Question E6E01:
Why is gallium arsenide (GaAs) useful for semiconductor devices operating at UHF and higher frequencies?  

Choices:  
A: Higher noise figures  
B: Higher electron mobility  
C: Lower junction voltage drop  
D: Lower transconductance    

Correct Answer is  B: Higher electron mobility  

Why the correct answer is correct:
- GaAs has much higher electron mobility than silicon (roughly 5–6× at room temperature), so electrons can move faster for a given electric field.  
- Drift velocity is $v_d = \mu E$, so higher mobility $\mu$ gives higher $v_d$.  
- Faster carriers mean shorter transit time across the device channel, which raises the frequency at which the device can provide gain:
  $$t_{\text{transit}} = \frac{L}{v_d} = \frac{L}{\mu E} \quad \Rightarrow \quad f_T \approx \frac{1}{2\pi t_{\text{transit}}}$$
- Equivalently, higher mobility boosts transconductance and thus the unity-gain frequency:
  $$f_T \approx \frac{g_m}{2\pi C_{gs}}$$
  where higher $\mu$ increases $g_m$.  
- Result: GaAs devices (MESFETs, HEMTs) work well at UHF and microwave.

Why the other choices are incorrect:
- A: Higher noise figures  
  GaAs typically offers lower noise figures at microwave frequencies due to high mobility and favorable material properties. “Higher noise” would be a disadvantage, not a reason for use at UHF/microwave.

- C: Lower junction voltage drop  
  GaAs diodes generally have a higher forward voltage (about 1.2 V) than silicon (~0.7 V). Junction drop is not the key reason for high-frequency performance.

- D: Lower transconductance  
  GaAs FETs generally have higher, not lower, transconductance for a given geometry, aiding high-frequency gain.

Useful formulas / rules of thumb:
- Drift velocity: $v_d = \mu E$
- Transit time: $t_{\text{transit}} = L / v_d$
- Cutoff (unity-gain) frequency: $f_T \approx 1/(2\pi t_{\text{transit}})$ or $f_T \approx g_m/(2\pi C_{gs})$
- Rule of thumb: Electron mobility in GaAs is several times that of silicon, enabling UHF and microwave operation.