Can you please compare the base current of 2 transistors:

T1 has Vbe = 0.7 V and Vce= 0.7 V T2 has Vbe = 0.7 V and Vce= 0.1 V

i believe that T1 is in active mode while T2 is in Saturation mode, so the only comparison i am noticing is that Ib(T2) > Ib(T1), if you have any other comments please tell me.

thanks in advance

  • \$\begingroup\$ I, personally, miss pictures here. You'd save me from drawing it on a piece of paper, if you were to provide some picture. \$\endgroup\$ – Dzarda Oct 30 '14 at 10:30
  • \$\begingroup\$ It's impossible to say without knowing the type of transistor (its parameters) and the emitter or collector current \$\endgroup\$ – clabacchio Oct 30 '14 at 10:36
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    \$\begingroup\$ What does the C tag have to do with transistors? \$\endgroup\$ – Matt Young Oct 30 '14 at 12:08
  • \$\begingroup\$ It is not possible to compare the base currents, because we don't know what they are. We only know Vbe. Since Vbe is the same for both, absent any other information, the most logical conclusion is that both base currents are equal. Since T2 has a lower Vce, it is more likely to be in saturation. T1 Vce is 0.7V, so I would guess that it may not be in saturation, but may be close to it. However, saturation voltage does vary between different types of transistors. Some are designed to have very low Vce(sat). \$\endgroup\$ – mkeith Oct 30 '14 at 18:57
  • \$\begingroup\$ Because all transistor nodes are on a defined potential (voltage) - and assuming the same transistor in both cases - we can, of course, give a qualitative answer. We are not able to give a value for the base currents but we can say if Ib(T1) is smaller or larger than Ib(T2). See my detailed answer. \$\endgroup\$ – LvW Oct 31 '14 at 9:03

1.) The collector-base voltage for T1 is Vcb=0V. Hence, you are just at the threshold for the active mode.

2.) In case of T2 the base-collector diode also is open allowing an additional current through the base node. The transistor cannot work as an amplifier and Ib(T2)>Ib(T1).

EDIT 1 : For clarification of some misunderstandings: The OP did ask for the base current Ib - that is the current through the base node, which NOT necessarily is identical with the current through the base-emitter path. In case the base-collector junction is forward biased (as in cas 2 with Vce=0.1 V) the base current Ib has two components: Ibe and Ibc.

EDIT 2 : This effect (current from base to collector) can be clearly verified for Vce=0. For this case, all detailed Ic=f(Vce) graphs show that for Vce=0 all curves start at NEGATIVE Ic values (they do not start at the origin). And negative Ic values describe a current out of the collector node (npn case). This is nothing else than Ibc as part of the current Ib=Ibe+Ibc.

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  • \$\begingroup\$ In a saturated NPN BJT, it is common for the base to be at higher potential than the collector. Vce(sat) of 0.2V is realistic. But that does not mean there is net current flow from base to collector. The net current flow is into the collector, into the base, and out of the emitter. \$\endgroup\$ – mkeith Oct 30 '14 at 19:05
  • \$\begingroup\$ mkeith - perhaps you can reconsider the downgrading of my answer? I didn`t mention "saturation" - I rather spoke about "amplifier" application. Do you really deny that there is a current from the base to the collector (npn case) if the base-collector diode is open (Vbc=+0.6 volts for the given values)? A simple simulation can proove that the current into the base does increase correspondingly. I repeat: Of course Ib(T2)>Ib(T1). \$\endgroup\$ – LvW Oct 31 '14 at 9:11
  • \$\begingroup\$ I undid my down vote, and hereby apologize. I deny that there is a NET current flow out of the collector when Vce is 0.2V and Vbe is 0.7V. But, assuming that the transistors are the same, I think you are right about the relative base currents and in general. Your answer should be considered correct, I think. \$\endgroup\$ – mkeith Nov 1 '14 at 4:52
  • \$\begingroup\$ Thanks you soooo much LvW and mkeith, i appreciate your contribution to my question \$\endgroup\$ – Mohammed Shlibek Nov 1 '14 at 11:24

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