# Connecting only emitter to collector-BJT

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?

• Your battery "sees" two diodes back-to-back. There's a very small current flowing because one of the diodes is driven in reverse direction. The current is slighty higher if it's the BE diode (that one usually leaks more.) Commented Dec 27, 2017 at 10:52

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.

• I doubt if this calculation - from the physical point of view - can be of any relevance. I suppose you have "done your math" correctly - however I think the technica/physical interpretation of the results cannot be correct. The Ebers-Moll equations are based on the following cause-and-effect sequence: The voltage Vbe causes the currents Ic and Ib. In your calculation, you have reversed the sequence: The (unknown) voltage Vbe is caused by a current into the collector (assuming Ib=0). I think such a mathematical manipulation (involving cause and effect) can have no physical meaning.
– LvW
Commented Dec 29, 2021 at 9:10
• @LvW Equations do not care for cause and effect. Either an equation gives correct answers whichever way we use it, or the equation is incorrect. Either the calculated voltage (in this case) gives zero $I_B$ or it doesn't. Commented Dec 29, 2021 at 12:16
• Quote: "Equations do not care for cause and effect" . Does this apply also to controlled sources (as used for the BJT models)? Do you think you can exchange controlling and controlled quantities? I am afraid that you are wrong. What you have calculated is the voltage Vbe that must be externally applied in order to make Ib=0 - but this was not the problem!! In contrary - the base node is open! As said - from the math point of view you are right.....however.....
– LvW
Commented Dec 29, 2021 at 19:32
• If a particular $V_{BE}$ can be externally applied and result in $I_B=0$, then the connection to the external voltage supply can be opened (i.e. the base left unconnected) and the currents and voltages within the transistor will remain unchanged. Commented Jul 27, 2023 at 2:46
• No - I dont think so. According to your own example calculation, the base current would be zero if and only if the externally applied voltage Vbe is app 89mV. Removing this voltage source, all currents within the device will be affected - and the voltage between B and E will not be 89mV anymore.
– LvW
Commented Jul 27, 2023 at 8:13

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.

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.

• Quote: "In other words forward biasing the B-E leads to a base current which makes the BJT to operate in the saturation region ". I think, to be more correct, it should read: ....forward biasing both the B-E junction as well as the B-C junction...... When only the B-E path is forward biased (and the B-C path reverse biased), we have the classical amplification mode.
– LvW
Commented Nov 28, 2021 at 9:47
• "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." No. WIth the base left open, the transistor will be squarely in the cutoff region. See my answer. Commented Dec 29, 2021 at 0:09