but in the transistor figure the current from the negative terminal flows directly to the emitter and does not flow through the rheostat
Yes current does flow through the rheotat.
You have over simplified things.
See my modified version of your diagram, I1 = I2 + I3
JimB
The rheotat DOES DROP THE VOLTAGE. Get a proper understanding of simple resistive voltage dividers before trying to understand transistors.according to ur figure how does the rheostat vary the voltage of the electrons which are going to the emitter
1) voltage of electrons (energy of electrons) going to the emitter does not drop at all according to ur figure but the function of that rheostat is to drop the voltage
then what is the function of the rheostat there.
It combines with I3 and flows back to the battery as I1.2)also where will that I2 end up after it passes through the rheostat
No, this is standard work. Google it.3)can u explain the entire functioning of common emitter transistor in terms of electron movement
No.also can u explain my doubt in transistor amplifier i have given the link
https://www.electro-tech-online.com/threads/transistor-amplifier-help.100696/
The rheotat DOES DROP THE VOLTAGE. Get a proper understanding of simple resistive voltage dividers before trying to understand transistors.
It combines with I3 and flows back to the battery as I1.
No, this is standard work. Google it.
No.
when the pot is turned up so its output voltage is 0.6v then the base-emitter of the transistor is forward-biased and begins to conduct current. If the pot is turned up more then the base current increases but the base voltage barely becomes higher than about 0.7V.
The current gain of the transistor causes a collector to emitter current that is a few hundred times higher than the base current.
'audioguru genius'
now u have given me some energy
what u r saying is that the rheostat is used to vary the base current
as the base current increases which means automatically the base emitter voltage must have increased due to which a larger no of electrons is now coming out of the base. i.e it gives an indirect meaning of varying the base emitter voltage by varying the base current.
it is the converse of Ohm's law(as voltage increases current increases)
i.e as current increases it means that voltage must have increased.
the same is applied here am i right????
A couple of posts back I took the time to draw an illustration of how the currents and voltages relate between the rheostat and base-emitter of the transistor. It is very explanatory. I think you are on the right track but please take the time to understand my illustration.
Not really.What you are saying is that the rheostat is used to vary the base current.
As the base current increases which means automatically the base emitter voltage must have increased due to which a larger no of electrons is now coming out of the base.
Not really.
The base-emitter junction is a diode, not a resistor. The voltage might increase from 0.65V to 0.68V when the current is increased a lot.
We do not talk about electrons and holes when we describe how a transistor amplifies. Electrons and holes describe the transistor without a signal (its bias currents).
the transistor has an input and an outpout coupling capacitors so it amplifies AC, not DC.
The emitter has a capacitor bypassing the emitter resistor so the transistor has an AC voltage gain that is RC and RL in parallel divided by its internal emitter resistance that is 0.026 divided by its collector current.
This transistor circuit has a fairly low input impedance so it must be driven by a low impedance for low loss.
The input signal changes the base-emitter voltage slightly that changes the base current a little. The collector current changes much more which is amplification.
my problem is not what u think of
my problem is that the electrons I3 go through the rheostat only at the end(after passing through the transistor) then how does rheostat vary the voltage of base emitter
If you already understood my illustration, then you already understood the answer to this question.
Look at the drawing on the left. Consider that the rheostat, when divided by the "wiper", becomes essentially two resistors.
Then consider the wiper is connected to the B-E of the transistor, so that the resistance of B-E is parallel to the resistance of the bottom half of the rheostat. You know that resistors in parallel can be treated as one resistor.
So, refer to the illustration on the left. You have a voltage divider.
So, if you move the rheostat wiper and change the voltage, you understand that two voltage drops add to be TOTAL VOLTAGE in a series circuit.
So, a change in the rheostat voltage will set up a voltage drop change in the B-E.
My illustration describes everything you ask. Use your analytical reasoning abilities to understand.
i understand ur explanation but this is valid only if u consider R2 and R3 together
if u consider them separately u can explain this am i right
i.e the voltage divider diagram can only explain my doubt and not the one on ur right
No, it is the SAME.
It explains it the SAME. Please consider that the bottom of the rheostat joins with B-E to become a parallel resistance.
As the voltage drop in the top half of the rheostat is varied, so will the voltage drop across B-E+bottom of rheostat(parallel)...IN A CONVERSE MANNER.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?