# How can we prove that collector current in active region is higher than the saturation region?

In the VI characteristics of BJT transistor its showing that the collector current in saturation region is less than the collector current in active region.

How can we prove it with a BJT circuit?

• – The Photon Jul 12 '17 at 2:28
• Or are you asking how to design a curve tracer? – The Photon Jul 12 '17 at 2:31
• @ThePhoton Sir i am asking that in the saturation region the voltage Vce is less like 0.2 that is less than the voltage in active region so the collector should be high in saturation region than the active region because Ic=Vcc-Vc/Rc ? – Rohit Jul 12 '17 at 2:44
• if you keep Ib constant and want to get the device into saturation region, you'd increase Rc or reduce Vcc. – The Photon Jul 12 '17 at 2:47
• When we say "current is lower in saturation mode" we mean "Keeping Ib constant, current is lower in saturation mode". It's also perfectly valid to consider keeping Vcc & Rc constant and push the device into saturation by increasing Ib. Then indeed Ic would be higher in saturation mode. This is exactly what we do when we want to use a BJT as a switch. – The Photon Jul 12 '17 at 2:48

First, let's look at the standard common-emitter circuit.

There's two ways to look at this

1. We can keep $I_b$ constant and increase $R_C$ or decrease $V_{cc}$ until the transistor enters saturation. In this case, the collector current is minimum at saturation, or anyway less than it was when we were operating the BJT forward-active.

2. We can keep $V_{cc}$ and $R_C$ constant and increase $I_B$ (by increasing $V_{BB}$ or decreasing $R_B$) until the BJT enters saturation.

In this case, the collector current increases until the circuit reaches saturation, and the current in saturation is the maximum.

In either case, saturation happens because the source (VCC and RC) isn't able to provide enough current to keep the BJT operating forward active. The saturation current is roughly $V_{CC}/R_C$ (ignoring the small $V_{CE}$ drop. We reached this maximum capability of the source by either increasing $I_B$ until $\beta I_B > V_{CC}/R_C$ or by reducing the current capability of the source.

We can also look at the load line analysis for your BJT:

Here I took your transistor characteristic curves and added (the red line) the load line for $V_{CC}=15\ \rm V$ and $R_C = 3\ \rm k\Omega$.

If we increase $R_C$ that means shifting the slope of the load line down, keeping the x-intercept at 15 V. If we reduce $V_{CC}$ that means shifting the load line horizontally to the left. If we do either of those while following one of the characteristic curves for a fixed $I_B$, we eventually move the intersection point into the saturation region at a lower current than it was for the initial forward active case.

If we increase $I_B$ we shift the intersection point up and to the left, until it reaches saturation with $I_B \approx 60\ \rm\mu A$ and $I_C \approx 5\ \rm mA$, the maximum available from the 15 V, 3 kohm source.

Unless the Early Voltage is infinity, the current in active region is guaranteed to be higher than in the saturation region, holding Ibase constant.

• That's the answer I was looking to see as I just now read downwards. Why is it that no one else mentioned the Early Effect?? Comes instantly to mind. – jonk Jul 12 '17 at 4:44
• Even if Early voltage is infinite, further decreasing Vcc after reaching saturation will result in saturation current less than forward active current. – The Photon Jul 12 '17 at 4:47

I will say it is the nature of BJT.

Simply while voltage goes up between collector and emitter, the current increase rapidly in the beginning (could use as amplifier), then hit saturation(could use as switch), increasing slows down.

If you want to verify it, give it voltage, use DMM to measure the current.

• Yes...Exactly thats my question can you prove your statement by taking a circuits because when i am doing its giving collector current high in saturation and low in active region – Rohit Jul 12 '17 at 2:11
• I will say check the type of your BJT, and put the resistors in the circuit, so it limits the current that you can observe. The schematic will look the similar as this post electronics.stackexchange.com/questions/51405/… – LVmiao Jul 12 '17 at 2:27