With the widespread application of self-recovery fuses in the electronics industry, we are paying more and more attention to the safety of electronic products. We need to know a kind of performance of the products.
Among the technical indicators of PPTC, one of the technical indicators is Vmaxi. Vmaxi represents: the maximum voltage that the self-recovery fuse protector can withstand in the blocking state. That is to say: in a circuit with a series of resettable fuses, when the current of the circuit is abnormal, the resettable fuse will change from low resistance to high resistance within a certain time range, thereby preventing the flow of abnormally large current , To protect the follow-up circuit from being damaged by high current. At this time, the voltage of the circuit is almost all 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, permanent damage will occur, and the self-recovery fuse cannot be recovered. In the long-term application process, there have been some problems that ignore the technical requirements of Vmaxi.
For example: when an engineer chooses a resettable fuse, the voltage of the circuit is 12V, and the selected resettable fuse is a 16V series. When performing experiments to verify the function and function of the resettable fuse, an external voltage stabilizer and steady current The power supply, in order to obtain the minimum current of the self-recovery fuse startup protection, under the condition that the load remains unchanged, slowly adjust the output voltage of the power supply to increase the load current. When the current reaches the minimum current of the self-recovery fuse startup protection, self-recovery The fuse enters the protection state, but at the same time the self-recovery fuse is burnt out. At this time, the voltage output of the stabilized power supply has reached 37V.
　　From the above situation: 1. The model selection is correct; 2. The experimental method is incorrect; 3. Vmaxi is ignored;
　　Analysis of the reason: During the experiment, the technical requirements of Vmaxi were ignored, causing the applied voltage to be far greater than Vmaxi, causing the self-recovery fuse to burn out.
　　This is because the engineer used the load resistance to remain constant and increased the power supply voltage to change the load current. 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 change the load current.
　　 When the current reaches the minimum current for the resettable fuse to start protection, within a certain time range, the resistance of the resettable fuse will change from low resistance to high resistance, reducing the load current to about 10mA, preventing the flow of large currents. Use a stabilized current power supply to verify the function of the self-recovery fuse:
1. Set the output voltage of the regulated and steady current power supply to the required voltage;
2. Short-circuit the positive and negative output lines of the stabilized current power supply;
3. Adjust the current output knob to make the output current reach the required current value;
4. Connect the two pins of the self-recovery fuse directly to the positive and negative terminals of the output line of the stabilized current power supply;
5. Turn on the stabilized and stabilized current power supply. When the output of the power supply jumps from the stabilized current state to the stabilized state, the self-recovery fuse has entered the protection state and the whole experiment is over.