Transistors - BJT & FET

Question 1

(a) In saturation the collector is nearly grounded: $I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_C}=\frac{12-0.2}{120}=\frac{11.8}{120}=0.0983\text{ A}\approx 98.3\text{ mA}.$ (Setup of loop 1 mark, value 1 mark.)

(b) Minimum base current: $I_{B(min)}=I_{C(sat)}/\beta_{min}=0.0983/50=1.967\text{ mA}$. (2) With overdrive factor 3: $I_B = 3\times1.967=5.90\text{ mA}$. (1) $R_B=\frac{V_{logic}-V_{BE}}{I_B}=\frac{5-0.7}{5.90\times10^{-3}}=\frac{4.3}{0.00590}=729\ \Omega.$ Choose nearest standard ≈ $680\ \Omega$ (increases overdrive slightly, safe). (2)

(c) Overdrive (forcing $I_B$ above the minimum) guarantees the transistor stays fully saturated despite the lowest possible $\beta$ (spread across parts/temperature), keeps $V_{CE(sat)}$ low (low power loss), and speeds turn-on. (3, one mark per valid reason.)

Question 2

(a) From $I_E=I_C+I_B$, divide by $I_C$: $\frac{I_E}{I_C}=1+\frac{I_B}{I_C}$. Since $\alpha=I_C/I_E\Rightarrow I_E/I_C=1/\alpha$, and $\beta=I_C/I_B\Rightarrow I_B/I_C=1/\beta$: $\frac{1}{\alpha}=1+\frac{1}{\beta}\Rightarrow \frac1\beta=\frac{1-\alpha}{\alpha}\Rightarrow \beta=\frac{\alpha}{1-\alpha}.$ (3) Invert: $\beta(1-\alpha)=\alpha\Rightarrow\beta=\alpha(1+\beta)\Rightarrow\alpha=\frac{\beta}{1+\beta}.$ (2)

(b) $\beta=\frac{0.995}{0.005}=199$. (1) At $\alpha=0.990$: $\beta=\frac{0.990}{0.010}=99$. (1) A 0.5% change in $\alpha$ roughly halves $\beta$ (199→99) — $\beta$ is extremely sensitive to $\alpha$ near unity. (1)

Question 3

(a) $I_D=\tfrac12 k_n\frac WL (V_{GS}-V_{th})^2$. Differentiate w.r.t. $V_{GS}$: $g_m=\frac{\partial I_D}{\partial V_{GS}}=k_n\frac WL (V_{GS}-V_{th}).$ (3) Since $(V_{GS}-V_{th})=\sqrt{2I_D/(k_n W/L)}$, $g_m=k_n\tfrac WL\sqrt{\frac{2I_D}{k_n W/L}}=\sqrt{2k_n\tfrac WL\,I_D}.$ (2)

(b) Overdrive $V_{ov}=2.0-0.8=1.2\text{ V}$. $I_D=\tfrac12(2\text{m})(1.2)^2=\tfrac12(2\times10^{-3})(1.44)=1.44\text{ mA}.$ (2) $g_m=k_n\tfrac WL V_{ov}=(2\text{m})(1.2)=2.4\text{ mA/V}.$ Check: $\sqrt{2(2\text{m})(1.44\text{m})}=\sqrt{5.76\times10^{-6}}=2.4\text{ mA/V}.$ ✓ (2)

(c) For fixed $I_D$, $g_m=\sqrt{2k_n(W/L)I_D}$ rises with $\sqrt{W/L}$; larger device means larger current at same $V_{ov}$ (or lower $V_{ov}$), giving more gain. Cost: larger gate capacitance / area → lower bandwidth and more power/chip area. (3)

