3 ms·
The I/O pins have a resistor and diode to ground as protection. I didn't see any diodes to VCC. The outputs use a chain of successively-larger transistors like
by kens 6y ago
The I/O pins have a resistor and diode to ground as protection. I didn't see any diodes to VCC. The outputs use a chain of successively-larger transistors like modern output cells, to get from the tiny internal current to the large external current.
The pins have a variety of implementations, depending on if they are tri-state, input only, output only, bidirectional, and other factors.
The 8086 has a two-phase clock. It's all one clock domain. The asynchronous input pins (I think reset and interrupt) have clocked latches to synchronize them.
What's DFT?
- egsmi 6y agoDFT = Design For Test https://en.m.wikipedia.org/wiki/Design_for_testing https://en.m.wikipedia.org/wiki/Design_for_testing They have a resistor to ground in parallel with the buffer? I get the diode but I can only assume the resistor is a weak pull to prevent charge building up at the pin. But it seems the diode would mostly do that job, to the point of preventing damage anyway. I guess they weren’t concerned with leakage power then. Protection resistors in series with the buffer are fairly common these days. Especially on GPIO type pins.
- kens 6y agoThe resistor is on series. I.e pin - resistor - diode to ground and circuitry. The resistor is just some silicon, probably around 30 ohms.
- egsmi 6y agoThat makes sense. That’s the current limiting resistor. It protects the protection. See page 12 of this deck. http://www.eng.biu.ac.il/temanad/files/2017/02/Lecture-9-IO.pdf http://www.eng.biu.ac.il/temanad/files/2017/02/Lecture-9-IO....