# When base is open, is the base-emitter junction reverse biased? How does leakage current from collector to emitter flow?

I've been studying photo transistors, but I'm rusty on leakage current in bjt. I don't get why CE leakage current is $$\ 1+\beta \$$ times CB leakage current. It seems like the bjt amplifies CB leakage current. I have two questions troubling me:

1. How can a thermally generated electron-hole pair in CB junction contribute to a current from the supply? Won't the BE junction stop it?(Because the charge carriers have to cross two reverse biased junctions)
2. The leakage current relation $$\I_{CEO} = (1+\beta)I_{CBO}\$$ seems to suggest that the common emitter configuration amplifies the leakage current from collector to base($$\CB\$$). How can leakage current from collector to base be amplified to produce a huge leakage current from collector to emitter?

simulate this circuit – Schematic created using CircuitLab

• Search “what is the Early Effect” May 13, 2021 at 14:36
• @TonyStewartEE75 it is base width modulation due to change in collector supply. Is Early effect responsible for the amplification in CB leakage current: $I_{CEO} = (1+\beta)I_{CBO}$ ? May 13, 2021 at 15:06
• It seems like the bjt amplifies CB leakage current. It does. The CB leakage current is the current flowing into the base. As long as the transistor is in active mode (BE in forward, BC in reverse with enough reverse voltage), which it is (in active mode). Then Ib is amplified. May 13, 2021 at 15:24
• @Bimpelrekkie I can understand if base supply is present, the bjt amplifies the base current. But CB leakage current is not base current, it is flowing from collector supply into the base. This is different from the base current right? May 13, 2021 at 15:27
• In phototransistor the base is open and the bjt amplifies CB leakage current as you said. But I don't get this, how bjt treats CB leakage current SAME as the base current from base supply.. May 13, 2021 at 15:28

First the $$\I_{CBO}\$$ current will flow into a "positive" direction in the collector (electrons current in NPN). Thus, $$\I_C = I_{C_F} + I_{CBO}\$$ where $$\I_{C_F}\$$ = "forward current". But for the base terminal, the $$\I_{CBO}\$$ current is a "negative current" (holes current) $$\I_B = I_{B_F} - I_{CBO} \$$

And now after we applied the voltage across the collector and emitter and left the base terminal "open". We "create" a "voltage gradient" across the BJT's. Thus, there will be some small voltage drop across the B-E junction, say 0.2V. Therefore the B-E junction will be forward-biased. But the base is left open thus:

$$\I_{CEO} =\alpha \: I_{CEO} + I_{CBO} \$$

And if we "solve" for $$\I_{CEO}\$$

We have this:

$$\I_{CEO} = \frac{ I_{CBO}}{ 1 - \:\alpha} = (\beta + 1) I_{CBO}\$$

• Ah when the base is open, part of the collector supply voltage drops across the BE junction and makes it "slightly" forward biased. This means with base open, there is only ONE reverse biased junction, CB ? May 13, 2021 at 16:18
• Yes, only one, CB junction.
– G36
May 13, 2021 at 16:21
• Never thought of this, thank you! May 13, 2021 at 16:23
• Might add that Icbo is also called 'collector-base leakage current' due to B-C junction. That leakage will bias the B-E junction slightly to forward conduction. Tying the base to emitter will shunt the leakage around the B-E junction, and so C-E leakage will be less. May 13, 2021 at 17:29
• But it looks like you understand it well. Because we have a BE junction in forward bias, and some part of emitter electrons (base terminal is open, thus Ie = Ic = Iceo) will "meet" the Icbo current "holes" at the base terminal. So, the rest of the emitter electrons ($\alpha \: I_{CEO}$) will flow to the collector. But at the same time, Icbo current "will give back " previously lost electrons. This is why we have $I_{CEO} =\alpha \: I_{CEO} + I_{CBO}$
– G36
May 13, 2021 at 19:16