non-CP/M Z80 board

On Jun 16, 2008, at 10:07 PM, David Kane wrote: Here, what the 6800 can drive in terms of capacitance is another way of stating its output drive current capability. I'll expound for the benefit of those who haven't gotten their head around it yet. I thank our very own Tony Duell for helping *me* get my head around this many years ago, when my nice clean square waves were turning into deformed triangles as I tried to drive a big FET for an IR transmitter in a commercial product. This is a simplification, but it gets the point across. Think of it in terms of a signal changing state from 0 to 1. CMOS chips are FET-based, and the input of a CMOS logic gate typically goes to the gate ("gate" in FET terms, not logic terms) of a FET. The FET's gate is essentially a tiny capacitor. To turn the FET on, one must charge that capacitor. So, you need to raise the voltage on the FET's gate from ~0V to ~2V (the 74HCT 0-1 threshold according to Fairchild's AN-368 app note). The higher the capacitance of that logic gate's input (which is the gate capacitance of the input FET, typically Ciss on transistor datasheets) the more current it'll take to charge it to a the 0-1 threshold voltage in a given amount of time. If the capacitance is lower, it'll take less current. Now, recall that capacitances in parallel add up...so if the input capacitance of (say) a 74HCT245 is 3.5pF (from the datasheet), if you put five 74HCT245 inputs on a bus line, the total capacitance (excluding PCB traces, etc) is 5*3.5 = 17.5pF...requiring more drive capability from whatever is driving that bus line. So if the drive capability is insufficient, the rise time of the pulse will lengthen, and turn that nice sharp rising edge into a long sweeping transition, screwing up the timing and confusing lots of things. -Dave -- Dave McGuire Port Charlotte, FL