IF you are after an amplifier, you need a collector resistor (or emittor depending on what you want) and a voltage source for the collector, all it does atm (for the info provided) is pull that point to GND
Ans it will do that for a wide voltage at the input as well
Yes, that's only part of a circuit - there's not enough there to say how it might perform - the missing parts and information will affect it far too much, to the extent that the transistor could well be destroyed!.
Yes, this is a stage to a supply voltage. I have another pin that has the + and when i put my stage input at 18V i hope that i´ll have in pin 11 the groud, making between pins my supply voltage. But when i have 0V in my stage input i don´t have to have the supply voltage.
My ideia it´s possible and correct?
It´s possible making this with a normal transistor (BC547)?
In this example if input A it´s 12V and i have a collector voltage of 24 what i´ll have in output 11 when A=0? I´ll have 24V? And when A=24? I will have 0V?
Regards
When the base receives 0 volts in an NPN transistor, the NPN basically becomes inactive, so the voltage at point 11 would have to be about Vcc (which is 24V) because it is fed through the resistor.
And because the output of the NPN in this mode is inverted with respect to the input, you should get a low value (probably 0 volts) when 24V is applied to the base.
My problem it´s that i´ll not apply 24V to the base. I´ll apply 12V. My question is that if with 12V i´ll have the same ideia?? The resistance values are correct?
Supply voltage=24V, load resistor=10k, this mean only 2,4mA collector current. The transistor beta minimum 200, so the required base current 1,2microamps. The 40k resistor give from 12V about 300microamps for base, so quite enough. This input can work from about 1,5V!