Bjt amp with 2 voltage sources bias point

Question

I'm trying to bias the bjt amp with two voltage sources and make the collector of bjt at 0.5V to maximize output swing but the simulation give me too much different result, why? The collector current I choose - 5ma, R4 voltage drop - 1V, XMM1 shows 3.6V

Answer 1

Can you use adaptive biasing, like this

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

There appears to be an issue with the circuit you have drawn. I recreated the circuit and I see 4V and 2.5mA across R3 which is what one should expect

If you want Vc to be .5V instead of 1V you need to decrease the value of your Emitter resistor to increase the current though the transistor. R4 of ~170R is about what you would need.

^{simulate this circuit – Schematic created using CircuitLab}

One thing to note is that if you modify the beta of the transistor up/down from 100 it can have a significant effect on the Q point of the transistor has only small effect on the Q point. This allows part-to-part variation (within reason) to be accommodated without having to individually trim the circuit to each transistor.

EDIT: I renumbered my resistors to match yours.

Answer 3

There are many methods to find \$V_C\$ voltage. But if BJT is in the active region we can replace the \$I_C\$ with a current source. And the circuit diagram will look like this:

^{simulate this circuit – Schematic created using CircuitLab}

1 - I replace the current source with an open circuit and solve for \$V_C\$.

$$V_{C1} = V_{CC}\frac{R_L}{R_L+R_C}$$ (Voltage divide rule)

2 - In this step, I "shot" the voltage source and solve for \$V_C\$

$$V_{C2} = -I_{R_C}*R_C $$

\$I_{R_C} = I_C * \frac{R_L}{R_L+R_C}\$ (current divider rule)

$$V_{C2} = -I_C * \frac{R_L}{R_L+R_C}*R_C $$

And finally we have:

$$V_C = V_{C1}+V_{C2} = V_{CC}\frac{R_L}{R_L+R_C}-I_C * \frac{R_L}{R_L+R_C}*R_C $$

Of Course, we could use Nodal analysis and solve for \$V_C\$

$$I_C+\frac{V_C}{R_L} - \frac{V_{CC} - V_C}{R_C} = 0$$

Therefore:

$$V_C = \frac{(V_{CC} - I_C R_C)R_L}{R_C+R_L}$$

In this type of a circuit, we can find voltage swing quite easy. The positive peak occurs when the BJT enters the cut-off region.

Hence

$$V_{Cmax} = V_{CC}\frac{R_L}{R_L+R_C} = 4.58V peak = 3.2V RMS $$

The negative peak is when the transistor is in the saturation region.

And the equivalent circuit looks like this:

^{simulate this circuit}

$$V_{Cmin} = \frac{R_L (R_E V_{CC} + R_C V_{cesat} + R_C V_{EE})}{R_C R_E + R_C R_L + R_E R_L} = -2.96Vpeak = -2.09VRMS $$

Or
$$V_{Cmin} \approx\frac{R_E}{R_E+R_C}(V_{CC}+|V_{EE}|)-V_{EE}+ V_{cesat}\approx(\frac{200\Omega}{200\Omega + 900\Omega}*10V)-5+0.2V \approx-2.98V peak $$

And the voltage gain will be around

$$A_V=\frac{R_C||R_L}{r_e + R_E}*\frac{\beta}{\beta + 1}\approx\frac{R_C}{R_E}\approx 4.5 V/V$$