Small and Large Signal Models: two different transistors, same device
For DC you use one model. For the wiggle riding on top, a completely different one.
Start here: two instruments on the same car
A car has a fuel gauge and a speedometer. Both describe the same car. Neither can answer the other's question — the fuel gauge cannot tell you how fast you are going, and it is not broken for failing to.
A transistor amplifier is doing two things at once, and each needs its own instrument. It sits at a DC operating point, and it wiggles a small signal around that point. The large-signal model describes where it sits; the small-signal model describes how it wiggles.
Why you cannot use one model for both
The DC question is "where is the Q-point?" — that involves the whole supply voltage, the bias resistors, β, and the 0.7 V junction drop. Big voltages, steady values.
The AC question is "if the input moves by 10 mV, how much does the output move?" — and for that, the supply voltage is irrelevant. It is not changing, so it contributes nothing to the change in output.
💡 That is why, in small-signal analysis, the supply rail is treated as ground. It looks alarming the first time — the +18 V rail drawn as 0 V — but it is correct. You are drawing a map of what changes, and the supply does not change. Anything steady is a fixed point, and a fixed point is as good as ground.
The rule for switching between them
🔀 Two circuits from one drawing
For DC (large signal)Capacitors become gaps — they block steady voltages. Work out the Q-point.
For AC (small signal)Capacitors become wires — they pass signals freely. Supply rails become ground. Work out the gain.
Notice the capacitors do opposite things in the two analyses. That is not a contradiction — it is precisely why they are there. A coupling capacitor is chosen to block DC and pass AC, so it separates the two jobs from each other, and the designer put it there for exactly that reason.
Why the small-signal parameters depend on the Q-point
The order matters: you must find the DC operating point first, because the AC behaviour depends on it.
A transistor's small-signal resistance is set by how much DC current is already flowing through it. Park it at a different Q-point and you get a different gain from the same circuit and the same components.
💡 So the two models are not independent alternatives — they are two stages of one calculation. DC first to find where the transistor sits, then AC to find how it responds there. A question that gives you bias resistors and asks for gain is asking you to do both, in that order.
A transistor amplifier does two things at once: it sits at a DC operating point, and it wiggles a small AC signal around that point. Trying to analyse both with one model is painful. So engineers split the problem — a large-signal model handles the DC bias, and a small-signal model handles the AC. The two analyses barely resemble each other, and knowing which to use is half the skill.
⚔️ Large signal vs small signal
Large signalFor DC biasing and switching. Uses the simple constants: V_BE ≈ 0.7 V, I_C = βI_B. Non-linear and approximate, but adequate because we only need the Q-point. Capacitors are treated as open circuits.
Small signalFor AC gain, input/output impedance. Linearises the device about the Q-point using h-parameters or the hybrid-π model. Valid only for signals small enough that the curve looks straight. DC sources become ground, capacitors become shorts.
The linkSmall-signal parameters depend on the Q-point: r_e = 26 mV/I_E. So you must do the DC analysis first to know what the AC parameters even are.
The hybrid-π (r_e) model
Key small-signal parameters, all from the Q-point:
r_e = V_T/I_E = 26 mV / I_E(mA) Ω (emitter resistance)
r_π = (β + 1) r_e ≈ β r_e Ω (base input resistance)
g_m = I_C/V_T = I_C(mA)/26 S (transconductance)
Also: g_m = 1/r_e (approximately, since I_C ≈ I_E)
The h-parameter model
Treats the transistor as a two-port black box:
v_be = h_ie·i_b + h_re·v_ce
i_c = h_fe·i_b + h_oe·v_ce
h_ie = input resistance (Ω) ≈ r_π
h_re = reverse voltage ratio (unitless, usually ignored)
h_fe = forward current gain (unitless) = β
h_oe = output conductance (S) → 1/h_oe = output R
Simplified (h_re ≈ 0, h_oe ≈ 0):
A_v = −h_fe·R_C/h_ie (the minus is the 180° inversion)
A_i = −h_fe
R_in = h_ie
The single most useful formula here is r_e = 26 mV / I_E. It links the DC bias directly to the AC gain. Bias the stage at 1 mA and r_e = 26 Ω. Bias it at 2 mA instead and r_e halves to 13 Ω — which doubles the gain. That's why changing the bias current changes the gain, and why the two analyses can't be separated entirely.
