Transistors - BJT & FET

Question 1

(a) $R_C$ and $R_E$ (4 marks)

KVL around output loop: $V_{CC} = I_{CQ}R_C + V_{CEQ} + V_{RE}$. $R_C = \frac{V_{CC} - V_{CEQ} - V_{RE}}{I_{CQ}} = \frac{12 - 6 - 1.2}{1.0\text{ mA}} = \frac{4.8}{1\text{ mA}} = 4.8\text{ k}\Omega$ (2 marks)

$R_E = \frac{V_{RE}}{I_{EQ}} \approx \frac{1.2}{1.0\text{ mA}} = 1.2\text{ k}\Omega$ (using $I_{EQ}\approx I_{CQ}$; strictly $I_E = I_C(1+1/\beta)=1.0067$ mA → $R_E \approx 1.19$ k$\Omega$). (2 marks)

(b) $R_1$, $R_2$ (4 marks)

$I_{BQ} = I_{CQ}/\beta = 1.0/150 = 6.67\ \mu\text{A}$. Divider current $I_{div} = 10 I_{BQ} = 66.7\ \mu\text{A}$. Base voltage $V_B = V_{BE} + V_{RE} = 0.7 + 1.2 = 1.9\text{ V}$.

$R_2 = V_B/I_{div} = 1.9/66.7\ \mu\text{A} = 28.5\text{ k}\Omega$. (2 marks) $R_1 = (V_{CC}-V_B)/(I_{div}+I_{BQ}) = (12-1.9)/(66.7+6.67)\ \mu\text{A} = 10.1/73.3\ \mu\text{A} = 137.7\text{ k}\Omega$. (2 marks)

(Accept $R_1\approx (V_{CC}-V_B)/I_{div} = 151.5$ k$\Omega$ if base current neglected — state assumption.)

(c) Gain proof (6 marks)

Small-signal model: replace BJT with $g_m v_{be}$ current source, $r_\pi$, and $r_o$ from output to emitter. With emitter fully bypassed, emitter is AC ground. (1) Output node: $v_{out} = -g_m v_{be}(R_C \parallel r_o)$, and $v_{be} = v_{in}$ (input across $r_\pi$ referenced to AC ground). (2) Hence $A_v = \frac{v_{out}}{v_{in}} = -g_m(R_C\parallel r_o).\qquad\blacksquare$ (1)

Numbers: $g_m = I_{CQ}/V_T = 1.0\text{ mA}/25.85\text{ mV} = 38.68\text{ mS}$. (1) $r_o = V_A/I_{CQ} = 80/1\text{ mA} = 80\text{ k}\Omega$. $R_C\parallel r_o = (4.8\cdot80)/(84.8) = 4.528\text{ k}\Omega$. $A_v = -38.68\text{ mS}\times 4.528\text{ k}\Omega = -175.1$. (1)

(d) Sensitivity to $\beta$ (4 marks)

Thévenin base: $V_{TH} = V_B$ (open-circuit) $= V_{CC}R_2/(R_1+R_2)$, $R_{TH}=R_1\parallel R_2$. Base loop: $V_{TH} = I_B R_{TH} + V_{BE} + I_E R_E$, with $I_E = (\beta+1)I_B$, $I_C=\beta I_B$: $I_C = \frac{\beta(V_{TH}-V_{BE})}{R_{TH} + (\beta+1)R_E}.$ (2) When $(\beta+1)R_E \gg R_{TH}$, $I_C \to (V_{TH}-V_{BE})/R_E$, independent of $\beta$ — this is the stabilising effect of degeneration. (1)

Compute $R_{TH}=R_1\parallel R_2 = 137.7\parallel 28.5 = 23.6$ k$\Omega$, $V_{TH}=V_B=1.9$ V. $I_C(150) = 150(1.9-0.7)/(23.6\text{k}+151\cdot1.2\text{k}) = 180/(23.6+181.2)\text{k}=180/204.8\text{k}=0.879$ mA. $I_C(300) = 300(1.2)/(23.6\text{k}+301\cdot1.2\text{k}) = 360/(23.6+361.2)\text{k}=360/384.8\text{k}=0.9356$ mA. Fractional change $= (0.9356-0.879)/0.879 = +6.4\%$ for a $100\%$ change in $\beta$. (1)

(e) Code (4 marks)

def q_point(beta, VCC=12, R1=137.7e3, R2=28.5e3, RE=1.2e3, VBE=0.7):
    VTH = VCC * R2/(R1+R2)
    RTH = R1*R2/(R1+R2)
    IC = beta*(VTH - VBE)/(RTH + (beta+1)*RE)
    return IC
 
print(q_point(150)*1e3, "mA")  # ~0.879 mA
print(q_point(300)*1e3, "mA")  # ~0.936 mA
frac = (q_point(300)-q_point(150))/q_point(150)
print(frac*100, "%")           # ~6.4 %

Marks: correct Thévenin (1), correct loop equation (2), correct confirmation of (d) (1).

