Switch debouncer uses only one gate

该电路能在每次按下S1时产生一个消除弹跳的脉冲,并且仅使用来自远程上拉电阻R2的逻辑电源。它适用于检测未通电设备(例如刚从待机状态启动的系统)中的按键按下事件。当S1未被按下且C1处于充电状态时,R1将IC1驱动至VSS(地),几乎不消耗功率,使VOUT保持接近5V。按下S1时,C1迅速放电并将IC1驱动至VDD,此时IC1大量导通,将VOUT拉低至接近0V,直至R1重新为C1充电,再次将IC1驱动至VSS。一旦C1充分充电,IC1回到VSS并停止消耗功率,从而减轻了对VDD的负载,使VOUT返回高电平状态。D1到D3与R3一起工作,将VOUT的电平移位,以提高与CMOS逻辑的兼容性。

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The circuit in Figure 1 produces a single debounced pulse each time you press S1. Moreover, the circuit uses only logic power from the remote pull-up resistor, R2. You can use the circuit to detect when a key is pressed in a nonenergized device, such as a device in a system that's just coming up from standby. The circuit operates as follows: Assume that you have not yet pressed S1 and that C1 is in a charged state. Under these conditions, R1 drives IC1 toward VSS (ground), causing the IC to consume virtually no power. This action allows VOUT to remain near 5V. However, when you press S1, C1 rapidly discharges and drives IC1 toward VDD. Under these circumstances, IC1 conducts heavily, pulling VOUT near 0V until R1 charges C1 enough to again drive IC1 toward VSS. Once C1 charges sufficiently, IC1 goes to VSS and stops drawing power. This action unloads VDD and causes VOUT to return to a high state. D1 to D3, in conjunction with R3, shifts the level of VOUT for improved compatibility with CMOS logic.

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