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I have to find the working zone of this BJT, for different values of P, by using a simulation software like Simulink. But I would like to know how to do it manually. The way I learned is sort of guessing if the collector and emitter voltages are higher or lower than the base voltage.
In this circuit the zener voltage is 5.1V and the LED forward bias is 2V. enter image description here

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    \$\begingroup\$ You have to look at the circuit, figure out what it does (how it works), analyze it. Do some calculations. You make this circuit a bit complex with that potmeter in place. I would replace it with a fixed resistor of 10 k and do the analysis. Then make it 5 K and analyze again. Simplest case: the pot is 0 k what then happens? Real engineers first do this analysis (it can be quick and dirty) and then confirm their findings using a simulation. \$\endgroup\$ – Bimpelrekkie May 8 at 11:11
  • \$\begingroup\$ @Bimpelrekkie what I'm having trouble with is calculating the voltage across the R1 resistor \$\endgroup\$ – Pedro May 8 at 11:16
  • \$\begingroup\$ What voltage do we need at the emitter of the NPN before any current can flow? What does that then mean for the voltage at the base of the NPN? Remember that the BE of an NPN behaves as a diode in forward. \$\endgroup\$ – Bimpelrekkie May 8 at 11:26
  • \$\begingroup\$ @Bimpelrekkie assume everything is ideal \$\endgroup\$ – Pedro May 8 at 11:33
  • \$\begingroup\$ What do you understand about this circuit? You didn't answer any of @Bimpelrekkie 's questions. \$\endgroup\$ – Elliot Alderson May 8 at 14:02
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Often, you don't know. Experience might help you make a guess, but you are not sure.

What you will do is make a guess as to what region it is and then try to solve the circuit. See what comes out and if it makes sense. You might find, in the case of a NMOS, that you assumed saturation, but then find a gate-source voltage of just a few tens of mV. You then know that assumption was wrong, and that you should probably re-do the calculations assuming the transistor is in sub-threshold.

You can itterate like this (which is what something like SPICE or simulink will do) - you assume one thing, do the math, and then update the assumptions (pick models that better fit the regions you seem to be in) and re-do the math, etc...

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You wrote in comments to assume "everything is ideal." This means different things depending upon the context. Since I don't read tea leaves well, I get to choose that context:

  1. The zener is ideal in the sense that it fires at exactly \$V_Z=5.1\:\text{V}\$ and there is no current-dependence of \$V_Z\$.
  2. The BJT is ideal in the sense that it's \$V_\text{BE}\$ must be forward-biased by exactly \$700\:\text{mV}\$ (silicon) and that its \$\beta=\infty\$. Also, the saturated \$V_\text{CE}=0\:\text{V}\$.

Given the above, all you need to do is work out when the voltage divider reaches a voltage needed to forward-bias the BJT's \$V_\text{BE}\$ while also reaching the zener's \$V_Z\$. Also, if you subtract these two values from the supply voltage, you also get the voltage across \$R_1\$ (and its current, if you want it.)

Just use the above ideal values and solve the following equation for \$P\$: \$\quad15\:\text{V}\cdot\frac{P}{P+R_1}=V_\text{BE}+V_Z\$.

Since \$\beta=\infty\$, the collector current will instantly be whatever is allowed by its collector load when the BJT is "saturated." Since \$V_\text{CE}=0\:\text{V}\$, the collector voltage will be the same as the emitter voltage (which you know is \$V_Z\$.) From this, you can work out the voltage difference across the collector's resistor and LED, if you want that.

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