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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.

TVV
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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.

jonk
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    @TVV: There is a [youtube BJT animation video here](https://www.youtube.com/watch?v=EDju2GTOFDM) 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 02 '16 at 05:53
  • @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 02 '16 at 08:17
  • @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 02 '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 02 '16 at 16:53
  • I got a lot out of what you wrote jonk, so +1 from me :) I really like the bit about "*permits*, does not *cause*". I knew that indirectly, but stating it like that sticks well in the mind. If you want the advice of a technical communicator, I'd say attend to definitions (I had to go look up hole diffusion currents), especially for beginners, and keep paragraphs short (like break that third one up into two or three or four). But either way I follow your answers with great interest :) – scanny Sep 02 '16 at 17:07
  • @scanny: I appreciate the nice comment and the suggestion (though I wonder whether or not getting into the meaning of _diffusion_ would overly complicate the answer.) I do know I have at least two audiences -- the questioner _now_, and others _whenever_. I wish I had more time to either write less or to write more. But I don't. So I just give things a shot and hope. Thanks for the kind words! – jonk Sep 02 '16 at 17:13
  • @jonk and scanny, I don`t know if you have seen some of my comments or answers in other threads - however, I am convinced that the base current neither "permits" nor is the "cause" for the collector current. Both are simply the result of splitting the emitter current into two (unequal) portions. And both are controlled by the exponenetial law involving the base-emitter voltage only. This can be (and has been) proven!. – LvW Sep 02 '16 at 18:15
  • @LvW: See my post here for the exponential laws of which I'm well aware: http://electronics.stackexchange.com/questions/252197/why-is-vbc-absent-from-bjt-equations/252199#252199 – jonk Sep 02 '16 at 18:18
  • @jonk, thank you for the link. But now I am really confused: After reading your linked contribution (Ebers-Moll equations) I cannot understand how you came to the conclusion (in this thread) that IB "permits" IC. For my understanding, the meaning of "permit" is practically the same as "control". So - which quantity does control IC in your view: IB or VBE? (I hope, you will not answer with the chicken-egg problem, which would be a bit too simple). Thank you. – LvW Sep 02 '16 at 19:01
  • @jonk, I forgot to mention (because YOU have mentioned I. Getreu): I like him and I like (and have) his book!. As far as I remember, he describes the BJT clearly as a voltage-controlled current source. – LvW Sep 02 '16 at 19:05
  • @LvW: I contacted Ian a few years ago, got him to secure the rights to his book from Tektronix, and helped him republish his book! He's still kicking around! Anyway, the DC currents in a BJT are voltage-driven as shown in the Ebers-Moll models (all three are just different viewpoints of the same thing.) However, there is also physics taking place in the device and it's interesting. Ian doesn't discuss the physics (much, anyway.) Just the mathematical modeling. The physics comes from books often titled "Microelectronics" and there you will find the description I wrote about here. – jonk Sep 02 '16 at 19:08
  • @jonk, I know many books dealing with this subject - also from the physical point of view. More than that (as I have mentioned already) several observable effects and circuit properties as well as working principles clearly show that the BJT is voltage-controlled. Therefore, my confusion about your wording "IB permits IC". Do you know that Barrie Gilbert qualifies the base current as a "defect" and P. Brokav as a "nuisance"? So - what is your opinion? Does IB "permit" IC or is IB simply a kind of an unwanted by-product? – LvW Sep 02 '16 at 19:29
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    @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 02 '16 at 19:47
  • @jonk, of course, the base current does exist. However, I think it is a kind of wasted effort to ask how the BJT would behave in case of IB=0. From AD (Barrie Gilbert) I have learned that they have produced working BJTs with a IC/IB ratio in the order of of 1E3 or even more. Hence, I am not sure if the base current really is necessary to "restore the neutral space charge". Nevertheless, it is more or less guesswork. Thank you. – LvW Sep 03 '16 at 08:34
  • @LvW: It is _always_ the case, even in the superBeta BJTs you mention, that the base current is required to restore the space charge. There are a LOT of other effects going on. It's not that simple. For _all_ the currents I'm aware of in the simplest DC case, you can look at a Figure 3-4 I provided in this response I earlier gave: http://electronics.stackexchange.com/questions/254391/terminology-for-bjt-regions-of-operation/254397#254397 – jonk Sep 03 '16 at 13:39
  • @jonk, yes - I am aware that there are many other effects going on. And - of course, everything is OK with the figure you have mentioned (showing all the currents). However, my only point was that - for my opinion - it is not more than unrealistic guesswork to say how the part would behave without a certain current. Without the current the part would be a complete other part (in our phantasy). And I doubt if there is anybody who could describe the properties of such a "virtual" part. – LvW Sep 03 '16 at 13:52
  • @LvW: Ah. So we agree on that last point, then. Thanks. – jonk Sep 03 '16 at 13:55