CE amplifier gain formulas
CE with emitter fully bypassed by a capacitor:
A_v = −R_C/r_e ← large gain, but unstable
R_in = R₁∥R₂∥(βr_e)
CE with unbypassed emitter resistor R_E:
A_v = −R_C/(r_e + R_E) ← lower gain, far more stable
R_in = R₁∥R₂∥(β(r_e + R_E))
If R_E >> r_e: A_v ≈ −R_C/R_E
— gain set by a resistor RATIO, independent of the
transistor entirely. This is negative feedback at work.
Worked numerical 1 — full small-signal analysis
Using the divider-bias circuit from the previous topic (V_CC = 18 V, R₁ = 39 kΩ, R₂ = 3.9 kΩ, R_C = 4 kΩ, R_E = 1.5 kΩ, β = 100), which gave I_E = 0.624 mA — find r_e, g_m, r_π, voltage gain (bypassed and unbypassed) and input impedance.
STEP 1 — small-signal parameters from the DC Q-point:
r_e = 26 mV/I_E = 26/0.624 = 41.7 Ω
r_π = βr_e = 100 × 41.7 = 4.17 kΩ
g_m = I_C/V_T = 0.624/26 = 0.024 S = 24 mS
STEP 2 — gain with R_E BYPASSED by a capacitor:
A_v = −R_C/r_e = −4000/41.7 = −95.9
(magnitude 96, inverted)
STEP 3 — gain with R_E UNBYPASSED:
A_v = −R_C/(r_e + R_E) = −4000/(41.7 + 1500)
= −4000/1541.7 = −2.59
Bypassing multiplies the gain by 37× — but the bypassed
gain depends on r_e, which depends on temperature and bias,
so it drifts. The unbypassed gain ≈ −R_C/R_E = −2.67 is set
by resistors alone and barely moves.
STEP 4 — input impedance (bypassed case):
R_in = R₁∥R₂∥r_π
= 39k ∥ 3.9k ∥ 4.17k
39k∥3.9k = 3.545 kΩ
3.545k∥4.17k = (3.545 × 4.17)/(3.545 + 4.17) = 1.916 kΩ
R_in ≈ 1.92 kΩ
STEP 5 — output impedance:
R_out ≈ R_C = 4 kΩ (ignoring h_oe)
Worked numerical 2 — the gain-stability trade-off
For the bypassed amplifier above, what happens to the gain if temperature rises so that I_E increases 20%?
I_E rises from 0.624 mA to 0.749 mA
New r_e = 26/0.749 = 34.7 Ω (was 41.7 Ω)
Bypassed gain:
A_v = −4000/34.7 = −115.3 (was −95.9)
→ gain changed by +20% ✗ unstable
Unbypassed gain:
A_v = −4000/(34.7 + 1500) = −2.606 (was −2.588)
→ gain changed by only +0.7% ✔ stable
The trade: 37× more gain, or 28× better stability. Real
designs often split R_E into two parts, bypassing only
one — getting moderate gain with usable stability.
Worked numerical 3 — h-parameters given directly
A CE amplifier has h_ie = 2 kΩ, h_fe = 120, h_oe = 25 µS, with R_C = 3 kΩ. Find A_v, A_i, R_in and power gain.
Output resistance of the transistor:
1/h_oe = 1/(25 × 10⁻⁶) = 40 kΩ
Effective load = R_C ∥ 40k = (3 × 40)/43 = 2.79 kΩ
Voltage gain:
A_v = −h_fe × R_L'/h_ie = −120 × 2790/2000
= −167.4
Current gain:
A_i = −h_fe = −120
Input resistance:
R_in = h_ie = 2 kΩ
Power gain:
A_p = |A_v × A_i| = 167.4 × 120 = 20 088
In dB: 10 log₁₀(20 088) = 43 dB
Note how including h_oe reduced the effective load from
3 kΩ to 2.79 kΩ — a 7% gain reduction. Ignoring h_oe
(the usual simplification) would have given A_v = −180.
💡 Exam angle: the reliable marks are (1) stating clearly that DC uses the large-signal model with capacitors open, while AC uses the small-signal model with capacitors shorted and DC supplies grounded; (2) computing r_e = 26 mV/I_E; and (3) A_v = −R_C/r_e with the minus sign explained as 180° phase inversion. If the question gives h-parameters, use them directly — don't try to convert to r_e.
Syllabus points
h-parameter / small-signal model
Large-signal model
Create a free account to tick topics off, take notes as you read, watch the video lessons and get a day-by-day study plan built around your exam date.