>>you can also remove the inverter for red led
>>and wire it to the input side of the green led's inverter.
Considering that the builder of the logic probe has no knowledge of the current sourcing capability of the input, the original configuration would be preferred. Putting the red LED input before the the inverter shows a diode load to the input. The OPs configuration shows a high impedence load.
actually the 10k resistor drop makes the input to the circuit not a high impedance input, 10k is usually consider neither high nor low impedance, >1MOhm is typically considered high impedance. But for a steady state input (non changing), which is what the circuit was designed to measure, the 10k resistor is sufficient to not effect the logic level of the circuit. I was going to mention this aspect (not a high impedance input) when he first posted the circuit, but realized that for the circuit's intended purpose this wasn't an issue, so I didn't.
Comments
and wire it to the input side of the green led's inverter.
>>and wire it to the input side of the green led's inverter.
Considering that the builder of the logic probe has no knowledge of the current sourcing capability of the input, the original configuration would be preferred. Putting the red LED input before the the inverter shows a diode load to the input. The OPs configuration shows a high impedence load.
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I think I know what I'm doing...
...Maybe...
Post Edited (bugg) : 2/16/2005 3:55:13 PM GMT