From what I understand, in the saturation mode of NPN transistor, both EB and CB junctions are forward biased. Now, if the voltage across CB causes it to be forward-biased, how does the current flow in the opposite direction of voltage applied? - i.e from collector to emitter. How can a PN junction be forward biased while allowing a current of opposite direction to flow through it?
How does the current flow through the collector-base junction of an NPN transistor operating in saturation?
Short answer: For the same reason that the severely reversed-biased BC diode passes large currents despite being reverse-biased when the npn is in active mode (and reverse-biased diodes should have negligible current)!
Long answer: Imagine the same npn transistor in common-emitter configuration, emitter is connected to ground, collector to Vcc = 10V through a resistor R and base to a variable voltage source. When Vb is below the threshold voltage Vt, say 0.5V, the transistor is off, therefore Ic = Ie ~ 0 and there is no voltage drop on the resistor R, therefore Vc ~ Vcc ~ 10V. Then gradually increase Vb, the BE diode becomes forward biased, and assuming the electron-driven devices, BE draws a large amount of electrons from the ground (going opposite to the conventional direction of current) and provides them to the p end of the BC diode, which is currently reverse-biased. However, BC is not an isolated diode here, therefore do not expect it to act as one. The junction field in the BC sweeps all the provided electrons across the junction, generating a large downward collector-emitter current, while the BC is reverse biased. Obviously this is contrary to what a normal diode should do, but then again BC is not a normal diode; you can think that it is "hacked" in a sense. For clarity, the E-field in the BC diode is always from n (collector) to p (base).
Still on the same story, by increasing Vb, Ic increases, the voltage drop across R increases which leads to Vc decreasing until it reaches Vb from above. Now you would expect Ic to be zero, because the potential difference across CB is zero, but then again, we are dealing with a hacked diode. The same E-field across its junction (which is also decreased but never changed direction) sweeps away all the provided electrons, continuously maintaining the downward current.
Continuing on, even further increase in Vb drags Vc lower than Vb, but the internal E-field still does not change direction (although waning in magnitude), acting as just like above. The back-to-back diode picture should never be taken literally, i.e. you cannot make a transistor by wiring up two discrete diodes back to back, because you cannot keep the bipolar nature of charge carriers (electrons and holes) using discrete components (all holes and electrons will become electrons when transmitted across the wire connecting the diodes).
As for the current flowing in opposite direction to the voltage, here you are dealing with active devices as opposed to passive. Active devices can exhibit negative conductances.
Current flows from base to emitter. That is straightforward. Current also flows from base to collector. Hopefully, that is also straightforward. In both cases, current is flowing across a forward biased junction when the transistor is in saturation.
Due to transistor action, current also flows from collector to emitter. If you have studied transistors, you already know about this. This is a bit more complicated to explain, and I won't even try. But it is the defining characteristic of a bipolar junction transistor (BJT). Anyway, the NET current flow is from collector to emitter. The emitter voltage in this situation will be lower than the collector voltage.
Another way to look at it. Charges entering the base can get to GND by two different paths. Path 1 is directly to emitter. Path 2 is collector first, then emitter.
So, in no case does current flow in reverse direction from low to high voltage, nor does it flow across a reverse biased diode junction.
Hope that helps. Transistors can regulate the flow of current, but they can't make current flow from a low voltage to a high one. To do something like that requires a switching regulator, or charge pump or some such.
Obviously you are confused because there's say 0,3V at the collector but 0,7V at the base and still the current flows from the collector to the base to get sinked to GND through the emitter.
This really seems to be a contradiction. The explanation comes from the small thickness of the base. Carriers have much wider thermal random motion than the width of the base. So, it's no problem to them to penetrate through a narrow potential wall. The diffusion due thermal motion is the key mechanism that makes transistors possible.
In NPN transistors the doping of the emitter is much higher than the doping of the base => +voltage at the base causes essentially an electron current, less a hole current. Those electrons get mainly diffused to the collector due it's big area compared to the base connection wire. That makes collector more negative than base if there's enough resistance between the collector and +voltage supply.
\$\begingroup\$ Your's is a very plausible answer. Sure enough, I figured it in the last 2 years. \$\endgroup\$– KrakenAug 11, 2018 at 19:26