen begins to light up. At this time, CY07 continues to charge through the resistance value of RY33 and VDY09 after the internal resistance is connected in parallel, and the charging speed is accelerated. The voltage of its negative end (right end) becomes more and more negative until the charging is balanced.
During the entire process from the beginning of CY07 charging to the end of charging, the node current of YC07 and RY33, VDY09, and RY31 remains at OmA. That is, the current flowing into the node is equal to the current flowing out of the node (Kirchhoff's current law). Therefore, the more negative the negative terminal voltage of VDY09 is, the smaller the charging current is, and the larger the conduction current of VDY09 is, and it will remain in the conduction state when CY07 is in the charge balance state, and VY10 will be cut off.
When the power is turned off, the +8v voltage disappears. CY07 discharges through the +8Y power supply internal resistance and RY33, but because the capacitor has the basic characteristics that the voltage polarity cannot change suddenly but the current direction can change suddenly, its discharge current is opposite to the charging current. At this time, VDY09 is cut off, and the discharge current passes through RY33 and VY10 emitter junction. At the same time, CY06 also discharges through the emitter junction of VY10, so that VY10 obtains forward conduction current, VDYOIA/02A/03A is turned on, and the cathode of the picture tube is cut off, which plays the role of eliminating the bright spot when the power is turned off.
Therefore, in the attached figure, VDY09 is turned off at the moment when CY07 starts to charge. When CY07 continues to charge so that the left positive and right negative voltages at both ends increase, VDY09 is turned on and VY10 is turned off. When CY07 discharges, VDY09 is turned off and VYIO is turned on. When a color TV set using this shutdown bright spot elimination circuit has a bright spot in the center of the screen after shutdown, the fault can be eliminated by replacing CY07 and CDY09.
However, it should be noted that in the OM8373 super chip color TV, due to the black current detection circuit, some models such as Changhong SF2539 omit this shutdown bright spot elimination circuit in the actual object, because the RCB output terminal of OM8373 will output a low level when in standby mode.