Connecting only emitter to collector-BJT

Question

I started studying the BJT Configurations and how they operate in active mode, saturation mode, and cutoff. And that every two neighboring junctions have either forward or reverse bias, But I wonder what happens when I only connect one battery between only collector and emitter (N.Terminal to the emitter and P.Terminal to the collector). what happens to the holes in the middle base. Does current flow? why? or why not?

Answer 1

Another answer says:

In other words its inconclusive to determine if the transistor is either in the active, saturation or cut-off region. Since there is no reference made with the base terminal.

However, the base being open does not prevent us from calculating \$V_{BE}\$.

If the base is disconnected, then \$I_{B}=0\$. The \$V_{BE}\$ that is present when the base is left unconnected is exactly the same as the \$V_{BE}\$ that if applied across the base and emitter terminals would cause there to be 0 base current. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$ when \$I_B=0\$.

We know that

$$V_{BC}=V_{BE}-V_{CE}$$

From the Ebers-Moll model,

$$I_{B} = I_{S}[\frac{1}{\beta_F}(e^{V_{BE}/V_T}-1) +\frac{1}{\beta_R}(e^{V_{BC}/V_T}-1)]$$

Setting \$I_B=0\$ and rearranging, gives

$$\beta_R(e^{V_{BE}/V_T} -1) + \beta_F(e^{V_{BC}/V_T} -1) = 0 $$

$$\beta_Re^{V_{BE}/V_T} + \beta_Fe^{V_{BC}/V_T} = \beta_R + \beta_F $$

$$\beta_Re^{V_{BE}/V_T} + \beta_Fe^{V_{BE}/V_T-V_{CE}/V_T} = \beta_R + \beta_F $$

$$e^{V_{BE}/V_T}[\beta_R + \beta_Fe^{-V_{CE}/V_T}] = \beta_R + \beta_F $$

So, if I have done my math correctly,

$$e^{V_{BE}/V_T}=\frac{\beta_R + \beta_F} {\beta_R + \beta_Fe^{-V_{CE}/V_T}} $$

If \$\frac{V_{CE}}{V_T}>6\$ then the above approximates to

$$e^{V_{BE}/V_T} \approx 1+\frac{\beta_F}{\beta_R}$$

or

$$V_{BE} \approx V_T \cdot ln\left(1+\frac{\beta_F}{\beta_R}\right)$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 89\$ mV. Consistent with our intuition, when the base is open-circuit, the transistor is in the cutoff region. \$V_{BE}\$ is too small for the transistor to be in the forward active region. There will be some leakage current through the emitter and collector, but it will be relatively small.

Answer 2

The voltage \$V_{CE}\$ applied has to drop across the two junctions: collector junction and emitter junction. Now the sign of the potential will be such that one junction will be forward biased and the other will be reverse biased. Since base terminal is open, same current has to flow through these junctions. So a small current having a magnitude in the order of reverse saturation current of the reverse biased junction will be flowing.

Because of the bias conditions prevailing, say forward biased emitter junction and reverse biased collector junction, electrons from emitter will be injected to base (npn transistor assumed). These electrons will be transported to collector and contribute to current. The holes injected from base to emitter also contribute to current. But the applied voltage will be dropping mostly in the collector (reverse biased) junction and hence the current will be very small.

Answer 3

First I agree with Nidhin

The explanations of flow of minority carriers leads to flow of reverse current.

In terms of the region of operation recall that in active region -transistor operates as an amplifier, cut off region- transistor acts as an open switch, saturation region- transistor operates as a closed switch.

Now the difference between saturation and cut-off is with the forward biasing of the Base-Emitter that allows the flow of base current. In other words forward biasing the B-E leads to a base current which makes the BJT to operate in the saturation region (closed switch)

I need to mention that BJT are current-controlled devices; and they rely on the base, emitter and collector currents. Base current from VBE

With that said applying the VCE means regardless of the Collector being at a higher potential compared to the Emitter. Neither is Base-Collector junction reverse-biased nor Base-Emitter junction forward biased which also explains the reverse current.

In other words its inconclusive to determine if the transistor is either in the active, saturation or cut-off region. Since there is no reference made with the base terminal.