Lab guide

BJT biasing: why a voltage-divider bias gets an emitter resistor

Working out the operating point by hand against 2N3904 PSpice results, and how far it moves when β changes.

To use a transistor as an amplifier you first park its operating point (I_C, V_CE) near the middle of the active region. These are PSpice results for a 2N3904 on a 6 V supply, set next to hand calculations to show in numbers why the emitter resistor is there.

1. Voltage-divider bias without an emitter resistor

R₁ = 120 kΩ and R_C = 1 kΩ, with only R₂ changed. Replacing the base side with its Thévenin equivalent gives V_TH = V_CC·R₂/(R₁+R₂), R_TH = R₁‖R₂ and I_B = (V_TH − V_BE)/R_TH.

R₂	V_TH	R_TH	I_B (hand)	I_C = β·I_B	I_C (PSpice)	V_CE	Mode
12 kΩ	0.545 V	10.9 kΩ	—	—	9.8 µA	5.99 V	nearly off
18 kΩ	0.783 V	15.7 kΩ	7.83 µA	1.1 mA	1.05 mA	4.95 V	active
27 kΩ	1.102 V	22 kΩ	18.7 µA	2.88 mA	2.86 mA	3.14 V	active
33 kΩ	1.294 V	25.9 kΩ	23.3 µA	3.64 mA	3.58 mA	2.42 V	active
100 kΩ	2.727 V	54.5 kΩ	—	—	5.73 mA	0.27 V	saturation

The hand calculation holds up: in the active rows β·I_B is within 4.4% of the PSpice I_C (using the simulated V_BE and β).
It is very sensitive to R₂: changing R₂ 1.8 times, from 18 kΩ to 33 kΩ, changed I_C 3.4 times. V_TH sits close to V_BE (about 0.7 V), so a small change in V_TH − V_BE is a large change in proportion.
Both ends leave the active region: with R₂ = 12 kΩ V_TH is only 0.55 V and the transistor is nearly off; with 100 kΩ V_CE falls to 0.27 V, saturation.

2. Too large an R_C saturates it

R_C	I_C (PSpice)	V_CE (PSpice)	V_CE if I_C stayed 2.86 mA
470 Ω	2.91 mA	4.63 V	4.66 V
1 kΩ	2.86 mA	3.14 V	3.14 V
1.8 kΩ	2.78 mA	1.00 V	0.85 V
2.7 kΩ	2.16 mA	0.18 V	−1.72 V

In the active region I_C barely depends on R_C; only V_CE = V_CC − I_CR_C falls. To keep I_C ≈ 2.86 mA here R_C must stay below (V_CC − 0.2 V)/I_C ≈ 2 kΩ; at 2.7 kΩ the formula would need V_CE = −1.72 V, which is impossible, so I_C dropped and the transistor saturated at V_CE = 0.18 V.

3. Adding an emitter resistor (self bias)

R₁ = 100 kΩ, R₂ = 33 kΩ and R_C = 1.2 kΩ, with R_E changed. V_TH = 1.489 V and R_TH = 24.8 kΩ. The full formula is I_B = (V_TH − V_BE)/(R_TH + (β+1)R_E); when (β+1)R_E ≫ R_TH it reduces to I_C ≈ (V_TH − V_BE)/R_E and β drops out.

R_E	(β+1)R_E / R_TH	I_C (full formula)	I_C ≈ (V_TH−V_BE)/R_E	I_C (PSpice)	V_CE
220 Ω	1.3	2.08 mA	3.68 mA (+78%)	2.07 mA	3.05 V
470 Ω	2.7	1.26 mA	1.74 mA (+38%)	1.26 mA	3.89 V
1 kΩ	5.4	693 µA	829 µA (+18%)	700 µA	4.46 V
2.2 kΩ	10.8	350 µA	386 µA (+10%)	350 µA	4.80 V

The full formula matched PSpice within 1.0% in every row. The short estimate, though, is 78% high at R_E = 220 Ω, where (β+1)R_E/R_TH is only about 1.3, and still 10% high at 2.2 kΩ, where the ratio is about 11. Use it only as a quick guess when the ratio is 10 or more, and the full formula in a report.

4. How far does the operating point move with β?

β can differ two to three times between parts with the same number. Here are both circuits above with only β changed.

Circuit	β = 100	β = 150	β = 300	β 100 → 300
No emitter resistor (R₂ = 27 kΩ)	1.87 mA	2.8 mA	5.61 mA	×3.0
R_E = 1 kΩ (self bias)	659 µA	707 µA	763 µA	×1.16

Without an emitter resistor I_C is simply proportional to β; with β = 300, V_CE falls to 0.39 V, right at the edge of saturation. With R_E, tripling β raised I_C by only 16%. That is why amplifier bias circuits have an emitter resistor (bypassed with a capacitor when the signal gain must be kept).

5. In your lab report

Write down V_TH and R_TH first, then work out I_B and I_C by hand from your measured V_BE and β and compare them with the readings.
Judge the region from the sign of V_BC: negative means active, positive (with V_CE around 0.2 to 0.3 V) means saturation.

The BJT bias calculator does all of this from the resistor values. Get V_TH from the voltage divider calculator, resistor power from the Ohm's law calculator and switch base resistors from the transistor base resistor calculator.