Need help with NPN

I didn't understand.... You mean how to discount the .7V drop in the simulation? I know u can edit the resistance.... But IRL it is described in the datasheet and can be different from manufacturer to manufacturer

When I do the calculations assuming a voltage drop of 0.7 volts it is not accurate. The voltage drop varies depending on the voltage applied to the base, it can be more or slightly less (based on experiments in the simulator). But this approximation allows me to calculate the equivalent replacement of a transistor by a resistor (I*R=Voltage Drop), and the results are consistent with what I see in the simulator. Now I am no longer sure that it is correct to try to calculate the parameters of the circuit, provided that the voltage at the base is higher than the voltage at the emitter. ChatGPT pointed out that this is not the correct use of the transistor - which means there is no point in trying to calculate the circuit in this mode. On the other hand, in the simulator, this does not cause "obvious" breakdowns.

you are driving an unlimited current through the base of your transistor into the emitter and only limiting is due to emitter resistor you would destroy the transistor. you need a base resistor base resistor is derived from collector current divided by beta stick a 100k resistor between base and v source and try your calculations again you Will be much closer

pick a collector voltage let's say half your supply voltage and you want 5ma to pass through your transistor and let's say you want 1 volt on your emitter you divide 1 volt by your collector 5ma and that's your emitter resistor value eg 200R base current will be 5ma divided by transistor hfe=100 that's 50ua so your base resistor would be in your eg 30v divided by 50ua so 600k now I've not included our 1v emitter voltage or the 0.7v vbe so I pick 560k approx would be close enough to get you going in right direction

I forgot to mention your collector resistor is half your supply voltage divided by 5ma as example in my previous comments so 1k with 10v supply

your emitter resistor will now pass collector current plus base current roughly 5ma + 50ua

pip, Thank you. The language barrier was overcome, and I was still able to make out what you were trying to tell)

your welcome glad I helped a little.