How can I calculate beta with only the data in the image?
Q is the quiescent point. I've tried with the basics Vcc formulas, but since it doesn't provide any value for Vcc I could not proceed.
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1\$\begingroup\$ Look, \$V_{\text{CE}}\$ reaches very close to \$24\:\text{V}\$ and only when \$I_{\text{C}}\$ nears \$0\:\text{mA}\$. That's a very huge clue about \$V_{\text{CC}}\$, don't you think? \$\endgroup\$– periblepsisCommented Oct 31, 2023 at 4:45
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1 Answer
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Beta is going to be the collector current divided by the base current at a specified operating point. Taking the operating point from the load line graph: $$\beta = \frac{I_C}{I_B}~ @~ V_{CE} = 9V~and~I_B = 50 \mu A$$ You have the base current, so you just need to find the collector current.
You can determine \$V_{CC}\$ from the load line graph, it will be the X-axis intercept, 24 V.
At the Q-point \$V_{CE}\$ is shown as 9 V. This will be the voltage across the transistor, so the rest of \$V_{CC}\$ must be dropped across the collector and emitter resistors. From that you should be able to work out the current.
Depending on how accurate your answer is required to be you can take into account that $$ I_C = I_B\cdot \beta\\~\\and\\~\\ I_E = I_B\cdot (\beta + 1) $$ The difference is small, and if you ignore the base current through the emitter and just assume that \$I_E = I_C\$ the answer comes out as a round number, which in a classroom setting is often what they're looking for.
