Unsaturated BJT

Question

BJT Question

I'm struggling for an answer to part (c), I can do the first two questions.

I know that \$i_C = \beta \times i_B\$, however, I don't know how to calculate the maximum current gain of the unsaturated transistor.

Do you understand what it means for a transistor to be "saturated"? — Ignacio Vazquez-Abrams, May 14 '16 at 18:27
but the current flowing freely means that it has the same effect of being a closed switch. — Connor Jardine, May 14 '16 at 18:36
Understanding what it means for a transistor to be saturated. — Ignacio Vazquez-Abrams, May 14 '16 at 18:57
For the purposes of this question, the transistor is "saturated" when the collector current is limited by the external circuit, rather than by the transistor itself. Is that enough of a hint to get you going? — Dave Tweed, May 14 '16 at 19:35
When a transistor is saturated, Ib*Beta is *greater* than Ic, e.g. a transistor with a beta of 100 needs 1mA of base current to handle 100mA of collector current right? If the same transistor had 10mA of base current and 100mA of collector current we would say that it was saturated as 10mA (Ib) * 100 (Beta) >> 100mA (Ic) — Sam, May 14 '16 at 23:59
i actually disagree with @IgnacioVazquez-Abrams. if you take what Dave and Tom say and add the condition \$v_{CE}=0\$ which is a simple, equivalent to Connor's "current flowing freely", and is just a little extreme of "saturated", but not far off, it's easy to calculate \$i_C\$ and the range of \$\beta\$ so that \$i_B \beta > i_C \$. — robert bristow-johnson, May 15 '16 at 03:00
@robertbristow-johnson: What I was trying to imply was that those points were *insufficient* to solve the problem, not that they couldn't *eventually* be used to solve it given other information. — Ignacio Vazquez-Abrams, May 15 '16 at 03:07
but i disagree with you that the points are insufficient. Connor needs to convert that condition into some kinda mathematical statement. — robert bristow-johnson, May 15 '16 at 03:14

Massimo Ortolano · Answer 1 · 2016-05-14T21:06:08.960

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Recall that a BJT is saturated when both the base-emitter (BE) and the base-collector (BC) junctions are forward-biased (and not when "current flows freely from the collector to the emitter", which is a meaningless definition). Recall also that the relationship \$I_\mathrm{C} = \beta I_\mathrm{B}\$ is valid only in the active region, that is, when the BE junction is forward-biased and the BC junction is reverse-biased.

In your circuit the BE junction is always forward-biased, and the base current is approximately constant whatever the value of \$\beta\$. The BC junction, instead, can be either reverse- or forward-biased according to the value of \$I_\mathrm{C}\$, which defines the collector-to-base voltage \$V_\mathrm{CB}\$.

A way to solve point c) is then the following:

Assume that the BJT is working in the active region, \$V_\mathrm{CB}>0\$, and that the relationship \$I_\mathrm{C} = \beta I_\mathrm{B}\$ holds.
Calculate the collector potential \$V_\mathrm{C}\$ as a function of \$\beta\$.
Find up to which value of \$\beta\$ the assumption of point 1 is met, that is, find the maximum value of \$\beta\$ for which \$V_\mathrm{CB} = V_\mathrm{C}-V_\mathrm{BE}>0\$.

But of course, you can also solve the problem the other way round:

Assume that the BJT is saturated.
Calculate \$I_\mathrm{B}\$ and \$I_\mathrm{C}\$ under the assumption of point 1 and find their ratio. According to what I wrote in the first two paragraphs, what's the meaning of this ratio, then?

