Why Does Current Flow from Emitter to Collector in an NPN Transistor?

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In an NPN transistor, current flows from the emitter to the collector due to the potential differences across the junctions. When in saturation, Vbe is approximately 0.75 V, while Vce is around 0.2 V, allowing for current flow despite the expected behavior of diodes. The base-emitter junction is forward-biased, enabling electrons from the emitter to flow into the base and then to the collector, where they are attracted by the positive voltage. The collector must be at a higher potential than the base for the base-collector junction to conduct, which typically doesn't occur in powered transistors. The discussion emphasizes the importance of understanding the potential differences and biases in transistor operation.
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current direction?

in transistor (npn) saturation, we know (in CE) Vbe=~0.75 V. and Vce is around 0.2 V.
In such a case according to characteristic plots, we know Ic flows to emitter (via base)

But current always flows from higher to lower potentials.
Why in this case it is not?

please solve this...
 
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You would expect the base current to flow from base to collector wouldn't you?

However this is still a diode and it needs 0.6 volts to make it conduct. So the base has to be 0.6 volts more positive than the collector before it would make the B - C diode conduct.

This would only happen if the collector was at zero volts which will not happen if the transistor is powered.
 


vk6kro said:
You would expect the base current to flow from base to collector wouldn't you?

However this is still a diode and it needs 0.6 volts to make it conduct. So the base has to be 0.6 volts more positive than the collector before it would make the B - C diode conduct.

This would only happen if the collector was at zero volts which will not happen if the transistor is powered.

if it is like a diode, then it is operating in reverse bias, which allows only small currents (usually in micros) .
Another factor is such that , how can current pass comfortably in reverse bias (C-B), where it could not in forward potiential barrier (B-C)?

Thank you
 


staraet said:
if it is like a diode, then it is operating in reverse bias, which allows only small currents (usually in micros) .
Another factor is such that , how can current pass comfortably in reverse bias (C-B), where it could not in forward potiential barrier (B-C)?

Thank you

hey for the CE junction it is reverse biased which means holes in p type can move to n type and vice versa .But the minority charge carriers in p type namely electrons can move to
n type in reverse bias
since electrons which have come from emitter and are now in the base they are attraced due the positive voltage applied at the n type so they flow easily
(Hey don't confuse with minority in p type and electrons from emitter both are independent but both can flow easily to the n (collector) since + field is applied at the collector.
 


If you test any NPN transistor with a multimeter, you will see that it tests like two diodes with their anodes joined at the base.

So, if the collector voltage is less than the base voltage, the junction is forward biassed, but not enough for it to conduct unless the collector is actually at zero volts.
 


sudar_dhoni said:
hey for the CE junction it is reverse biased which means holes in p type can move to n type and vice versa .But the minority charge carriers in p type namely electrons can move to
n type in reverse bias
since electrons which have come from emitter and are now in the base they are attraced due the positive voltage applied at the n type so they flow easily
(Hey don't confuse with minority in p type and electrons from emitter both are independent but both can flow easily to the n (collector) since + field is applied at the collector.

""""since electrons which have come from emitter and are now in the base they are attraced due the positive voltage applied at the n type so they flow easily """"

as you said above, due to 0.2 V they get attracted easily;
but a much higher 0.65 V is pulling them to base region...
 


vk6kro said:
If you test any NPN transistor with a multimeter, you will see that it tests like two diodes with their anodes joined at the base.

So, if the collector voltage is less than the base voltage, the junction is forward biassed, but not enough for it to conduct unless the collector is actually at zero volts.

i agree with you that it can't be forward...
what about when transistor is in saturation? where Vbe is around 0.8 V...


also why will electrons (presently in p material) go to a attracting 0.2 V leaving 0.65-0.75 attraction towards base?

thank you
 


what about when transistor is in saturation? where Vbe is around 0.8 V...

If current does start to flow into the collector from the base, then this will drop the normal base current so the collector voltage will not drop any further.
 


vk6kro said:
what about when transistor is in saturation? where Vbe is around 0.8 V...

If current does start to flow into the collector from the base, then this will drop the normal base current so the collector voltage will not drop any further.

mr vk6kro
leave alone our battle with potentiometer if u think about what i am asking u can answer it
u can't understand what I'm asking
ok let's go further
could u explain how base current can be kept constant for output characteristics
there u vary Vce.but if u vary Vce base current should decrease as the +voltage of collector has increased as a result it will attract more electrons towards collector as a result base current should decrease but its said that base current is constant how is that

p.s-how to become a science advisor like u i.e how to get that tag that u have it in ur name
 
  • #10


ok let's have another approach...

suppose all values Vbe, Vce, Ib, Ic, Ie... are Constant...i,e.now we are at anyone single point in characteristic curves.

now if an electron, starting from earth(emitter ground) , travels all it's way to Vce;

can anyone say how the potiential changes for that electron, to all its journey to Vce?
initial potiential of Earth is obviously zero.

lets talk as if a pure science; in terms of voltages, attractions,collisions(if) etc...
...as if someone asked you to plot the 'potiential Difference' graph.
value of PD on Y-axis and distance from emitter Earth to Vce on X-axis.

you can also tell in terms of force(NET)-distance if you are comfortable..

remember that Kinetic energy of electron is not considered when we talk on electron movements. it is considered constant.
also we know force inside any matertial (irrespective of concentrations of any charges inside) is zero. (vectors cancel each other)

only at junctions(or ends) that any electrons get into troubles.
 
  • #11


staraet said:
ok let's have another approach...

suppose all values Vbe, Vce, Ib, Ic, Ie... are Constant...i,e.now we are at anyone single point in characteristic curves.

now if an electron, starting from earth(emitter ground) , travels all it's way to Vce;

can anyone say how the potiential changes for that electron, to all its journey to Vce?
initial potiential of Earth is obviously zero.

lets talk as if a pure science; in terms of voltages, attractions,collisions(if) etc...
...as if someone asked you to plot the 'potiential Difference' graph.
value of PD on Y-axis and distance from emitter Earth to Vce on X-axis.

you can also tell in terms of force(NET)-distance if you are comfortable..

remember that Kinetic energy of electron is not considered when we talk on electron movements. it is considered constant.
also we know force inside any matertial (irrespective of concentrations of any charges inside) is zero. (vectors cancel each other)

only at junctions(or ends) that any electrons get into troubles.

what i think is that since Vce is higher tha Vbe
the electric field is larger
hence according to work energy theorm change in K.E = work
i.e the field of Vce has increased its energy
since voltage is nothing but energy the voltage of the electron has increased
 
  • #12


sudar_dhoni said:
what i think is that since Vce is higher tha Vbe
the electric field is larger
hence according to work energy theorm change in K.E = work
i.e the field of Vce has increased its energy
since voltage is nothing but energy the voltage of the electron has increased

"""since Vce is higher tha Vbe """

not always is Vce is higher than Vbe...

i already mentioned in starting that Vbe=0.7-0.8 in saturation;Vce=0.2-0.25;
 
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