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Electronics · Ch 2 — Transistor Biasing

DC Load Line, Q Point and Transistor Biasing Methods

2.1

DC Load Line, Q Point and Transistor Biasing Methods

DC load line

The DC load line is a straight line drawn on the transistor's output characteristics that represents every possible pair of output DC voltage VCEV_{CE} and output direct current ICI_C under no-signal conditions. For any given circuit the maximum output current (the saturation current) and the maximum output voltage (the cutoff voltage) are fixed by the circuit elements; these two extremes are plotted on the vertical ICI_C axis and the horizontal VCEV_{CE} axis respectively (figure 2.1.1). Because the values are fixed for the applied DC source, the straight line joining them is the DC load line, and its usefulness is that it contains every operating point the circuit can take.

Finding the end points of the load line

Consider a common-emitter circuit driven by two DC sources — VBBV_{BB} in the base loop and VCCV_{CC} in the collector loop (figure 2.1.2). Applying Kirchhoff's voltage law to the base loop gives VBB=IBRB+VBEV_{BB} = I_B R_B + V_{BE}, so the base current is

IB=VBB−VBERB.I_B = \frac{V_{BB} - V_{BE}}{R_B}.

Applying KVL to the collector loop gives VCC=VCE+ICRCV_{CC} = V_{CE} + I_C R_C, which rearranges to VCE=VCC−ICRCV_{CE} = V_{CC} - I_C R_C and to

IC=(−1RC)VCE+VCCRC.I_C = \left(\frac{-1}{R_C}\right)V_{CE} + \frac{V_{CC}}{R_C}.

This has the form y=mx+cy = mx + c, confirming that the load line is a straight line. Its two end points are found by taking each variable to zero:

  • setting IC=0I_C = 0 gives the X-axis (cutoff) point VCE(cutoff)=VCCV_{CE(cutoff)} = V_{CC};
  • setting VCE=0V_{CE} = 0 gives the Y-axis (saturation) point IC(sat)=VCC/RCI_{C(sat)} = V_{CC}/R_C.

Joining these two points gives the DC load line (figure 2.1.3).

The Q point (operating point / quiescent point)

When the input signal is zero, the intersection of the load line with the transistor's output characteristic curve fixes a single point called the Q point, also known as the operating point or quiescent point. It is fixed by the applied DC voltages. Using the fixed-bias circuit of figure 2.1.7 — with RC=1 kΩR_C = 1\,\text{k}\Omega, RB=10 kΩR_B = 10\,\text{k}\Omega, β=200\beta = 200 and VCC=10 VV_{CC} = 10\,\text{V} — the source VBBV_{BB} is varied to set the base current. For IB=25 μAI_B = 25\,\mu\text{A} the collector current is IC=βIB=5 mAI_C = \beta I_B = 5\,\text{mA} and VCE=VCC−ICRC=5 VV_{CE} = V_{CC} - I_C R_C = 5\,\text{V}, giving point Q1Q_1 at the middle of the load line. A larger base current (IB=40 μAI_B = 40\,\mu\text{A}, giving IC=8 mAI_C = 8\,\text{mA}, VCE=2 VV_{CE} = 2\,\text{V}) pushes point Q2Q_2 near saturation, while a smaller base current (IB=10 μAI_B = 10\,\mu\text{A}, giving IC=2 mAI_C = 2\,\text{mA}, VCE=8 VV_{CE} = 8\,\text{V}) pushes point Q3Q_3 near cutoff (figure 2.1.8). The operating point can therefore be placed anywhere along the load line depending on the application; the point carrying the zero-signal collector current ICQI_{CQ} and voltage VCEQV_{CEQ} has coordinates Q(VCEQ,ICQ)Q(V_{CEQ}, I_{CQ}).

Selecting the operating point on the load line

Where the operating point is placed depends on the job the transistor must do:

  • For switching, the operating points are chosen in the saturation region (a closed switch) and the cutoff region (an open switch).
  • For amplification, the operating point is placed at the centre of the load line so that the signal is amplified faithfully.

For a transistor to work as an amplifier it must operate in the active region, which requires the emitter-base junction to be forward biased and the collector-base junction to be reverse biased. Biasing is usually done with two power supplies, but it can also be arranged with a single supply. The most commonly used biasing methods are: (i) fixed bias (base bias), (ii) collector-to-base feedback bias, (iii) emitter feedback bias, and (iv) voltage divider bias (universal bias). Of these, voltage divider bias is the most widely used.

