# Why Ic increases with increase in Ib?

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First up I don't know if the direction of Ib is right for the pnp transistor in the circuit ( I need to know the active region relation so, as the emitter-base forward current will be more than the collector-base reverse current, I have given this direction for Ib ) . I want to know how Ic increases with increase in Ib. I have seen a lot of explanations based on equations but I need to know what causes the relation between Ic and Ib.

Then let's just ignore the equations, then.

The base current doesn't cause collector current. It permits collector current.

In the PNP BJT you show in your circuit (I'm ignoring all of your circuit except for that detail), there is a hole diffusion current from the emitter into the base. This hole diffusion current can't get to the collector without crossing the base. (There is also an electron diffusion current from the base into the emitter, but let's ignore it for now.) The emitter is highly doped, so it has quite a concentration of acceptor holes. But the base is lightly doped, so it has a much smaller concentration of donor electrons. As the emitter hole diffusion current of holes attempts to cross the base to get to the collector, a small proportion of these forward injected holes combine with the lightly doped donor electrons in the base material and disappear, so to speak. This is called recombination and it causes the base in a PNP to begin to accumulate towards a more positive space charge and if that is allowed to continue, the hole diffusion current stops (it gets repelled, given time.) The solution here is for the base lead to supply more electrons to replace the donor electrons that were lost to recombination in the base material. Doing so restores the base's space charge and allows the forward injected hole diffusion current to continue as before on towards the collector. The base lead must continue to re-supply these electrons in the PNP transistor so that the space charge doesn't build up positively until it blocks that current.

In short, you MUST supply electrons into the PNP base in order to maintain the base space charge balance so that the injected hole diffusion currents can continue to cross from the emitter, across the base region, and then to the collector.

There are more details. I mentioned the electron diffusion current from the base to the emitter and there are also minor base-collector currents across that reverse biased junction, too. But the above gets most of the gist across. And does so without equations.

• @TVV: There is a youtube BJT animation video here that illustrates some of jonk's points and might be of some help. There are a couple of errors the narrator makes, but I think it gives the gist in a visual form that's fairly accessible. Note he describes an NPN transistor rather than a PNP as you have in your circuit. – scanny Sep 2 '16 at 5:53
• @Jonk, Trying to illustrate your explanation with some figures (assuming a beta value of 200): 10 additional electrons injected into the base node (replacing the "electrons lost to recombination") will allow the "forward injected hole diffusion current" to increase by 2000 holes. Does such a scenario sound convincing and logical? And the assumed current ratio of 200 is rather conservative. In your explanation, you have concentated on the diffusion current only and you have forgotten the influence of the E-field caused by VBE. – LvW Sep 2 '16 at 8:17
• @scanny, I must admit that the given link (a small 4 min. video) is more correct and more enlightening than many other explanations. It is rather short and, of course, neglects some minor effects - but it concentrates on the most important effect: The width reduction of the depletion layer caused by the applied base-emitter voltage. – LvW Sep 2 '16 at 11:40
• @LvW: It wasn't my intent, at all, to cover all the details here. And I certainly did NOT want to discuss the barrier potentials. II was merely trying to provide a simple discussion about why $I_B$ is a recombination current for $I_C$. Some criticize me for writing too much. None for too little. Here I tried to strike some balance and to directly approach the question without wandering about -- and to do it without equations, as asked. – jonk Sep 2 '16 at 16:53
• @LvW: It's not a bright line I see. The BJT is voltage driven, but there is a necessary recombination current that must be supplied by the base. A lot of people write that this is unwanted, but the fact is that there is a very useful BJT behavior that is wanted and you don't get to pick and choose some parts of reality without other attending parts. $I_B$ is required to be injected in order to restore the neutral space charge. Without it, the base would charge up and block further flow. Supplying it keeps the base's space charge from accumulating and thereby preventing further flow. – jonk Sep 2 '16 at 19:47