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.

  • \$\begingroup\$ Do you understand what it means for a transistor to be "saturated"? \$\endgroup\$ May 14, 2016 at 18:27
  • \$\begingroup\$ yeah current flows freely from the collector to the emitter \$\endgroup\$ May 14, 2016 at 18:29
  • 1
    \$\begingroup\$ That definition is too oversimplified to be of use. \$\endgroup\$ May 14, 2016 at 18:29
  • \$\begingroup\$ but the current flowing freely means that it has the same effect of being a closed switch. \$\endgroup\$ May 14, 2016 at 18:36
  • \$\begingroup\$ Knowing that won't help you calculate ß. \$\endgroup\$ May 14, 2016 at 18:40

2 Answers 2


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??)


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:

  1. 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.
  2. Calculate the collector potential \$V_\mathrm{C}\$ as a function of \$\beta\$.
  3. 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:

  1. Assume that the BJT is saturated.
  2. 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?).

  • \$\begingroup\$ 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 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". \$\endgroup\$ May 15, 2016 at 1:39
  • \$\begingroup\$ @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. \$\endgroup\$ May 15, 2016 at 8:13
  • \$\begingroup\$ 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". \$\endgroup\$ May 15, 2016 at 16:43
  • \$\begingroup\$ 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. \$\endgroup\$ May 15, 2016 at 16:55
  • \$\begingroup\$ @robertbristow-johnson Referring to this diagram, Q2 can never saturate, because its base to collector junction is always reverse-biased. \$\endgroup\$ May 15, 2016 at 17:04

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