What will happen if we give voltage to a transistor (BJT) across emitter and collector without connecting the base? Will it conduct current?
And what will happen to the barrier potential which is across emitter-base and base-collector?
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\$\begingroup\$ It might help, in this case, consider the transistor as two back to back diodes (even if you won't get transistor action from two diodes). One diode will be reverse biased, letting through only a current of the order of its reverse saturation current, while the other will be weakly forward biased with that tiny current flowing through and a correspondingly small voltage developed across it. KCL requires both currents to be equal in magnitude. \$\endgroup\$– Sredni VashtarFeb 28 at 15:30
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2 Answers
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There will be some small current from collector to emitter due to leakage.
First of all, linear models of a BJT (like hybrid pi model) often include a resistance in parallel with the collector to emitter path.
Second there is some small leakage from collector to base. That leakage will then be amplified by the base (as any other base current would be) resulting in additional collector to emitter current.
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\$\begingroup\$ "That leakage will then be amplified by the base".....amplified by the base? What does this mean? What is "the base"? Please, can you further explain and give some reference to this statement? \$\endgroup\$– LvWFeb 28 at 8:06
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\$\begingroup\$ @LvW As you are almost certainly aware, a BJT has three connection points that are typically labeled Collector, Base, and Emitter. The amount of amplification (current wise) is the ratio of the collector and base currents (usually called B or Hfe). Ic = B * Ib should still be valid even if the base current is due to internal leakage from the collector to base junction. See this stackexchange post electronics.stackexchange.com/questions/124668/… \$\endgroup\$– user4574Feb 28 at 21:27
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\$\begingroup\$ So you think that in the case under discussion the leakage current is amplified by the factor B ? Are you aware that the BJT is a device which is controlled by the voltage Vbe? More than that, in the expression Ib=Ic/B the factor B relates the base current Ib (between B and E) with the current Ic. And the equation tells us only that Ib is a certain fraction of Ic. Nothing else! \$\endgroup\$– LvWMar 1 at 8:57
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1\$\begingroup\$ @LvW But won't any leakage current into the base automatically generate base-emitter voltage. How can you have one without the other? As long as the Base-Emitter junction is forward biased, the equation Vbe = n * Vt * ln(Ie/Is + 1) should hold for any applied current, and Ie = Is*(e^(Vbe/vt/n)-1) should hold for any applied base-emitter voltage. Note that Ie = emitter current, Is = base-emitter reverse saturation current, n = diode ideality factor, Vbe =base emitter voltage, Vt = thermal voltage. \$\endgroup\$– user4574Mar 1 at 16:16
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\$\begingroup\$ Sorry, but I cannot agree. You are speaking about an "applied base-emitter voltage". But there is no (externally) applied Vbe. More than that, as a simplification we assume that a current could "generate" a voltage . But physically this is wrong. Of course, there will be a certain (small) voltage across the B-E junction - but not as a result of the leakage current. In contrast - the applied voltage Vce drives a leakage current through both pn junctions - and we have a voltage division between the C-B and the B-E path. However, without the classical amplifying effect. \$\endgroup\$– LvWMar 2 at 10:37
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In datasheets this is defined at limits as $$ _{Collector-Emitter ~Breakdown ~Voltage ~~~ V_{(BR)CEO} = 40~ @ ~ Ic=1mA,Ib=0}$$ for a typical general purpose transistor like the MMBT3904. This is leakage resistance due to the “Early Effect”.
