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In the following circuit:

enter image description here

I know the expression for the collector current is $$I_C =\frac{V_{CC}-V_C}{R_L}$$ Say I have a potentiometer for \$R_L\$ and I decide to set it to 0 Ω. What would \$I_C\$ be in this case?

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  • \$\begingroup\$ You should be able to quickly derive this active mode result:$$I_\text{C} = \frac{\beta}{\beta+1}\cdot \frac{V_\text{CC}\frac{R_2}{R_1+R_2}-V_\text{BE}}{R_\text{E}+\frac{R_1\cdot R_2}{\left(R_1+R_2\right)\cdot\left(\beta+1\right)}}$$ and take note of its independance of the collector resistor value. \$\endgroup\$ – jonk Oct 25 '18 at 5:54
  • \$\begingroup\$ By the way, you will know if the mode is actually active by using the above equation to compute the voltage drop across \$R_\text{L}\$. If it is excessively high and forces the collector voltage below the base voltage, then it's not in active mode and is instead in saturation and the above equation no longer holds. \$\endgroup\$ – jonk Oct 25 '18 at 8:00
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Let me show the voltages and collector current (IC).

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, \$V_B = V_{BE} + V_E = V_{BE} + I_E \cdot R_E\$.

Since \$I_C \approx I_E\$, the equality above turns into \$V_B \approx V_{BE} + I_C \cdot R_E\$. Finally, $$I_C = \frac{V_B - V_{BE}}{R_E}$$ (I will not show what VB is since it can be found with simply voltage divider rule.)

See? Collector current is independent from collector load!

NOTE: Placing a resistor to emitter creates a constant current source. It also guarantees the thermal stability.

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Lets assume Beta = infinity. And assume the Ic is near 1mA, so the Vbe is approximately 0.6 volts (0.5 volts near 10uA, 0.4 volts near 100 nanoAmps).

Ic = (Vbase - 0.6) / Re

Where Vbase = VDD * the voltage divider ratio

thus Vbase = VDD * R2 / (R1 + R2)

and now, finally,

Ic = { [VCC * R2/(R1 + R2)] - 0.6} / RE

Again, assuming beta = infinity, Vbe = 0.6, and Vearly = infinity

With all 3 resistors being 1KOhm, and VCC = 3 volts, we have

Ratio = 0.5,

Vbase = 3*0.5 = 1.5 volts,

Vemitter = Vbase - 0.6 = 1.5 - 0.6 = 0.9 volts

Ic = 0.9 / 1K = 0.9 milliAmps

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