![]() If you can find that information that would help. Unfortunately we dont know the load at the collector of the transistor so it's a bit impossible to specify any exact values for the two resistors that would ensure both proper turn on and proper turn off. What you could do is assume the output of the gate goes from 0.5v for a low to 3.5v for a high and go from there. The other thing though is that you have to ensure that the transistor can turn on fully too, and if you swap the two resistors the only resistor that can turn it on would be the 4.7k so that would be a reduction in base emitter current of about 1/5 what it was, roughly, plus the 1k would pull it up a lot so there's a chance that it may not turn on at all which is also not good. ![]() Any higher and the transistor will begin to conduct. If the supply is 5v that would mean the base should be about 4.7 volts as a minimum i would say. To turn that transistor completely off, the base voltage has to be close the +Vcc. ![]() The combination of 4.7 and 1k like that only raises the output by about 0.26 volts, which does not seem like enough. Hi, That's the same impression i get when i consider the max output of the LS02 gate for a logic high. ![]()
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