Difference between emitter and collector?

[iScience]

Hi, i might have asked this previously but don't remember.in a BJT, could one treat the emitter as the collector and vice versa? if not, why not? what about JFETs? could one treat the drain as the source and the source as the drain?

 

[Baluncore]

When BJTs are operated with emitter and collector swapped they have a reduction in beta from about 100 to only about 2. They also have poor voltage breakdown characteristics.

When FETs are operated with source and drain swapped, they still behave like FETs but the gate threshold voltage is offset. Where there is an intrinsic substrate diode, the FET is effectively in parallel with a diode which prevents normal operation.

 

[iScience]

can i guess then that this might imply that the emitter is more heavily doped than the collector? and in FETs the level of doping from source to drain also varies? not sure what you mean by this though:

Where there is an intrinsic substrate diode, the FET is effectively in parallel with a diode which prevents normal operation.

 

[AlexCaledin]

Of course emitter of an usual planar BJT is more heavily doped, because it's formed with double diffusion. So the thin emitter-base junction has also much greater capacitance than collector-base.

 

[Baluncore]

Power devices have large area collectors or drains because of heat dissipation requirements. If you operate them with polarity reversed, the specifications will have to be reduced for thermal reasons.

Any protection systems that use diode clamps will prevent operation with the component reversed.

not sure what you mean by this though:

A reverse biassed pn junction is used as an insulator in a MOSFET. When the device is reversed, that pn junction becomes forward biassed and ceases to be an insulator. It then becomes a forward biassed diode that prevents operation.

 

[NascentOxygen]

If memory serves me well, the only attraction in operating a BJT inverted is that it can give a smaller V_ECsat when used as a low current switch.

But better not quote me on that.

 

[Baluncore]

the only attraction in operating a BJT inverted is that it can give a smaller VECsat when used as a low current switch.

With a beta below 6, it would need to be a small signal being switched. I guess that BJT application has now been taken by the mosfet.

NascentOxygen; you appear to be correct, but not by much. The spice model for Si NPN shows that with base connected to collector we get Vce = 752mv, when the E and C terminals are reversed the voltage falls by 2.5% to Vec = 734mv. That is with a 10k pull-up resistor to 5V.

 

[NascentOxygen]

NascentOxygen; you appear to be correct, but not by much. The spice model for Si NPN shows that with base connected to collector we get Vce = 752mv, when the E and C terminals are reversed the voltage falls by 2.5% to Vec = 734mv. That is with a 10k pull-up resistor to 5V.

Base returned to collector is not a typical switch configuration.

I'm expecting V_ECsat of around 10mV at low current. It may be dependent upon transistor type.

 

[iScience]

more of a physics question: for BJTs, either cases of PNP or NPN have most current in the emitter, so how does the collector become the heat dissipator? i know these N and P pieces are all in close proximity, but still, why the collector and not the emitter?

wild guess: for JFETs, heat is produced towards the drain because an existing depletion region provides resistance to oncoming (electron) current, and the current moves toward the drain bringing the heat with it.

let the criticisms begin!

 

[thankz]

from semiconductor physics 2nd ed by sze, "the neutral base width is reduced to zero at sufficient Vcb and the collector depletion region is in direct contact with the emitter depletion region. at this point, the collector is effectively short-circuited to the emitter, and large current can flow."

that's as much as I'm going to quote, the math is still way over my head, for now. time for pics:

 

[Baluncore]

i know these N and P pieces are all in close proximity, but still, why the collector and not the emitter?

For BJTs the base current I_B is small compared with I_CE so we can ignore it and assume that the same current flows through the emitter as the collector. In the linear mode, V_C voltage is dropped across the collector relative to V_B and V_E so the collector is where the W = V_CE * I_CE heat is generated. The same goes for the drain of a FET.

 

[iScience]

VC voltage is dropped across the collector relative to VB and VE so the collector is where the W = VCE * ICE heat is generated.

confirm this for me please let's see if i have this right
during both saturation and active modes, you have the effect where electrons cross the base collector junction in reverse bias, moving them from a region of higher potential to a region of lower potential, heating them up. during saturation you have an even greater voltage drop across the collector and emitter than you would have in active mode, but because there's more resistance in the transistor during saturation, there's simply more current (I_C) during active mode to dissipate more total energy caused by the single effect of lowering in potential.

 

[Baluncore]

during saturation you have an even greater voltage drop across the collector and emitter than you would have in active mode,

Saturation is defined as the collector voltage being somewhere between the base and emitter voltages.
There is little power dissipated in saturation because the V_CE is less than V_BE.

Power is the product of voltage and current. Switching power economically requires that either the voltage or the current be close to zero.

 

[LvW]

from semiconductor physics 2nd ed by sze, "the neutral base width is reduced to zero at sufficient Vcb and the collector depletion region is in direct contact with the emitter depletion region. at this point, the collector is effectively short-circuited to the emitter, and large current can flow."
:

I know a lot of books dealing with transistor physics - however, up to now, I never have seen a statement like "the collector is effectively short-circuited to the emitter, and large current can flow".
My recommendation: Forget this sentence (and change the book?); in my view: Nonsense.
In many gain stages, this "large" current is in the order of some milliamps only.
The BJT acts as a voltage-controlled (non-ideal) current source with an output resistance (at the collector) in the order of 20..100 kOhms.

 

[NascentOxygen]

I know a lot of books dealing with transistor physics - however, up to now, I never have seen a statement like "the collector is effectively short-circuited to the emitter, and large current can flow".
My recommendation: Forget this sentence (and change the book?); in my view: Nonsense.
In many gain stages, this "large" current is in the order of some milliamps only.

The author here is talking about avalanche breakdown. The wording is a large current "can" flow, not "will" flow. It's perfectly valid, but is not something OP should be concerned with until the basic operation of the BJT has been mastered.

 

[LvW]

Thank you for the information - I did not know (I didn`t try to read the bad book reproduction); and it seems that the OP also didn`t understand the issue.

 

[Baluncore]

Base returned to collector is not a typical switch configuration.
I'm expecting VECsat of around 10mV at low current. It may be dependent upon transistor type.

Attached is your expectation ... 5mV reversed as opposed to 20mV normal.