3 ms·
The output pins are pulled to Vdd (-9) by a transistor driven by a bootstrap load circuit. So you'll lose one threshold voltage, not two. (You can see these boo
by kens 6y ago
The output pins are pulled to Vdd (-9) by a transistor driven by a bootstrap load circuit. So you'll lose one threshold voltage, not two. (You can see these bootstrap load capacitors in the die photo near the pads.)
Inconveniently, the 8008 datasheet doesn't specify how low the low outputs are; it just says they are at most 0.4 volts.
- egsmi 6y agoThanks for the information. It so cool that you can just examine the die and know the answer. There are definitely trade offs to consider so I can understand why the Intel engineers took that route. On the one hand, one gets a very negative output level, which I suppose no one really cares about in the 70s so maybe that’s not even a disadvantage. But to level shift more toward ground would consume more die area per IO and increase the output impedance of the pin, unless one did another buffer stage, which would be even more die area. It’s interesting how we can just absorb this extra complexity in modern circuits without even batting an eye. It’s barely even considered as extra complexity.