# Help With Simulating Common Collector Amplifier In PSPICE

I have included as much information as possible, so it should just be a matter of pointing out something simple (hopefully).

I am trying to model a biasing circuit for a Common Collector BJT Amplifier. I am running a Transient Analysis on the circuit and am receiving the following simulation results (attached images), which I am having trouble interpreting. I have also listed my minimum specs for the circuit and my theoretically computed values. As far as I can tell, I am doing everything correctly, although I am not seeing how to choose the Collector Resistance since it does not appear in any of my equations. I am also not seeing the voltage gain nor the Q-point as computed by PSPICE in the simulation Output File.

MINIMUM SPECS:

Input Resistance >= $5$k$\Omega$
Output Resistance = $50\Omega$
Voltage Gain $\geq 0.95$ with $5$k$\Omega$ load
Output Voltage Swing of at least $2$Vpp across $5$k$\Omega$ load
$V_{cc} \leq 20$V

FORMULAS USED:

Input Resistance: $R_1 || R_2 || (\beta+1)R_E$
Output Resistance: $R_E || r_{\pi}/(\beta+1)$
Voltage Gain: $$\frac{R_E(\beta+1)}{r_{\pi} + R_E(\beta+1)}$$
$G_m$: $I_c/0.026$
$r_{\pi}$: $\beta/G_m$

COMPUTED VALUES:

$R_1$: $36k\Omega$
$R_2$: $50k\Omega$
$R_e$: $10k\Omega$
$R_c$: ?
$r_{\pi}$: $5075$
$g_m$: $0.0197$
$I_c$: $5.122 \times 10^{-4}$
Voltage Gain: $0.995$
Vb: $5.772$V
Input Resistance >= $20.5$k
Output Resistance = $50$
Output Voltage Swing of at least $2$Vpp across $5$k load
$Vcc \leq 10$V

• As you can see, I do not have a value for $R_c$, as I do not see how it affects any of my equations.

Circuit Image: Simulation Waveform Image: Simulation Waveform Cursor Values Image (Corresponds To Previous Waveform Image): Simulation Output File: • Where is the Q-POINT?
• For a common collector the collector should be shorted to $V_{CC}$. Hence, $R_{C} = 0$. – Null Nov 17 '14 at 1:50
• My schematic suggests an unknown collector resistor, so my best guess was that it would not matter. But then why include it? Surely it must affect something. – user58446 Nov 17 '14 at 1:56
• I suppose it's there to be general, but it is 0 for a true common collector circuit, by definition. You would include a collector resistor for a common emitter (output is at the collector) but for a common collector you don't care about the collector node (it's not in the signal path, and just needs to be at a high enough voltage that the transistors isn't in saturation) so you can tie it to $V_{CC}$. – Null Nov 17 '14 at 2:00
• So if Vb is 5.772V, my only concern with Rc is that Vc is greater than 5.722V, to function in forward-active mode? I do know that C>B>E is forward-active and C<B>E is saturation and C<B<E is cutoff. But these voltages we desire conflict with my understanding that the voltage across Rc is determine by the voltage bias across R1. How does this all come together? This may its own question... – user58446 Nov 17 '14 at 2:17
• $R_1$ and $R_2$ set $V_B$, and then to make sure the transistor is in forward-active rather than saturation you need $V_B < V_C$. The voltage across $R_C$ is determined by $I_C$ since $V_C = V_{CC} - I_CR_C$. To set $I_C$, assume a diode drop ($0.7$V) for the BE junction, giving you $V_E$ from your $V_B$. Then set $R_E$ to give you the desired $I_E \approx I_C$. – Null Nov 17 '14 at 3:51