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When talking about BJT transistors, a very common graph is the one shown below, that plots (VCE,IC) as an ascending function. As far as I know, the BJT manages VCE so that IC is beta times IB, being
VCE = 10V - IC 10K
in the example circuit. Which can be rewritten as
IC = (10V-VCE)/10K

But that equation doesn't fit the graph at all. I understand that it behaves in a different way around 0 since the voltage can not go under 0, but I don't understand why it grows.

enter image description here

enter image description here

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    \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$ – Qmechanic Mar 19 '19 at 5:41
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    \$\begingroup\$ As far as I know, the BJT manages VCE so that IC is beta times IB That is not how things work. It would be better to say: When in active mode, the BJT manages its conductivity such that IC is beta times IB. The way that the plot of Ic vs Vce is measured forces a Vce across the BJT. So the transistor cannot influence Vce at all. Also note that the plot cannot be measured with the common emitter circuit that you show. So you should not try to understand the plot while assuming the circuit you show is used to measure that plot. Both are different situations! \$\endgroup\$ – Bimpelrekkie Mar 19 '19 at 7:28
  • \$\begingroup\$ Just for a better understanding: Your circuit diagram contains two variables: Ic and Vce. However, of course the input voltage (5V) is an (independent) variable and the base-emitter voltage is another (dependent) variable - not a constant with Vbe=0.7V. \$\endgroup\$ – LvW Mar 19 '19 at 8:29
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The equation you say is the one imposed by the external components that accompany the BJT, not an equation that describes inner behavior of the transistor. That graph shows the relation between Vce and Ic, for particular Ib.

The equation you show to calculate Ic can be drawn as a straight line, therefore, the point where the transistor will work is the point where the both plots, the straight line and the other curve, intersect.

I recommend you to ask this questions on the Stack Exchange forum dedicated to Electrical and Electronics Engineering for a better range of answers.

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Something is wrong with the graph. Ten volts cannot push more than 1 mA through a 10 k resistor even if there is zero volts drop collector to emitter.

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  • \$\begingroup\$ This is not really an answer, better to make it a comment. \$\endgroup\$ – Bimpelrekkie Mar 19 '19 at 7:22
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The transistor configuration shown is that of common emitter, which is normally used as a voltage amplifier. That is, voltage variations at the input (base of the transistor) are amplified at the output (collector).

The graph above it showing \$I_C\$ vs \$V_{CE}\$ shows the collector current (\$I_C\$) increasing until it reaches \$\sim 5.5\ mA\$, where it reaches saturation and then plateaus, with the base current being noted as \$I_B=20\ \mu A\$. This suggests that a variation in base current between $0\ \mu A$ and $20\ \mu A$ results in a variation in collector current between \$0\ mA\$ and \$\sim 5.5\ mA\$, which seems more representative of a current gain transistor configuration than voltage gain. Given the \$10V\$ source at the collector, I would think that the circuit does not match the graph.

What the graph shows, though, is the operating range of a transistor (input up to \$I_B = 20\ \mu A\$, output up to \$I_C=\sim 5.5\ mA\$), and that any further increase in input voltage/current will not result in any further increase in the output -- it has already reached its maximum.

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