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A BJT is a N-P-N device taking advantage of the PN Junction that is key here. it isn't the fact that a piece of P-type and N-type are next door to each other, they are soo close (ie at atomic level) that there is migration at the Junction

A Diode as a P-N junction very, very quickly becomes P-dep-N as charge migrates. It is this depletion region and thisthat is what producesresponsible for producing the required voltage drop of about 0.6V. Reverse bias a diode and this depletion region grows UNTIL too much voltage has been applied and the entire device breaks down.

Now take a BJT: N-P-N. This VERY quickly becomes N-dep-P-dep-N which is again where the 0.6Vbe requirement comes from.
If you inject charge into the P-doped base, these minority carriers are swept up into the depletion region and fwd bias the B-E (which is akin to a diode), but in doing tothis the other P-E junction (C-B) is equally influenced.

Now if you take two diodes, two PN diodes and place them on a PCB what you get is:

--trace--N-dep-P---trace---P-dep-N--trace You are not producing a required P-N junction between the two attempts to produce a base. There is no miniority charge mobility.

Now shrink it down, get them closer and closer and closer, at substrate level and at some point you will start to have some form of charge migration occurring BUT the geometries are such that you are making an extremely poor doped N-P-N

A BJT is a N-P-N device taking advantage of the PN Junction that is key here. it isn't the fact that a piece of P-type and N-type are next door to each other, they are soo close (ie at atomic level) that there is migration at the Junction

A Diode as a P-N junction very, very quickly becomes P-dep-N as charge migrates. It is this depletion region and this is what produces the required voltage drop of about 0.6V. Reverse bias a diode and this depletion region grows UNTIL too much voltage has been applied and the entire device breaks down.

Now take a BJT: N-P-N. This VERY quickly becomes N-dep-P-dep-N which is again where the 0.6Vbe requirement comes from.
If you inject charge into the P-doped base, these minority carriers are swept up into the depletion region and fwd bias the B-E (which is akin to a diode), but in doing to the other P-E junction (C-B) is equally influenced.

Now if you take two diodes, two PN diodes and place them on a PCB what you get is:

--trace--N-dep-P---trace---P-dep-N--trace You are not producing a required P-N junction between the two attempts to produce a base. There is no miniority charge mobility.

Now shrink it down, get them closer and closer and closer, at substrate level and at some point you will start to have some form of charge migration occurring BUT the geometries are such that you are making an extremely poor doped N-P-N

A BJT is a N-P-N device taking advantage of the PN Junction that is key here. it isn't the fact that a piece of P-type and N-type are next door to each other, they are soo close (ie at atomic level) that there is migration at the Junction

A Diode as a P-N junction very, very quickly becomes P-dep-N as charge migrates. It is this depletion region that is responsible for producing the required voltage drop of about 0.6V. Reverse bias a diode and this depletion region grows UNTIL too much voltage has been applied and the entire device breaks down.

Now take a BJT: N-P-N. This VERY quickly becomes N-dep-P-dep-N which is again where the 0.6Vbe requirement comes from.
If you inject charge into the P-doped base, these minority carriers are swept up into the depletion region and fwd bias the B-E (which is akin to a diode), but in doing this the other P-E junction (C-B) is equally influenced.

Now if you take two diodes, two PN diodes and place them on a PCB what you get is:

--trace--N-dep-P---trace---P-dep-N--trace You are not producing a required P-N junction between the two attempts to produce a base. There is no miniority charge mobility.

Now shrink it down, get them closer and closer and closer, at substrate level and at some point you will start to have some form of charge migration occurring BUT the geometries are such that you are making an extremely poor doped N-P-N

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source | link

A BJT is a N-P-N device taking advantage of the PN Junction that is key here. it isn't the fact that a piece of P-type and N-type are next door to each other, they are soo close (ie at atomic level) that there is migration at the Junction

A Diode as a P-N junction very, very quickly becomes P-dep-N as charge migrates. It is this depletion region and this is what produces the required voltage drop of about 0.6V. Reverse bias a diode and this depletion region grows UNTIL too much voltage has been applied and the entire device breaks down.

Now take a BJT: N-P-N. This VERY quickly becomes N-dep-P-dep-N which is again where the 0.6Vbe requirement comes from.
If you inject charge into the P-doped base, these minority carriers are swept up into the depletion region and fwd bias the B-E (which is akin to a diode), but in doing to the other P-E junction (C-B) is equally influenced.

Now if you take two diodes, two PN diodes and place them on a PCB what you get is:

--trace--N-dep-P---trace---P-dep-N--trace You are not producing a required P-N junction between the two attempts to produce a base. There is no miniority charge mobility.

Now shrink it down, get them closer and closer and closer, at substrate level and at some point you will start to have some form of charge migration occurring BUT the geometries are such that you are making an extremely poor doped N-P-N