What happens when collector-base voltage is 0 in transistor?

[goodphy]

Hello

I'm now studying current mirror and have a question about replacement of diode to transistor as shown in the attached image.

The upper image is original current mirror circuit shown in study document. Collector current I_C is very similar to current through R_bias thus current to load can be controlled by adjusting R_bias.

Since this scheme requires prefect matching of saturation current between diode and transistor base-emitter junction, 2nd circuit is suggested (Let's focus left circuit) where diode is replaced to identical transistor in 1st circuit with base-collector wiring.

This wiring ensures that collector-base voltage becomes zero and...It is wired for me as I have known that V_EB typically 0.2 V in normal operation. (so called active mode maybe?) As base is very thin layer thus current can flows from emitter to collector via diffusion (not drift by voltage from base to collector) but...Is this normal condition?

Does this wiring simply transform transistor to diode?

 

[meBigGuy]

Yes, That wiring essentially transform the leftmost transistor to a diode. If the transistor is matched to the output transistor the current will be mirrored fairly accurately.
I won't try to explain the secondary effects and errors that occur in this circuit. You can research that easily enough.

Vbe is ~0.7V in normal operation, why do you say 0.2V ?

 

[goodphy]

Yes, That wiring essentially transform the leftmost transistor to a diode. If the transistor is matched to the output transistor the current will be mirrored fairly accurately.
I won't try to explain the secondary effects and errors that occur in this circuit. You can research that easily enough.

Vbe is ~0.7V in normal operation, why do you say 0.2V ?

Oh..I wrote the wrong. 0.2 V is V_CB. Thanks for answering me!

 

[LvW]

Oh..I wrote the wrong. 0.2 V is V_CB. Thanks for answering me!

VCB=0 V (short circuit between C and B)!

 

[goodphy]

VCB=0 V (short circuit between C and B)!

Am...so...total voltage across the transistor is always V_BE regardless of its connections in circuit?

 

[meBigGuy]

The saturation current from collector to emitter can be 0.2V if the base is 0.7. But, since you connected the collector to the base it can't get below 0.7 or it starts turning itself off. Because of the physical construction of the device Vce can be less than Vbe, but you stop that from happening by shorting C to B (diode connecting the transistor).

 

[LvW]

The saturation current from collector to emitter can be 0.2V if the base is 0.7. But, since you connected the collector to the base it can't get below 0.7 or it starts turning itself off. Because of the physical construction of the device Vce can be less than Vbe, but you stop that from happening by shorting C to B (diode connecting the transistor).

* Current of 0.2 V ?
* Of course, it can be lower tha 0.7V. Why not 0.65 V or 0.6 V?

 

[meBigGuy]

OK on the typo. The saturation voltage (that the OP observed) can indeed be 0.2V if Vbe has driven the transistor into saturation.

But if the collector is shorted to the base it can't go much lower than 0.7. You can quibble that 0.6 is different than 0.7, but the point is the same. Less than around 0.7 and the transistor is turning off. So it stabilizes at a collector voltage and current that provides a stable operating point, behaving much like a diode.

 

[LvW]

You can quibble that 0.6 is different than 0.7, but the point is the same..

My only point was that there is no sharp cut-off. It`s like a "normal" pn diode with a - more or less - "smooth" transition to the off-state.

 

[meBigGuy]

Yeah, there is nothing magic about 0.7V. Diodes and transistors conduct at all voltages. I guess it was sloppy of me to say "turning off" or imply something special happens at around 0.7V. Your point is taken.

 

[thankz]

or it can be .2v if it is a germanium transistor

 

[stedwards]

Yeah, there is nothing magic about 0.7V. Diodes and transistors conduct at all voltages. I guess it was sloppy of me to say "turning off" or imply something special happens at around 0.7V. Your point is taken.

Ever wonder, with nothing special about 0.6 to 0.7 volts showing up in the transistor equations, why does this voltage range show up so often, as a rule of thumb, for effective cut-off?

 

[LvW]

Ever wonder, with nothing special about 0.6 to 0.7 volts showing up in the transistor equations, why does this voltage range show up so often, as a rule of thumb, for effective cut-off?

That`s a good question, I think.
Let me try the following answer:
Undoubtly, the BJT is a voltage controlled device which requires a certain voltage across the pn junction (base-emitter) to let a collector current flow.
For collector currents between some hundreds of µA and some mA (common range of currents) the base-emitter voltage, normally, must be in that range (0.6...0.7V).
The exact value is (a) not known fore each specific part (without measuring, due to large tolerances) and (b) not required because each good transistor-based amplifier should have feedback stabilization against tolerances and parameter uncertainties. And it is one of the benefits of dc feedback that - in practice - it does not matter if you are using 0.6V or 0.7V for calculating the resistors for the desired dc operating point.
The difference/uncertainty is app. in the tolerance range of the resistors used (some %).
Please note that I didn`t use the term "cut-off". Such a term is used only for application of the transistor as a switch (non-linear operation).

 

[meBigGuy]

One of the definitions of saturation is the point at which Vbe is greater that Vce. That seems to happen around 0.7V.

 

[stedwards]

Thanks for the help.

I tried to follow up on the question of the value of $V_{BE}$. Effectively 0.6 to 0.7 volts is the diode junction potential, so I looked at this nearly identical question.

The diode current is $I = I_S (e^{(V_D/nV_T)}-1)$, where $n\simeq 1$, and $V_T=kT/q$ is the thermal voltage. $T$ is the absolute temperature, $q$ is the electron charge, and $k$ is Boltzmann's constant.

$I_S$ is the reverse saturation current (or reverse leakage current), $I$ is the cathode current, and $V_D$ is the anode to cathode voltage.

I lean toward the theory that the potential is fairly independent of rated power for constant, DC current operation, where the devices are each operated at about the same percent of full rated power. This seems to indicate that the value is dependent upon the minimum junction deterioration temperature, which sets in above, about 175 C due to atomic migration, destroying the junction.

Some useful information is the thermal resistance $R_{\theta J A}$. $\Delta T = R_{\theta J A}W$ which is used to calculate the temperature of the junction. $W$ is the dissipated power. $\Delta T$ is the temperature difference between junction and ambient temperature (actually, the case or heat sink pad, or leads), and $W$ the power lost in the die, bonding and leads.

To good approximation, $W=I V_D.$

 

[meBigGuy]

I don't quite see how temperature has anything to do with deciding what Vbe value is declared as "ON". It really depends on the transistor. At some point the transistor is declared as saturated. Maybe that's when beta becomes 10, or when BE and BC are both forward biased.

These datasheets drive the base current to 1/10th the collector current for the saturation specs.
For example the 2N3904 (https://www.fairchildsemi.com/datasheets/2N/2N3904.pdf) (Vbe sat for 50ma = 0.95 max) (Vbe sat for 10ma is .65 to .85)
For the 2N2222 (https://www.fairchildsemi.com/datasheets/MM/MMBT2222A.pdf) . (Vbe sat for 500ma = 2V max ) (Vbe sat for 150ma = 0.6 to 1.2)

Both of these data sheets also have graphs for Vbe vs collector current when not saturated.