Small-signal: the circuit, differentiated
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| small-signal | small-signal: tín hiệu nhỏ |
Small-signal: the circuit, differentiated
- Differentiate the device law at the Q-point: BJT $g_m = I_C/V_T$, $r_\pi = \beta/g_m$; MOSFET $g_m = 2I_D/V_{ov}$. DC sources become shorts, coupling capacitors shorts at signal frequency — the hybrid-π model is one linear circuit for every small-signal question.
- Bypassed emitter: CE gain $-g_m R_C$. Degenerated: $-g_m R_C/(1 + g_m R_E) \to -R_C/R_E$ once $g_m R_E \gg 1$ — the validity condition made visible.
At $I_C = 0.99$ mA: $g_m = 38.1$ mS, $r_\pi = 2.6$ kΩ; gain with $r_o = 50$ kΩ: $-38.1\mathrm{m} \times 2.11\mathrm{k} = -80$ ($r_o \to \infty$ gives $-84$). Lift the bypass: $-84/(1 + 38.1) = -2.14$ against $-R_C/R_E = -2.2$ — 2.7% apart because $g_m R_E = 38 \gg 1$. MOSFET at the same current scale: $g_m = 1.5$ mS, CS gain $-4.5$; the BJT's 43.3 mS at 1.125 mA is 29× — the fundamental trade.
{exercise:uni-ee5x-smallsignal}
At IC = 0.99 mA, beta 100: gm and r-pi are
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
gm = IC/VT = 0.99m/26m = 38.1 mS; r_pi = beta/gm = 100/38.1m = 2.6 kOhm - both from differentiating at the Q-point.
Un-bypassing RE = 1 kOhm drops the CE gain from -84 to
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
-gm RC/(1 + gm RE) = -84/39.1 = -2.14, and -RC/RE = -2.2: within 2.7% because gm RE = 38 >> 1 - the validity condition of the textbook form.
Per milliamp of bias current, the BJT transconductance is about
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
BJT gm = I/VT = 43.3 mS at 1.125 mA; the MOSFET square law gives 1.5 mS. Transconductance for input resistance - the fundamental trade.