Why is the collector current at saturation less than the collector current in the forward active mode? Specifically,
Why is \$ I_C < \beta I_B \$ ?
Electronic – (NPN) Why is the collector current at saturation, less than the collector current in the forward active mode
npntransistors
Related Solutions
Image taken from maplesoft.com
If you look at the Ebers-Moll model, you can see that each junction is desctibed by a diode and a voltage controlled source. So the collector current will be, as in the first equation, dependent on Vbc (not Vbe) with the Shockley equation, but also on Ib, like in the second equation.
The second factor will be of some meaning only if Vb > Vc, which happens in the saturation region, where the B-C junction is forward biased and current can flow from the base to the collector. In the forward-active (also called linear) region, the B-C junction is reverse biased and the collector current is basically only given by Beta.
It's a small signal AC model, so the DC currents don't matter. Since \$ \frac{dI_{C}}{dV_{BE}}\$ is the same polarity for both, we can use the same model (i.e. a larger b-e voltage results in a larger load current, just they are both negative for the PNP version (which equates to the same result as the NPN)
Related Topic
- Electronic – “pull the collector below ground” and saturation
- Electronic – How to calculate the current and voltage when the emitter is grounded for saturation mode
- Electronic – does the collector current direction remain the same in saturation and active region
- Electrical – Collector current decreases with the increase of base current in saturation mode of bjt
- Electronic – How can we prove that collector current in active region is higher than the saturation region
- Electrical – Why collector current is small in value in saturation mode
Best Answer
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.