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Normally such diode resistor network is used to have fast turn-on of the transistor (as current goes through the diode), and slower turn-off as base charge is removed via the resistor.
In your situation it looks strange. The emitter is at 0V. when the IC outputs 3.3V (just an example), the base current is determined by almost the short circuit current of the CPU output. Can it be an open collector/drain output with internal pull-up resistor? If so, then it makes sense.
Normally such diode resistor network is used to have fast turn-on of the transistor (as current goes through the diode), and slower turn-off as base charge is removed via the resistor.
In your situation it looks strange. The emitter is at 0V. when the IC outputs 3.3V (just an example), the base current is determined by almost the short circuit current of the CPU output. Can it be an open collector/drain output with internal pull-up resistor? If so, then it makes sense.
Looks to me like it's the clock output from the ULA to the Z80. It needs fast clock edges so a speed-up diode wouldn't be a bad idea but unless the ULA has something limiting it's output current it looks as though it would be overloaded. These were very much cut price machines so if the schematic is correct, maybe they do it just to cut costs even though it's considered bad practice.
Icbo is the transistor leakage current in it's unbiased state so I can't see that would have any effect. The problem is the output of the ULA would see the diode and B-E of the transistor as two Si junctions to ground and would try to sink excesive current into them. Either the ULA has an internal current limited output or more likely, the schamatic is wrong and the diode should face the other way. If the diode is reversed, it would have the desired effect of speeding up the clock edges. Z80 devices need fast clock edges to work properly. The Z80 data sheet shows a similar transistor assisted clock drive.
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