transistor basics ( as buffer )

[fady232]

Hello, i had a question about the circuit below, basically it shows the VE is equal to 8.3V and i understood that, because VE=VB-0.7V (diode drop)
but why the signal goes from 0 to 8.3V and not from 0.7V to 8.3V? can someone explain? there is parasitic capacitor that cancels the dc component or where is my misunderstanding? thanks.

 

[barry]

can’t see your circuit

 

[betwixt]

The image is too small to read but I think you might be reasoning it backwards. If working in linear fashion, the base voltage will be about 0.7V higher than the emitter voltage, so if the output at the emitter is 8.3V the base would be ~9V. The output will drop to zero when the transistor stops conducting, in other words when the base voltage drops below 0.7V. The upper output voltage will always be about 0.7V lower than the input voltage, even if the collector is disconnected because the B-E junction will behave like a forward conducting diode.

Brian.

 

[fady232]

The image is too small to read but I think you might be reasoning it backwards. If working in linear fashion, the base voltage will be about 0.7V higher than the emitter voltage, so if the output at the emitter is 8.3V the base would be ~9V. The output will drop to zero when the transistor stops conducting, in other words when the base voltage drops below 0.7V. The upper output voltage will always be about 0.7V lower than the input voltage, even if the collector is disconnected because the B-E junction will behave like a forward conducting diode.

Brian.

oh alright, thanks a lot i understood.

 

[FvM]

You owe us a readable circuit schematic.

 

[fady232]

You owe us a readable circuit schematic.

yeah sorry XD