1. The method of distinguishing speed-up turnouts on the distribution board to indicate circuit failure
When positioning, you can judge the fault and range of the indicated circuit by measuring the AC and DC voltage between terminals X1 and X2 (or X1 and X3 when the position is reversed) and the current on terminal 3 of BD1-7. It means that when the circuit is working normally, the voltage between terminals X1 and X2 of the distribution board can be measured at about 60V AC, about 22V DC (X1+, X2-), and the current on terminal 3 of BD1-7 is about 50mA. Positioning to measure the voltage and current of X1 and X2, no voltage and no current, indoor fault. If the AC voltage of X1 and X2 is about 70v, it proves that X4 is open circuit, and then test that X2 and X4 have 70v voltage, it proves that X4 is open circuit indoors, otherwise, X4 is open circuit outdoors. If X1 and X2 have 110v voltage and no current, it proves that the outdoor X1 and X2 are disconnected, and then test X1 and X4 have 110V, which means that X1 is open outdoor, and X1 and X4 have no voltage, which means that X2 line is open outdoor). X1 and X2 have 110V and a large current (115mA), which means that the 1000Ω resistance in the room is short-circuited. Another judgment method: the test shows the current of the transformer BB3/4. If there is current, it is judged as indoor short-circuit; if there is no current, it is judged as Indoor open circuit, the reverse is the same. The following takes the positioning indication fault as an example for analysis.
1. When the current is 0, and the voltage of X1 and X2 is about AC105V, it can be judged that the outdoor X1 or X2 is open. If X1 and X4 are powered, X1 is turned on, otherwise, X2 is turned on. If there is no voltage between X1 and X2, it can be judged that it is an open circuit in the room (3 to X2 of BD1-7 or 4 to X1 of BD1-7 is open).
2. When the current is 40mA, the voltage of X1 and X2 is AC70V, DC35V, it can be judged that the branch X4 is open. Then test that X2 and X4 have AC70V voltage, it proves that X4 is open circuit indoors, otherwise X4 is open circuit outdoors.
3. When the current is 50mA, the voltage of X1 and X2 is AC60V, DC-22V (X1-, X2+), then it can be judged that the polarity of the outdoor diode is reversed.
4. When the current is 78mA, and the voltage of X1 and X2 is AC28V, DC0V, it can be judged that the outdoor diode is broken down (short circuit).
5. When the current is 100mA, and the voltage of X1 and X2 is AC5V, DC0V, it can be judged that it is a mixed line fault.
2. Briefly describe the function of AC five-wire turnout circuit
Taking the S700K turnout circuit as shown in the box diagram, the functions of the five wires are:
X1: A-phase power line of the action circuit and the common return line indicated by the reverse position of the switch;
X2: The transmission line of the B-phase power supply and the positioning indication line when the switch is moved to the positioning;
X3: The transmission line of the C-phase power supply and the reverse position indication line when the switch is pulled to the reverse position;
X4: The transmission line and positioning indication line of the B-phase power supply when the turnout is pulled to the reverse position;
X5: The transmission line of the C-phase power supply and the reverse position indication line when the switch is moved to the positioning position.
When the turnout is lost, it indicates a fault (1/3 closure is positioning as an example), as shown in the figure, at this time, the positioning indicates that the circuit is communicated by X1, X2, and X4.
First of all, the nature of the fault can be determined by measuring the current of No. 3 or No. 4 terminals of the transformer indicated by BD1-7.
When the turnout is an open-circuit fault (open circuit and semi-open circuit), the current is lower than the normal value (the current is much lower than 45mA); when the turnout is a short-circuit fault (short circuit and semi-short circuit), the current is higher than the normal value (the current reaches 90 mA above), it is mostly a short-circuit fault.
3. When the turnout fault is open circuit (the current value is less than the normal value)
Measure the voltage between X1 and X2 on the side of the combination to further judge the fault range. If the measured voltage between X1 and X2 is about 110 V, it can be determined that there is an open circuit between X1 and X2. At this time, you can use X2 to measure the voltage between X4 and X4: if the voltage is slightly smaller than the voltage between X2 and X1, then the X1 branch is normal, and the fault point is located in the X2 branch. Use X1 to find the fault point of the X2 branch; if X2 If there is no voltage between X4 and X2, the X2 branch is normal, and the fault point is located in the X1 branch. Just use X2 to find the fault point of the X1 branch. Measure the voltage between X1 and X2 to be 70-110 V, then it is a semi-open circuit, and the search method is similar to the open circuit fault search method.
If the measured voltage between X1 and X2 is about 10 V higher than the normal value, it can be determined that the X4 branch is open, and X2 can be used to find the fault point of the X4 branch. When the test value is between the open circuit value and the normal value, it is half For open circuit faults, the search method is the same as above.
If the measured voltage between X1 and X2 is 0 V, it can be determined that there is an open circuit between X1 and X2 in the combination. At this time, use BD1-7 to indicate the No. 3 terminal of the transformer to measure the voltage of side 05-1. If the voltage is about 110 V, it is determined that the X2 branch is open. At this time, use X1 to find the fault point on the X2 branch; if there is no 110 V Left and right voltage, it can be determined that the X1 branch is open. At this time, use BD1-7 to represent the No. 3 terminal of the transformer, and find the X1 branch to determine the fault point.
4. When the turnout is short-circuited (the current value is greater than the normal value)
A2: In the S700K display circuit, the display relay and rectifier box (also known as diode or rectifier) are connected in parallel. The direct current rectified by the alternating current through the rectifier box acts on the coil of the indicating relay to make it excited. If the two ends of the rectifier box are short-circuited and the DC cannot be rectified, it means that the relay cannot operate. Therefore, it means that the circuit short circuit fault is essentially a short circuit at both ends of the rectifier box.
As shown in the figure above, one end of the rectifier box is connected to BD1-7, which means the No. 3 terminal of the transformer, that is, the X2 branch. There are 24 contacts in 1 contact group and 2QDJ-132 contacts to the coil of DBJ 4) There is a short circuit point, which is called uniqueness here. Determine the fault point by measuring the position of the current change on X2.
The other end of the rectifier box is the No. 4 terminal of the BD1-7 transformer, which is the X1 branch circuit. From the above figure, it can be seen that the power of X1 is divided into three circuits through the motor. The first path returns to X4 through terminals B11 and A11, the second path returns to X3 through terminals B10 and C10, and the third path returns to the other end of the rectifier box through B10 and A10.