Question E6C04:
Which of the following is an advantage of BiCMOS logic?  

Choices:  
A: Its simplicity results in much less expensive devices than standard CMOS  
B: It is immune to electrostatic damage  
C: It has the high input impedance of CMOS and the low output impedance of bipolar transistors  
D: All these choices are correct    

Correct Answer is  C: It has the high input impedance of CMOS and the low output impedance of bipolar transistors  

Correct answer: C — It has the high input impedance of CMOS and the low output impedance of bipolar transistors.

1) Why C is correct
- BiCMOS combines CMOS gates (for inputs) with bipolar transistor drivers (for outputs).
- CMOS inputs draw essentially no DC current, giving very high input impedance. The main input limitation is capacitive:
  $$Z_{in} \approx \frac{1}{j 2\pi f C_{in}}$$
- Bipolar output stages provide strong drive with low output impedance, enabling fast edges and larger fanout. For a BJT emitter follower:
  $$Z_{out} \approx \frac{1}{g_m}, \quad g_m = \frac{I_C}{V_T} \ (\text{with } V_T \approx 25.9\ \text{mV at room temp})$$
- Result: easy-to-drive inputs (CMOS) and strong, fast outputs (bipolar).

2) Why the other choices are incorrect
- A: “Much less expensive” — False. BiCMOS requires a more complex fabrication process than standard CMOS and is typically more expensive, not less.
- B: “Immune to electrostatic damage” — False. CMOS gates are vulnerable to ESD due to thin oxides. BiCMOS often includes ESD protection, but it is not immune.
- D: “All these choices are correct” — False because A and B are not correct.

3) Useful rules of thumb
- CMOS input current is near zero; input impedance is effectively limited by gate capacitance.
- Bipolar outputs can source/sink higher currents with low $Z_{out}$, improving edge rates by charging/discharging load capacitances faster.
- BJT vs MOSFET drive strength: BJTs have higher transconductance at a given current:
  - BJT: $g_m = I_C / V_T$
  - MOSFET (approx.): $g_m \approx 2 I_D / V_{ov}$
  Higher $g_m$ → lower $Z_{out}$ → faster transitions.