保护放大器输出免受过载电压影响的电路

　　During normal operation, the amplifier outputs’ dc components are at one-half of the VCC supply—2.5V in this case, for which VCC is 5V. The 11V gate drive fully enhances the MOSFETs, and the shunt-regulator output is off

because its feedback input, Pin 5, is below its internal 0.6V threshold. If either output exceeds 5V, current flows through D3 into the R5/R6 divider, pulling the feedback terminal above its threshold. The shunt-regulator output then pulls the MOSFET-gate voltage from 11V almost to ground, which blocks high voltage from the amplifier by turning off the MOSFETs. The MOSFETs easily withstand the continuous output voltage, and the circuit returns to normal operation when you remove the short. Because the circuit does not respond instantaneously, zener diodes D1 and D2 provide protection at the beginning of a fault condition.

　　The waveforms of Figure 2 represent an operating circuit. One of the amplifier’s outputs (Trace 1) is a 1-kHz sine wave biased at a dc voltage of 2.5V. Trace 2 is the signal on the wire harness. It also starts as a 1-kHz sine wave biased at a 2.5V-dc voltage, but, at 200 µsec, it shorts to an 18V supply. Trace 3 is the shunt regulator’s output, initially biased at 11V but pulled to ground in response to the overvoltage condition. Trace 4 is current in the wire harness. Initially a sine wave, this current drops to zero in response to the overvoltage condition.

　　The components in Figure 1 optimize this circuit for 5V operation. For other voltages, you can adjust the R5/R6 resistor values. The shunt regulator must be able to function in saturation and, therefore, requires a separate supply pin in addition to the shunt output pin. The circuit repeatedly withstands 28V shorts without damage.

　　英文原文地址：http://www.edn.com/article/CA6455601.html