Question 4

(a) Thevenin: $V_{TH}=V_{CC}\frac{R_2}{R_1+R_2}=10\cdot\frac{10}{57}=1.754\text{ V}$. (2) $R_{TH}=R_1\|R_2=\frac{47\cdot10}{57}=8.246\text{ k}\Omega$. $I_E\approx\frac{V_{TH}-V_{BE}}{R_E+R_{TH}/\beta}=\frac{1.754-0.7}{500+8246/150}=\frac{1.054}{500+54.97}=\frac{1.054}{554.97}=1.899\text{ mA}.$ (2) $I_C\approx I_E=1.90\text{ mA}$ (or $\approx1.90$ mA neglecting $R_{TH}/\beta$: $1.054/500=2.11$ mA acceptable if stated). (1)

(b) $g_m=I_C/V_T=1.90\text{m}/0.025=0.0760\text{ S}=76\text{ mS}$. (2) $A_v=-g_m R_C=-(0.076)(2000)=-152$. (2)

(c) Bypassing $R_E$ removes AC negative feedback so full $g_m R_C$ gain appears (large gain). But without AC feedback the gain now depends directly on $g_m$ (hence $I_C$), which varies with temperature and $\beta$; so the AC gain becomes far more sensitive to bias-point drift than the emitter-degenerated case. (3)

Question 5

(a) (1 each)

• Cutoff: $V_{GS}<V_{th}$ → $I_D\approx0$.
• Triode (linear): $V_{GS}>V_{th}$ and $V_{DS}<V_{GS}-V_{th}$.
• Saturation: $V_{GS}>V_{th}$ and $V_{DS}\ge V_{GS}-V_{th}$.

(b) (4: correct region logic 2, correct equations 2)

def id_nmos(vgs, vds, vth, k):   # k = kn*(W/L)
    vov = vgs - vth
    if vov <= 0:
        return 0.0                        # cutoff
    if vds < vov:
        return k * (vov*vds - 0.5*vds**2) # triode
    return 0.5 * k * vov**2               # saturation

(c) Subthreshold leakage is the small drain current that flows when $V_{GS}<V_{th}$ (weak inversion), varying exponentially with $V_{GS}$. Lowering $V_{th}$ shifts that exponential curve so more current flows at $V_{GS}=0$, so off-state leakage rises exponentially as $V_{th}$ is scaled down. (3)

Question 6

(a) $V_{th}=0.5+0.4\left(\sqrt{0.7+1.5}-\sqrt{0.7}\right)=0.5+0.4(\sqrt{2.2}-\sqrt{0.7}).$ $\sqrt{2.2}=1.4832$, $\sqrt{0.7}=0.8367$, difference $=0.6465$. $V_{th}=0.5+0.4(0.6465)=0.5+0.2586=0.7586\text{ V}\approx0.76\text{ V}.$ (4)

(b) A positive $V_{SB}$ reverse-biases the source–body junction, widening the depletion region under the channel. More depletion charge must be supported by the gate before inversion occurs, so a larger $V_{GS}$ (higher $V_{th}$) is needed to form the channel. (4)

[
  {"claim":"Q1a Ic_sat = 98.3 mA", "code":"Ic=(12-0.2)/120; result = abs(Ic-0.09833)<1e-4"},
  {"claim":"Q1b RB ~ 729 ohm at overdrive 3", "code":"Ib=3*((12-0.2)/120)/50; RB=(5-0.7)/Ib; result = abs(RB-729)<5"},
  {"claim":"Q2b beta=199 at alpha=0.995", "code":"a=Rational(995,1000); b=a/(1-a); result = b==199"},
  {"claim":"Q3b Id=1.44mA and gm=2.4mS", "code":"Id=Rational(1,2)*2e-3*(1.2)**2; gm=2e-3*1.2; result = abs(Id-1.44e-3)<1e-9 and abs(gm-2.4e-3)<1e-9"},
  {"claim":"Q3a gm equals sqrt(2k*Id)", "code":"k=2e-3; Id=1.44e-3; gm1=k*1.2; gm2=sqrt(2*k*Id); result = abs(float(gm1-gm2))<1e-9"},
  {"claim":"Q6a Vth ~ 0.7586 V", "code":"Vth=0.5+0.4*(sqrt(2.2)-sqrt(0.7)); result = abs(float(Vth)-0.7586)<1e-3"}
]