Did you do the calculation I suggested in post #3?still not get good answer why the out is cutted
Actually, it's the correct answer.still not get good answer why the out is cutted
so the correct answer is no asnwer???Actually, it's the correct answer.
Hi,
you missed to care for Rload.
The 1k resistor to -10V and the 1k load resistor form a voltage divider.
Klaus
Did you do the calculation I suggested in post #3?still not get good answer why the out is cutted
first thanks dana for your effort you did a fantastic job with the simulationsAs the base approaches - 5 the transistor presents less load on the 1K 1K
V divider, so its V approaches -5 of the divider when it is unloaded by transistor.
So with base at -5 and emitter (R divider junction) at -5, the transistor is off
and the V stays at the divider - 5 = "cutted".
Notice the Vcut is when Vbase + Vbe = -5. Or Vbase =~ -4.5
View attachment 176030
View attachment 176031
Regards, Dana.
continuing with the emitter follower complications why do you think vbe is never 0.6 it is always 1.5 vdc or more in the shown circuitYour conclusion partially correct.
The divider does not "hold" / determine the voltage until the transistor is off.
Up to that point it affects but is only partial reason for divider node V. Due
to transistor, when conducting, dumps current into the node. But once the
Vbe collapses and transistor turns off then divider sets the node V.
Regards, Dana.
You are driving base hard to get the 800 mA in the collector, and this parts
base spreading R significant. If the below curve was extended to see higher
currents you would see the high Vbe....
View attachment 176180
Regards, Dana.
that is a completely new piece of info thanks manEither your teachers are wrong, or you misunderstood. Vbe is nominally .6 to .7 volts, but as the curves plainly show, it increases with collector current. It is not "always" 0.6V, that's just a rough estimate.
BUT in almost all text book it is always assumed that diode only drops its usuall forward diode drop voltage and "toss" the left voltage on the resistor ? so it is not like that even for the diode and the diode is supossed to accept diffrenet voltage across it as currnt increase or decrease through itEmitter Base junction looks very similar to a diode response.
Classic diode -
View attachment 176181
LED Diode -
View attachment 176182
Its not unusual to exceed .6V when driving a transistor into saturation (using it as a switch), 2N2222 example -
View attachment 176185
Here is a power transistor 2N3055 -
View attachment 176186
Regards, Dana.
BUT in almost all text book it is always assumed that diode only drops its usuall forward diode drop voltage and "toss" the left voltage on the resistor ? so it is not like that even for the diode and the diode is supposed to accept different voltage across it as current increase or decrease through it
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