In principle, the two ways are equivalent: you assume a condition, saturation or active region, and then you check for which values of \$\beta\$ the condition holds. In practice, however, one way is much shorter (which one and why?).

edited May 14 '16 at 21:06

answered May 14 '16 at 20:47

Massimo Ortolano

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i don't think "current flows freely from the collector to the emitter", is a meaningless definition for transistor saturation, but maybe it's not perfectly accurate. for instance [this NPN curve](http://www.sentex.ca/~mec1995/tutorial/xtor/xtor2/2fig5.gif) has a constant collector-emitter resistance of nearly 200Ω when the transistor is saturated. 200Ω is not "current flow[ing] freely", but the resistance is rather low for most electronic contexts. it's meaningful and as accurate as "200Ω is a conductor of current". – robert bristow-johnson May 15 '16 at 01:39
@robertbristow-johnson That definition is not only inaccurate, but wrong because the implication is in the other direction: _if_ the BJT is saturated, i.e., both the BE and BC junctions are forward biased, _then_ \$I_\mathrm{C}\$ is not constrained by the relationship \$I_\mathrm{C}=\beta I_\mathrm{B}\$. In addition, such misleading definition doesn't let one understand that, e.g, in a Darlington pair the second transistor never saturates. – Massimo Ortolano May 15 '16 at 08:13
i am not saying that \$i_C\$ is or is not constrained by the relationship of \$i_C = \beta i_B \$ regarding saturation. i am saying that saturation implies that \$ v_{CE} \approx 0 \$ which is saying that the collector to emitter path is acting approximately the same as a wire (or a resistor of very low resistance) in the context of the rest of the circuit (and the load line determined by the rest of the circuit). in **some** sense of semantic, a wire replacing the collector-emitter terminals is similar in meaning to "current flows freely from the collector to the emitter". – robert bristow-johnson May 15 '16 at 16:43
dunno which transistor you mean byt the *"second transistor"*, but in a Darlington pair, both transistors will saturate if you force enough current into the Darlington base. – robert bristow-johnson May 15 '16 at 16:55
@robertbristow-johnson Referring to [this diagram](http://hades.mech.northwestern.edu/images/0/03/Darlington_pairs.gif), Q2 can never saturate, because its base to collector junction is always reverse-biased. – Massimo Ortolano May 15 '16 at 17:04
what happens once Q1 saturates and you increase the base current into Q1 further? Q2 is on the edge of saturation and \$v_{CE}\$ for Q2 is close to zero. essentially the same place on the load line as it would be for saturation. – robert bristow-johnson May 15 '16 at 17:09
@robertbristow-johnson Saturation has a _unique_ definition that can be found in any electronics textbook: both junctions should be forward biased. According to this definition Q2 _never_ saturates, whatever current you inject into the base of Q1. That's why definitions like "current flows freely from the collector to the emitter" should be avoided. – Massimo Ortolano May 15 '16 at 17:21
listen Mass, i've been doing this in 1974 and i know what saturation means. you have identified something that the OP (Connor) has said "meaningless", but it is not meaningless. it is not **perfectly** accurate, but if *"current flows freely from the collector to emitter"* is taken to mean *"replace collector-emitter connection with hypothetical wire because \$v_{CE} \approx 0\$."*, the degree of accuracy is sufficient for me regarding the behavior of the circuit, unless \$R_C\$ is not exceeding the apparent saturation resistance by a large factor. – robert bristow-johnson May 15 '16 at 17:32
@robert bristow-johnson: Look at http://electronics.stackexchange.com/questions/254391/terminology-for-bjt-regions-of-operation please. – Incnis Mrsi Aug 27 '16 at 17:48

robert bristow-johnson · Accepted Answer · 2016-05-15T06:09:09.740

so, from part (a), you know what \$i_B\$ is and what \$i_C\$ is.

from the result of (a) and from part (b), you know what \$v_{CE}\$ and then \$R_C\$ are.

now using your definition of saturation, which is "current flows freely from collector to emitter", translate that to \$v_{CE} = 0\$, which might be overstating the case, but not by too much. is \$i_B\$ affected by this? calculate what \$i_C\$ is in this case.

from that and what Tom said about saturation: \$ \beta \ i_B \ge i_C \$, you can calculate your minimum \$ \beta \$. (or is it maximum \$ \beta \$ you want??)

Unsaturated BJT

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