hi,

"BTW, my tests show that 6k8 or above resistor should be used between MCU output pin and NPN transistor`s base if you need open it with high logical level"

How did you arrive at this value if I my ask?

As I said: it was my tests I did on hardware I have here. Let me explain using the datasheet of AT89C51RD2. Assume V_cc=5V.

According the datasheet, the minimum value of Input High Voltage V_IH=1,9V (let say, ~2V). This voltage on a pin is required to keep weak pullup (pFET P3) opened via internal loop-back link. So, drop voltage on weak pullup must not exceed ~3V. Such limit is reached when current through pFET P3 is about I_OH~=120μA.
Now we know that maximum current must not be above 120μA. As well we know that output voltage must not be below 2V. Base-Emitter Saturation Voltage depends on collector current; anyway assume it is V_BE(sat)=0,7V for standard transistor. It means, that drop voltage on resistor should be 2-0,7=1,3V. With current 120μA it requires: 1,3/0,00012~=10 kOhms. My tests with real hardware show that 6k8 is enough. Anyway I preffer to use 10k.
As about Darlington, so for such transistors V_BE(sat)=1,5V. Now (2-1,5)/0,00012~=4 kOhms. So resistor 4k7 may be suggested.

There is another point you need to pay attention at: the maximum output source current for a pin is about 120μA (see above). It means that if a simple NPN transistor used for "high current" load then you need to select transistor with very high h_FE. For example, TIP120 has h_FE=1000. And even with such good value it may drive only I_c 0,00012*1000=120mA.

Summary: use second pre-gain transistor or PNP Darlington which opens by low level on base. MCU accepts sink current better than produces source one.

Regards,
Oleg