Electronics > Beginners

Z80 Data, Address & Control signal buffering/driving design. Is this design ok?

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TomS_:

--- Quote from: castingflame on September 26, 2019, 12:08:15 pm ---Great Info thank you Tom!

I have since found out that the Z80 only has a capability of 'Standard TTL Load' on each pin which is 1.6mA (apparently!) so I 100% will need something to drive the LEDs. <<< This statement is for my benefit :)

--- End quote ---

You dont need much to drive modern LEDs - a couple of mA is plenty. Ive even used 3.3K resistors on a 3.3V supply to drive some LEDs with very respectable output. It will all depend on the LEDs you have. Play around with some values (1K+), you'll probably find that you dont need anywhere near 10's of mA or any special drivers.

But I guess you dont really want to be loading your signals down just to drive an LED, so buffering them through something else is still a good idea.

Ian.M:
On 'classic' ZX speccys, the clock output on the edge connector is very low level, IIRC under 1V pk-pk.   I *think* its also inverted with respect to the Z80 clock pin.  If you need it for any peripheral chips (e.g. Z80 PIO, CTC etc.), you'll have to regenerate it.   The easiest option is to AC couple it into a carefully biassed fast inverter, then put it through a couple more direct coupled inverter stages to square it up.


Due to the Z80's fanout limitations, and the fact its already driving the ULA and two banks of DRAM (assuming 48K model) its preferable to use 74HCT logic, not LS TTL, for its lower loading on the signals.

TomS_:

--- Quote from: Ian.M on September 26, 2019, 09:54:33 pm ---On 'classic' ZX speccys, the clock output on the edge connector is very low level, IIRC under 1V pk-pk.   I *think* its also inverted with respect to the Z80 clock pin.  If you need it for any peripheral chips (e.g. Z80 PIO, CTC etc.), you'll have to regenerate it.   The easiest option is to AC couple it into a carefully biassed fast inverter, then put it through a couple more direct coupled inverter stages to square it up.


Due to the Z80's fanout limitations, and the fact its already driving the ULA and two banks of DRAM (assuming 48K model) its preferable to use 74HCT logic, not LS TTL, for its lower loading on the signals.

--- End quote ---

I dont really know anything about the ZX Spectrum, so I went and had a look and it does indeed look like the clock signal on the edge connector will be inverse to what the CPU sees. The CPU clock pin is driven by an open collector circuit, and the signal from the ULA which drives that transistor (PHICPU) is presented on the edge connector.

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