Biasing

YOU WILL BE ABLE TO

Explain why the bias point must not depend on β.
Design a voltage-divider bias with emitter degeneration.
Choose a divider stiffness relative to base current.
Verify the operating point lands in the active region.

ME-2 L4 biased a MOSFET and found the source resistor to be the hero: negative feedback that made the operating point a property of the resistors rather than of the transistor. The BJT needs the same medicine more urgently, because β varies more wildly than any MOSFET parameter. The standard circuit is the same shape, with an emitter resistor in place of the source resistor.

Voltage-divider bias with emitter degeneration: R_1 and R_2 set the base voltage, R_E sets the current, R_C sets the output swing.

Read it in design order. The divider sets $V_B$ . Subtract the $0.7$ V base-emitter drop to get $V_E$ , and $R_E$ then fixes the emitter current — which, since $\alpha$ is within a percent or two of one, is essentially the collector current. Notice what did NOT appear anywhere in that chain: β. The current is set by a voltage divided by a resistance, and the transistor merely follows.

V_B = V_{CC}\frac{R_2}{R_1 + R_2}

V_E = V_B - 0.7\ \text{V}

I_C \approx I_E = \frac{V_E}{R_E}

V_C = V_{CC} - I_C R_C

the design chain

WORKED EXAMPLE

A bias design, start to finish

Target $I_C = 2$ mA from $V_{CC} = 12$ V, with a transistor whose β is nominally 100 but guaranteed only 100–300.
Choose $V_E = 2$ V — enough for the feedback to be effective, cheap in headroom. Then $R_E = 2\ \text{V}/2\ \text{mA} = 1$ kΩ.
Base voltage: $V_B = V_E + 0.7 = 2.7$ V.
Divider stiffness: $I_B = I_C/\beta = 2\ \text{mA}/100 = 20$ µA, so aim for a bleed current of $200$ µA. Then $R_1 + R_2 = 12\ \text{V}/200\ \mu\text{A} = 60$ kΩ, and $R_2 = 60\ \text{k}\Omega \times (2.7/12) = 13.5$ kΩ.
Stock values: $R_2 = 13$ kΩ, $R_1 = 47$ kΩ. Recheck: $V_B = 12 \times 13/60 = 2.6$ V, so $V_E = 1.9$ V and $I_C \approx 1.9$ mA — within 5% of target, using ordinary parts.
Choose $V_C \approx 7$ V for symmetric swing: $R_C = (12 - 6.9)/1.9\ \text{mA} \approx 2.7$ kΩ.
VERIFY the region: with $R_C = 2.7$ kΩ, $V_C = 12 - (1.9\ \text{mA})(2.7\ \text{k}\Omega) = 6.87$ V, so $V_{CE} = 6.87 - 1.9 \approx 5.0$ V. Far above the $0.2$ V saturation floor, and B-C is reverse biased. Active. ✓

R_1 = 47

kΩ,

R_2 = 13

kΩ,

R_E = 1

kΩ,

R_C = 2.7

kΩ, at

I_C \approx 1.9

WORKED EXAMPLE

Testing the β-independence claim

Take the design above and swap in a part with $\beta = 300$ instead of 100.
$I_B$ falls to about $6$ µA, so the divider is loaded even more lightly and $V_B$ rises by a few tens of millivolts at most.
$V_E$ , and therefore $I_C$ , barely move — a few percent.
Now imagine the same transistor swap with NO emitter resistor and the base fed directly: $I_C = \beta I_B$ would triple. The single resistor is the difference between a 3% shift and a 200% one.

A 3× change in β produces a few percent change in

I_C

PRACTICE

A bias circuit has $V_B = 3.7$ V and $R_E = 1.5$ kΩ. Find $V_E$ and the approximate collector current.

For a target $I_C = 1$ mA with $\beta = 150$ , what divider bleed current would you choose, and what total divider resistance does that imply from a $9$ V supply?

Continuing problem 2: with $V_E$ chosen as $1.5$ V, find $R_E$ , $V_B$ , and $R_2$ as a fraction of the total.

A design biases the base with a stiff divider but omits $R_E$ entirely, tying the emitter to ground. Explain in two or three sentences what goes wrong.