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

Question

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?

Possible duplicate of [(NPN) Why is the collector current at saturation, less than the collector current in the forward active mode?](https://electronics.stackexchange.com/questions/51405/npn-why-is-the-collector-current-at-saturation-less-than-the-collector-curren) — The Photon, Jul 12 '17 at 02:28
@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 02:44
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 02:47
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 02:48
@ThePhoton sir may be its a silly question but i am really confused....in this link below its saying that in saturation mode saturation occured when Ic>Icmax and active when Ic — Rohit, Jul 12 '17 at 03:27
My way of saying it is the BJT enters saturation when the load (connected to the collector) isn't able to provide enough current to keep the BJT in active mode. More detail in my answer to the question I linked. — The Photon, Jul 12 '17 at 03:31
For the same base current Ib the collector current in the active region will always be larger versus saturation due to base-width modulation (boost of beta) — sstobbe, Jul 12 '17 at 03:31
>@ThePhoton I read your answer and what i understood you said " Then if the base current were high enough that the forward active operating condition would give a collector current more than 10 mA" so here you saying the this 10ma is the saturation current isn't it? then again u said this is possible because Bjt cant deliver power "To do this, the BJT would have to actually be delivering power to the rest of the circuit, which it simply can't do".....so thats mean we cant get Ic greater in active mode than in saturation mode ...? am i correct sir???? — Rohit, Jul 12 '17 at 04:11
@ThePhoton also i did'nt get.....you said..... "It's also perfectly valid to consider keeping Vcc & Rc constant and push the device into saturation by increasing Ib."if we increse ib there will be increse in ic in active mode also then how can ic will be maximum in saturation mode ???? from the vi graph i am.saying this......its a humble request sir please clear these doubts..... — Rohit, Jul 12 '17 at 04:25
If you increase Ib (keeping Rc and Vcc constant), Ic will increase until the source (Vcc and Rc) can't supply any more current. This is the maximum current Vcc can supply through Rc. So it's more current than we got at any of the intermediate steps as we increased Ib. — The Photon, Jul 12 '17 at 04:27
After we get into saturation, further increasing Ib will not (very much) increase Ic. Again, because the source basically can't supply any more current. — The Photon, Jul 12 '17 at 04:28
@ThePhoton sir i have one doubt u said in your first point that you answered by keeping Ib constant incresing Rc and decresing Vcc is it possible to keep Ib constant by incresing Rc and Vcc???// — Rohit, Jul 12 '17 at 08:53
@Rohit, yes. Look at the load line chart I put in my answer. If you shift the red line left (reducing VCC) while staying on the same Ib=x curve for the transistor, the current of the intercept doesn't change (much), until you reach saturation. — The Photon, Jul 12 '17 at 14:38

score 3 · Answer 1 · answered Jul 12 '17 at 04:19

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

answered Jul 12 '17 at 04:19

analogsystemsrf

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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 04:44
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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 04:47

score 1 · Answer 2 · answered Jul 12 '17 at 02:05

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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.

answered Jul 12 '17 at 02:05

LVmiao

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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 02:11
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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 https://electronics.stackexchange.com/questions/51405/npn-why-is-the-collector-current-at-saturation-less-than-the-collector-curren?rq=1 – LVmiao Jul 12 '17 at 02:27

score 1 · Accepted Answer · answered Jul 12 '17 at 04:44

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

There's two ways to look at this

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.
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.

Sir please see this my doubt https://electronics.stackexchange.com/questions/326487/confusion-in-excess-hole-concentration — Rohit, Aug 30 '17 at 09:29

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

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