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I'm fairly new to transistor usage and determining stages of operation for a BJT. However, from what I've learned, in order for a BJT to be saturated, both junctions need to be forward biased, and that in saturation, an increase in base current does not equal an increase in collector current.

However, looking at this particular datasheet, and specifically this graph, they have current gain versus different collector-emitter voltages.

  1. Can I assume that saturation on this graph occurs when the DC current gain starts dropping, or below the knee of each line? For example, at 200 mA with a VCE of 10V, is this transistor saturated since the current gain is dropping?

  2. If #1 is true, how is it possible that the transistor is in saturation with a VCE that is greater than 1 V? I thought since in saturation, both diodes are forward biased, so that the emitter-collector junction has a relatively low voltage drop (below 0.7V). Am I missing something here in how I am interpreting this graph?

Help greatly appreciated! Been bothering me all day.

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  • \$\begingroup\$ The transistor going into saturation isn't a property of the transistor itself. The operation point for saturation cannot be found on a datasheet. Instead, it is a property of the circuit surrounding the transistor (plus the transistor as part of it.) You are still correct, though, that shallow saturation just starts to happen when the base-collector junction just starts to become forward-biased. Read some here: electronics.stackexchange.com/questions/276146/… \$\endgroup\$ – jonk Oct 4 '17 at 22:56
  • \$\begingroup\$ Hi jon, thank you for the quick answer. So, from what you're saying, the transistor saturated based around the circuit around it. So...is the VCE they're giving me the voltage of the circuit...not the VCE of the transistor itself? If I were to probe the collector-emitter junction of the transistor, will it give me less than 1V? Still a little confused on the curve they're giving me... \$\endgroup\$ – Confusius Oct 4 '17 at 23:02
  • \$\begingroup\$ When I first heard the term "saturation" related to transistors, I imagined it was all about a physical condition of the transistor that could exist as a "state of being." And if you take the idea to mean that \$V_{BC}\$ is forward biased while \$V_{BE}\$ is also forward biased (meaning that \$\mid V_{CE}\mid\le 700\:\textrm{mV}\$ for silicon devices at typical collector currents), then I suppose it is a condition of the BJT. But this condition is only imposed on it (forced) by the circuit around it. It is better to think of it as when the circuit forces the condition on the BJT. \$\endgroup\$ – jonk Oct 4 '17 at 23:09
  • \$\begingroup\$ I suggest you look up gummel plot, saturation isn't applicable to your graph \$\endgroup\$ – sstobbe Oct 4 '17 at 23:50
  • \$\begingroup\$ Are you the same person who asked Is possible to get Q2 saturated with the VI voltage? Because in that case you used a pnp transistor, whereas if you don't specify, like in this question, people will assume you are asking about an npn transistor. \$\endgroup\$ – The Photon Oct 5 '17 at 4:33
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It looks as though you haven't received a direct answer to your questions. There are a number of question/answer cases on this site to examine. But I'll just drirectly answer, instead.

  1. Can I assume that saturation on this graph occurs when the DC current gain starts dropping, or below the knee of each line? For example, at 200 mA with a VCE of 10V, is this transistor saturated since the current gain is dropping?

That chart has nothing at all to do with saturation. It's actually a good chart you picked to help me make this point, because they include some curves for three different values of \$V_{CE}\$ (must be a PNP device): \$-1.0\:\textrm{V}\$, \$-2.0\:\textrm{V}\$, and \$-10.0\:\textrm{V}\$.

What is important to note that in all three cases here, is the magnitude of all of these values: \$\mid V_{CE}\mid\quad\ge\: 1.0\:\textrm{V}\$. This means, by definition, that the transistor is not in saturation.

If it isn't clear to you why this is, then please think very closely about this. Suppose you have a forward-biased \$V_{BE}\$ junction. But also suppose that the \$V_{BC}\$ junction is not forward-biased. (So that there is little to no current due to it being forward-biased.) Then it must be the case that the collector is further away from the emitter than the base is, right? Has to be. By itself, this says that \$\mid V_{CE}\mid\quad\ge\: 0.7\:\textrm{V}\$. And this is what is meant when someone says that a BJT is in its active region and not in saturation. Saturation just starts right at the moment when \$\mid\: V_{CE}\mid\:\lt\: \mid V_{BE}\:\mid\$ and therefore the \$V_{BC}\$ junction is moving into forward-biased mode. The saturation starts out very shallow, because the forward biased BC junction only adds very, very tiny currents. When still very shallow, the value of \$\beta\$ isn't badly impaired. But as the collector gets closer and closer to the emitter, these currents rapidly increase and the transistor is soon into deep saturation where the \$\beta\$ is quite poor.

So, the curve you show is everything about active mode and nothing at all about saturation. All of the curves are everywhere just active mode curves. The decline that you see taking place is happening for other reasons (current crowding, for example), but not because of saturation. Also, you'll notice that the \$\beta\$ is declining "sooner" when the magnitude of \$V_{CE}\$ is smaller. (This can be a reason why you want to operate a BJT with more \$V_{CE}\$, than less. But that's another story for later.)

  1. If #1 is true, how is it possible that the transistor is in saturation with a VCE that is greater than 1 V? I thought since in saturation, both diodes are forward biased, so that the emitter-collector junction has a relatively low voltage drop (below 0.7V). Am I missing something here in how I am interpreting this graph?

Well, now you know your answer here.

Because saturation is a matter of "designer choice" and not so much a matter of the device itself (though, yes, the device does impact that choice), the datasheet folks will usually provide a couple of different graphs.

This one can be used to work out the value of \$V_{BE}\$ when saturated, for example. Note that they include curves where \$\frac{I_C}{I_B}=10\$:

enter image description here

That's a big clue about what they consider to be a possible saturation value for \$\beta\$ (10.)

Also, you can see still more detail here:

enter image description here

Where they provide nice curves that show saturation "occurring." This curve actually shows you a lot more useful information about the saturation process for the BJT, at a few of the useful collector current cases. You can extract some representative saturation \$\beta\$ values by looking over these curves.

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Here is the collector characteristics for a bipolar

enter image description here

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