Electronic – How to a ring oscillator work without burning

digital-logicoscillator

I built this ring oscillator and it works fine. After a while I wondered how can it work without breaking.

I am powering it with 5 volts so at start let's say capacitor is charged with 0 volts and a blue node is at 0V. So capacitor starts to charge via R1. When it reaches Vih min (~3.2 V) for IC2A the voltage level is flipped (so voltage at blue node will become high) and now is a red node at Voh + Vih (5 + 3.2 = 8.2 V) which is much higher than allowed from datasheet. This voltage is connected via R2 to input pin.

I've used this oscillator for maybe a year now and it's working fine. The IC is an SN74HC00N.

What I am missing?

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Best Answer

red node at Voh + Vih (5 + 3.2 = 8.2 V) which is much higher than allowed from datasheet.

Resistors R1 and R2 restrict the current that can flow into the TTL output and input nodes. It won't burn. Input current is limited to much lower than the 1 mA limit specified in the data sheet. It can't happen. I mean, why else would R2 be present. The internal protection diodes are perfectly adequate with such a small current injection.

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