bjtcommon-emittersaturationtransistors
Why is the collector current smaller in value when a transistor is in saturation mode of operation? My guess is that although both the junctions are forward biased, the (forward) currents are 'opposing' each other and hence the net current is a small value. Is this explanation true? And if not, why is it so?
(The comparison of current in saturation mode is being made with the current in active mode for an NPN transistor in CE configuration)
One way to look at this is, the BJT enters saturation when the supply is no longer able to provide enough current to keep it in forward active operation. For example, let's look at the basic common-emitter amplifier:
Let's say V273 is 5 V and R277 is 500 Ohm. Then if the base current were high enough that the forward active operating condition would give a collector current more than 10 mA, the collector would have to actually be pulled below ground (not indicated on the schematic, but let's say ground is the node connected to the emitter of the BJT).
To do this, the BJT would have to actually be delivering power to the rest of the circuit, which it simply can't do.
Of course, the BJT isn't even so close to ideal that it can pull the collector all the way down to ground, so saturation sets in at some higher collector voltage, which we normally take as about 0.2 V for typical silicon devices.
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.
Best Answer
In an NPN in active mode Vcb must be so large (with collector positive) that the maximum number of electrons from the emitter, coming through the base, are sucked into the collector. Increasing Vcb hardly changes the current as the electron flow is at maximum already. The electron flow is determined by the BE junction!
In an NPN in saturation mode Vcb is smaller, so small that the flow of electrons is influenced by Vcb. This is the red part of the graph in Andy's answer, a small change in Vce (which is just Vcb + Vbe) will cause a large change in Ic. In saturation many electrons make it to the base instead of being pulled into the collector. That makes the base current larger and the collector current smaller. When the base current is kept constant that would result in a smaller collector current.
Note that there is no "opposing" current, the BC junction does not need to be fully forward biased to already influence Ic. It is more the fact that the BC junction isn't so far into reverse mode for it to be able to "pull out" all the electrons it possibly can out of the base region. So less electrons reach the collector resulting in a smaller Ic.