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Common-mode voltage breakdown of op amp input stage and its protection measures

Source: InternetPublisher:方世玉223 Keywords: Op amp protection circuit Updated: 2024/09/09

Common-mode voltage breakdown of op amp input stage and its protection measures

The breakdown of common-mode input mainly refers to the breakdown of the collector junction of the input stage. Here, the μA301 integrated operational amplifier is used to illustrate its failure and protection measures, as shown in the figure. In Figure (a), the two input terminals of the μA301 operational amplifier are connected to the base of the NPN transistor. When the collector is connected to the power supply voltage Vs and works normally, the input terminal is always negative for Vs, so there will be no forward bias between the collector and the base of the two transistors in the input stage. However. If the power supply voltage Vs is removed from the circuit, and pins ② and ③ still maintain a positive potential, that is, there is still a common-mode input voltage. The collector junction of the transistor will have a current flowing from the base to the collector. If the source resistance from pin 2 or pin ③ to the power supply is very low, the collector junction current formed at this time will be very large, so that it can damage the input stage transistor. This phenomenon occurs when the power supply voltage is removed first, but the signal source is not removed. Common-mode input will cause the integrated operational amplifier to fail and be damaged. Another situation that makes the common-mode input of the integrated operational amplifier fail is that if the signal source contains a capacitor element, when the capacitor is charged to the high potential of the signal source, the power supply voltage Vs and the signal are removed at the same time. Due to the existence of the charging voltage of the signal source capacitor, it is equivalent to the signal source voltage still existing. If this capacitor is greater than 0.1μF, the discharge current of the capacitor is enough to burn out the transistor.

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