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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\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

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 \$V_{CE}\$ is "large" relative to \$V_T\$, and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

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

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ 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.

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.

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\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

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 \$V_{CE}\$ is "large" relative to \$V_T\$, and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

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

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ 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.

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\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

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 \$V_{CE}\$ is "large" relative to \$V_T\$, and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

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

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ 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.

If the base is disconnected, then \$I_{B}=0\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

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 \$V_{CE}\$ is "large", and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

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

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ 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.

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\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

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 \$V_{CE}\$ is "large" relative to \$V_T\$, and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

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

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ 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.