Question 2

(a) $g_m$ derivation (5 marks)

Saturation: $I_D = \tfrac12\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th})^2$. $g_m = \frac{\partial I_D}{\partial V_{GS}} = \mu_nC_{ox}\frac{W}{L}(V_{GS}-V_{th}).$ (2) Using $I_D$: $(V_{GS}-V_{th}) = \sqrt{2I_D/(\mu_nC_{ox}(W/L))}$, so $g_m = \sqrt{2\mu_nC_{ox}\tfrac{W}{L}\,I_D}.$ (1)

Numbers ($V_{ov}=1.0-0.5=0.5$ V): $I_D = \tfrac12(200\mu)(10)(0.5)^2 = \tfrac12(2000\mu)(0.25) = 250\ \mu\text{A}$. (1) $g_m = (200\mu)(10)(0.5) = 1.0\text{ mS}$. (1)

(b) Body effect (5 marks)

$V_{th} = 0.5 + 0.4(\sqrt{0.7+1.5} - \sqrt{0.7}) = 0.5 + 0.4(\sqrt{2.2}-\sqrt{0.7})$ $= 0.5 + 0.4(1.4832 - 0.8367) = 0.5 + 0.4(0.6466) = 0.5 + 0.2586 = 0.7586\text{ V}$. (2)

New $V_{ov} = 1.0 - 0.7586 = 0.2414$ V. $I_D = \tfrac12(2000\mu)(0.2414)^2 = (1000\mu)(0.05829) = 58.3\ \mu\text{A}$. (2)

Implication: in stacked (series) transistors, the upper device has $V_{SB}>0$, raising its $V_{th}$ and drastically reducing drive current (here $250\to58\ \mu$A, a $4.3\times$ drop) — degrades speed; designers must upsize $W$ or minimise body bias. (1)

(c) Subthreshold swing (5 marks)

$I_D \propto e^{(V_{GS}-V_{th})/(nV_T)}$ (for $V_{DS}\gg V_T$). Swing $S = \left(\frac{d(\log_{10}I_D)}{dV_{GS}}\right)^{-1}$. $\log_{10}I_D = \frac{(V_{GS}-V_{th})}{nV_T\ln10}+$const, so $S = n V_T \ln 10.$ (2) At 300 K: $S = 1.5 \times 25.85\text{ mV}\times 2.3026 = 89.3\text{ mV/decade}$. (2)

Physical floor: even for ideal $n=1$ (perfect gate coupling, no depletion-cap loading), $S_{min}=V_T\ln10 = 59.5\approx 60$ mV/dec at 300 K. It stems from the Boltzmann tail of the carrier energy distribution — thermionic injection over the barrier means $I_D$ can rise at most $e$-fold per $V_T$. Sub-60 requires non-thermionic mechanisms (tunnel FETs, negative-capacitance). (1)

(d) Code (5 marks)

import numpy as np
kT_q = 0.02585           # V
n = 1.5
Vth = 0.5
kp = 200e-6; WL = 10     # µCox * W/L
I0 = 1e-9                # subthreshold prefactor (assumed)
 
Vgs = np.linspace(0.0, 1.2, 500)
Id = np.where(
    Vgs < Vth,
    I0*np.exp((Vgs-Vth)/(n*kT_q)),            # subthreshold
    0.5*kp*WL*(np.maximum(Vgs-Vth,0))**2 + I0  # above-threshold (continuity offset)
)
 
# extract S from subthreshold region
mask = (Vgs > 0.1) & (Vgs < Vth-0.05)
slope = np.polyfit(Vgs[mask], np.log10(Id[mask]), 1)[0]  # decades per volt
S = 1000/slope    # mV/decade
print("S =", S, "mV/dec")   # ~89.3
 
# import matplotlib.pyplot as plt
# plt.semilogy(Vgs, Id); plt.xlabel('Vgs'); plt.ylabel('Id')

Stitching: the two regions are joined near $V_{GS}=V_{th}$; slope extraction uses only the exponential region so the fit yields $S\approx89.3$ mV/dec. Marks: subthreshold model (2), above-threshold (1), slope→S extraction (2).