Small-signal: the circuit, differentiated
Introduzido| English | Significado |
|---|---|
| small-signal model | small-signal model |
| transconductance | transconductance |
| emitter degeneration | emitter degeneration |
A whisper made loud
- A microphone gives a few millivolts riding on nothing; a speaker needs volts. The transistor in between is biased at a steady point, and the signal is a small wiggle around it.
- For small wiggles, a curved device law looks straight. This lesson replaces the transistor with that straight-line version and reads the gain off it.
Linearise at the Q-point
- The small-signal model 小信号模型 replaces each device by its slopes at the bias point. DC supplies become shorts to ground, and coupling capacitors become shorts at signal frequency.
- The key slope is the transconductance 跨导 $g_{m} = \dfrac{\partial i_{out}}{\partial v_{in}}$: how much output current a small input voltage produces.
In the small-signal model, a DC supply rail becomes signal ground.
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
A constant voltage has no wiggle, so for the signal it behaves as a short to ground. The same holds for large coupling and bypass capacitors at signal frequency.
The BJT's hybrid-π model
- Differentiate $I_{C} = I_{S}e^{V_{BE}/V_{T}}$: $g_{m} = \dfrac{I_{C}}{V_{T}}$. The base looks like $r_{\pi} = \dfrac{\beta}{g_{m}}$, and $r_{o}$ models the slight slope of the output curves.
- At $I_{C} = 0.99\,\text{mA}$ with $\beta = 100$: $g_{m} = \dfrac{0.99}{26} = 38.1\,\text{mS}$ and $r_{\pi} = \dfrac{100}{38.1\,\text{mS}} = 2.6\,\text{k}\Omega$.
At I_C = 0.99 mA and β = 100: g_m and r_π are
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
$g_{m} = \dfrac{I_{C}}{V_{T}} = \dfrac{0.99}{26} = 38.1$ mS; $r_{\pi} = \dfrac{\beta}{g_{m}} = 2.6$ kΩ. Both come from differentiating the exponential at the Q-point.
What is a BJT's g_m at I_C = 2.6 mA, in mS (V_T = 26 mV)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
$\dfrac{2.6\,\text{mA}}{26\,\text{mV}} = 100$ mS. Transconductance is proportional to bias current, whatever the transistor.
Worked example: the common-emitter gain
- With $R_{E}$ bypassed by a capacitor, the output current $g_{m}v_{\pi}$ flows through $R_{C} \parallel r_{o}$: $A_{v} = -g_{m}(R_{C} \parallel r_{o})$.
- $R_{C} = 2.2\,\text{k}\Omega$, $r_{o} = 50\,\text{k}\Omega$: $A_{v} = -38.1\,\text{mS} \times 2.11\,\text{k}\Omega = -80$. Without $r_{o}$ it would be $-84$. The minus sign means the stage inverts.
A common-emitter stage has g_m = 40 mS and R_C = 5 kΩ, with R_E bypassed and r_o very large. What is the gain?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
$A_{v} = -g_{m}R_{C} = -0.04 \times 5000 = -200$: inverting, with a large but β- and temperature-dependent gain.
Emitter degeneration
- Remove the bypass so $R_{E} = 1\,\text{k}\Omega$ carries signal too: $A_{v} = -\dfrac{g_{m}R_{C}}{1 + g_{m}R_{E}} = -\dfrac{84}{39.1} = -2.14$.
- When $g_{m}R_{E} \gg 1$ this tends to $-\dfrac{R_{C}}{R_{E}} = -2.2$, set by resistors alone. Emitter degeneration 发射极负反馈 trades gain for a stable, linear one.
Un-bypassing R_E = 1 kΩ drops the CE gain from −84 to
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
$-\dfrac{g_{m}R_{C}}{1 + g_{m}R_{E}} = -\dfrac{84}{39.1} = -2.14$, and $-\dfrac{R_{C}}{R_{E}} = -2.2$: within 2.4% because $g_{m}R_{E} = 38 \gg 1$.
The MOSFET's version
- Differentiate the square law: $g_{m} = \dfrac{2I_{D}}{V_{GS} - V_{t}}$. At 1.125 mA with 1.5 V of overdrive: $g_{m} = 1.5\,\text{mS}$, and a common-source stage with $R_{D} = 3\,\text{k}\Omega$ gives $-4.5$.
- A BJT at the same current has $g_{m} = 43.3\,\text{mS}$, 29 times more. The MOSFET's gift is its gate: no input current at all.
At the same 1.125 mA bias, the BJT's transconductance is about
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
$\dfrac{1.125}{26} = 43.3$ mS for the BJT; the square law gives $\dfrac{2 \times 1.125}{1.5} = 1.5$ mS for this MOSFET. The MOSFET pays in $g_{m}$ for its zero gate current.
Three common slips
- $g_{m}$ from the estimated current. Use the exact Q-point once you have it.
- $-\dfrac{R_{C}}{R_{E}}$ without checking. It needs $g_{m}R_{E} \gg 1$.
- Treating the model as a new circuit. It is the same circuit, differentiated at its bias point.
Match each small-signal parameter to its formula.
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
Each formula is a derivative of the device law, taken at the Q-point.
Recap
- BJT: $g_{m} = \dfrac{I_{C}}{V_{T}}$, $r_{\pi} = \dfrac{\beta}{g_{m}}$; MOSFET: $g_{m} = \dfrac{2I_{D}}{V_{GS} - V_{t}}$.
- Common-emitter gain $-g_{m}(R_{C} \parallel r_{o})$; degenerated, about $-\dfrac{R_{C}}{R_{E}}$.
- The BJT gives more $g_{m}$ per milliamp; the MOSFET draws no gate current.