battery distribution fuse block.Usage and precautions for self restoring fuses

　　With the widespread application of self restoring fuses in the electronics industry in Shenzhen, we are paying more and more attention to the safety of electronic products. We need to know the performance of the products used

　　Among the technical indicators of PPTC, one is Vmaxi, which represents the maximum voltage that the self recovery fuse protector can withstand in a blocked state. That is to say, in a circuit where the self recovery fuse is connected in series, when the current of the circuit is abnormal, the self recovery fuse will jump from low resistance to high resistance within a certain time range, thereby preventing the flow of abnormal large current and protecting the subsequent circuit from being damaged by large current. At this time, almost all of the voltage of the circuit is added to the self recovery fuse.

　　If the voltage applied to the self recovery fuse exceeds Vmaxi at this time, it is easy to cause damage to the self recovery fuse, causing permanent damage and making it impossible to recover. In the long-term application process, there have been some issues that overlook the technical requirements of Vmaxi.

　　For example, when an engineer selects a self recovery fuse, the voltage used in the circuit is 12V, and the self recovery fuse used is a 16V series. When conducting experiments to verify the function and function of the self recovery fuse, an external voltage stabilizing and current stabilizing power supply is connected. In order to obtain the minimum current for the self recovery fuse to start protection, while the load remains unchanged, the output voltage of the power supply is slowly adjusted to increase the load current, When the current reaches the minimum current for the self recovery fuse to start protection, the self recovery fuse enters the protection state, but at the same time, the self recovery fuse burns out. At this time, the voltage output of the stabilized power supply has reached 37V.

　　From the above situation, it can be seen that: 1. The model selection is correct; 2. Incorrect experimental method; 3. Neglecting Vmaxi;

　　Analysis of the reason: During the experiment, the technical requirements of Vmaxi were ignored, resulting in an applied voltage far greater than Vmaxi, causing the self recovery fuse to burn out.

　　This is because the engineer used the method of keeping the load resistance constant and increasing the power supply voltage to change the current of the load. The correct method: Set the voltage of the power supply to 12V, connect the self recovery fuse in series in the circuit, change the load size of the circuit, and cause the load current to change.

　　When the current reaches the minimum current for the self recovery fuse to start protection, within a certain time range, the resistance of the self recovery fuse will jump from low resistance to high resistance, reducing the load current to about 10mA and preventing the flow of large currents. The method of verifying the function of the self recovery fuse using a stable voltage and current supply:

　　1. Set the output voltage of the stable voltage and current supply to the required voltage;

　　2. Short circuit the positive and negative output lines of the voltage stabilizing and current stabilizing power supplies;

　　3. Adjust the current output knob to achieve the desired current value;

　　4. Connect the two pins of the self recovery fuse directly to the positive and negative terminals of the output lines of the voltage stabilizing and current stabilizing power supplies;

　　5. Turn on the stable voltage and current supply, and when the output of the power supply jumps from the stable current state to the stable voltage state, the self recovery fuse has entered the protection state, and the entire experiment is over.

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