BJT biasing: the Q-point and the β shield
Введено| English | Значение |
|---|---|
| bipolar junction transistor | A three-layer transistor in which a small base current controls a larger collector current. |
| active region | активный режим |
| bias point | точка смещения |
| negative feedback/ˈneɡətɪv ˈfiːdbæk/ | A control process in which a change triggers responses that oppose that change. |
Same part number, twice the gain
- Buy a hundred transistors of one part number and their current gain β can range from 100 to 200. Each one still has to work in the same circuit.
- An amplifier needs its transistor at a steady operating point. This lesson sets that point so that β hardly matters.
Three regions of the BJT
- The bipolar junction transistor 双极型晶体管 has three regions. Cut-off: no current. Active region 放大区: $I_{C} = \beta I_{B}$ with $V_{BE} \approx 0.7\,\text{V}$. Saturation: $V_{CE}$ below about 0.2 V, a closed switch.
- An amplifier lives in the active region. Its DC operating point, the bias point 工作点 or Q-point, is the pair $(I_{C}, V_{CE})$.
Match each BJT region to its use.
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
Amplifiers bias into the active region; digital switches swing between cut-off and saturation.
The BJT stays in the active region as long as
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Saturation starts at about $V_{CE} = 0.2$ V, with both junctions forward biased. Checking $V_{CE} > 0$ lets a saturated transistor pass as an amplifier.
Voltage-divider bias
- Two resistors $R_{1}$, $R_{2}$ set the base voltage; $R_{C}$ sets the collector voltage; $R_{E}$ in the emitter ties the current to the base voltage.
- Replace the divider by its Thévenin equivalent, $V_{B}$ behind $R_{B} = R_{1} \parallel R_{2}$. Then solve one base loop: $V_{B} = I_{B}R_{B} + 0.7 + (\beta + 1)I_{B}R_{E}$.
A bias divider has R1 = 47 kΩ to a 10 V supply and R2 = 10 kΩ to ground. What is V_B, in volts, ignoring base current?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
$10 \times \dfrac{10}{57} = 1.75$ V. Its Thévenin resistance is $47\,\text{k} \parallel 10\,\text{k} = 8.25\,\text{k}\Omega$.
Worked example: find the Q-point
- $V_{CC} = 12\,\text{V}$, $R_{1} = 68\,\text{k}\Omega$, $R_{2} = 12\,\text{k}\Omega$, $R_{C} = 2.2\,\text{k}\Omega$, $R_{E} = 1\,\text{k}\Omega$, $\beta = 100$. $V_{B} = 12 \times \dfrac{12}{80} = 1.8\,\text{V}$ and $R_{B} = 10.2\,\text{k}\Omega$.
- $I_{B} = \dfrac{1.8 - 0.7}{10.2\,\text{k} + 101\,\text{k}} = 9.9\,\mu\text{A}$, so $I_{C} = 0.99\,\text{mA}$ and $V_{CE} = 12 - 2.18 - 1.0 = 8.8\,\text{V}$: well inside the active region.
The 12 V divider-bias stage (68 k/12 k, R_C 2.2 k, R_E 1 k, β = 100) has I_C and V_CE of
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
$V_{B} = 1.8$ V, $R_{B} = 10.2$ kΩ; the exact base loop gives $I_{B} = 9.9\,\mu$A, $I_{C} = 0.99$ mA, $V_{CE} = 8.8$ V. 1.1 mA is the quick estimate that ignores base current.
The quick estimate, and its error
- Ignoring base current: $V_{E} = 1.8 - 0.7 = 1.1\,\text{V}$, so $I_{E} \approx 1.1\,\text{mA}$. That is 11% above the exact 0.99 mA.
- The error is base current loading the divider. A stiffer divider (smaller $R_{1}$ and $R_{2}$) shrinks it, at the cost of more supply current.
In a divider-bias stage V_B = 2.5 V and R_E = 1.8 kΩ. Ignoring base current, estimate I_E, in mA.
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
$V_{E} = 2.5 - 0.7 = 1.8$ V, so $I_{E} = \dfrac{1.8}{1.8\,\text{k}} = 1.0$ mA. The exact loop gives slightly less.
Worked example: double β
- Same circuit, $\beta = 200$: $I_{B} = \dfrac{1.1}{10.2\,\text{k} + 201\,\text{k}} = 5.2\,\mu\text{A}$, so $I_{C} = 1.04\,\text{mA}$ and $V_{CE} = 8.7\,\text{V}$. Doubling β moved $I_{C}$ by only 5%.
- $R_{E}$ is the reason. If $I_{C}$ rises, $V_{E}$ rises, which squeezes $V_{BE}$ and cuts the base current: negative feedback 负反馈 inside the bias. With $R_{E} = 0$, $I_{C}$ would double.
Doubling β from 100 to 200 in the worked-example stage moves I_C by
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
$I_{C}$ goes from 0.99 to 1.04 mA. The emitter resistor turns any current rise into a smaller $V_{BE}$, which throttles the base current.
Three common slips
- Checking $V_{CE} > 0$. Saturation starts at about 0.2 V; check $V_{CE} > 0.2\,\text{V}$, with margin.
- Mixing the estimate and the exact chain. Carry one set of numbers through, then check.
- Crediting the divider for β-independence. $R_{E}$ does that work; the divider just holds $V_{B}$.
Removing R_E (shorting the emitter to ground) makes the Q-point less sensitive to β.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Without $R_{E}$ there is no feedback, so $I_{C} = \beta I_{B}$ follows β directly; doubling β doubles $I_{C}$ and may push the transistor into saturation.
Recap
- Thévenin the divider, solve the base loop, then check the region.
- The quick estimate ignores base current; the exact loop includes it.
- $R_{E}$ makes the Q-point nearly independent of β through negative feedback.