Voltage divider bias (universal bias)

Voltage divider bias is the most widely used arrangement in the initial stages of amplifier circuits (figure 2.1.9, drawn with an npn transistor). Biasing is set by a proper choice of R1R_1, R2R_2 and RER_E: resistors R1R_1 and R2R_2 form a potential divider across the supply VCCV_{CC}, and the voltage developed across R2R_2 forward-biases the emitter-base junction. The emitter resistor RER_E provides stability for the operating point.

How the emitter resistor stabilises the Q point: if the collector current ICI_C rises because a temperature change alters β\beta, the emitter current IEI_E also rises, so the drop across RER_E increases. This larger emitter-resistor drop reduces VBEV_{BE}, which in turn lowers the base current IBI_B and hence brings ICI_C back down — automatically opposing the original increase.

Circuit analysis: the divider voltage across R2R_2 is V2=VCCR2/(R1+R2)V_2 = V_{CC}R_2/(R_1 + R_2). Applying KVL to the base loop, V2=VBE+IEREV_2 = V_{BE} + I_E R_E, so IE=(V2−VBE)/REI_E = (V_2 - V_{BE})/R_E; since IE≈ICI_E \approx I_C, IC=(V2−VBE)/REI_C = (V_2 - V_{BE})/R_E. The collector loop gives VCE=VCC−IC(RC+RE)V_{CE} = V_{CC} - I_C(R_C + R_E). Because β\beta does not appear in these expressions, the Q point is essentially independent of β\beta, which gives excellent stabilisation. The one drawback is that RER_E also provides AC feedback, which reduces the voltage gain; this is corrected by connecting a bypass capacitor CEC_E in parallel with RER_E.

End points and Q point for voltage divider bias (approximate analysis)

Starting from VCE=VCC−IC(RC+RE)V_{CE} = V_{CC} - I_C(R_C + R_E), the load-line end points are:

  • IC=0I_C = 0 gives the X-axis point VCE=VCCV_{CE} = V_{CC};
  • VCE=0V_{CE} = 0 gives the Y-axis point IC=VCC/(RC+RE)I_C = V_{CC}/(R_C + R_E).

The straight line joining them is the DC load line for voltage divider bias (figure 2.1.10). For the Q point, the base loop gives V2=VBE+IEREV_2 = V_{BE} + I_E R_E, so with IE≈ICI_E \approx I_C,

ICQ=V2−VBERE,V2=VCCR2R1+R2,I_{CQ} = \frac{V_2 - V_{BE}}{R_E}, \qquad V_2 = \frac{V_{CC}R_2}{R_1 + R_2},

and the collector loop gives

VCEQ=VCC−ICQ(RC+RE).V_{CEQ} = V_{CC} - I_{CQ}(R_C + R_E).

Together VCEQV_{CEQ} and ICQI_{CQ} define the operating point.

Advantages of voltage divider bias

Voltage divider bias offers three main advantages:

  1. The Q point does not shift, giving excellent stabilisation, because the stabilisation does not depend on β\beta.
  2. It is used in almost all amplifier circuits.
  3. It provides better amplification when used in amplifiers.

Leakage currents

Current conduction in a BJT is carried by both majority and minority charge carriers. The emitter region emits majority carriers that are collected by the collector to form the main collector current IC(MAJORITY)I_{C(MAJORITY)}; in addition, a small minority-carrier component flows because the collector-base junction is reverse biased — this is the collector-to-base leakage current ICBO(MINORITY)I_{CBO(MINORITY)}. The total collector current is therefore

IC(TOTAL)=IC(MAJORITY)+ICBO(MINORITY).I_{C(TOTAL)} = I_{C(MAJORITY)} + I_{CBO(MINORITY)}.

The leakage current ICBOI_{CBO} is very small compared with the majority-carrier current. In general, the flow of current due to minority carriers under reverse-bias conditions is called the leakage current. Two leakage currents are important:

(I) Collector-to-base leakage current ICBOI_{CBO}: with the emitter terminal open and the collector-base junction reverse biased (figure 2.1.14), the current that still flows in the collector is ICBOI_{CBO}. It depends on temperature (thermal generation of electron-hole pairs — it roughly doubles for every 10 °C rise in a silicon transistor) and on the reverse-bias voltage.

(II) Collector-to-emitter leakage current ICEOI_{CEO}: with the base terminal open and the collector reverse biased (figure 2.1.15), the collector current is ICEOI_{CEO}. It too depends on temperature and on the reverse voltage.

The two are related. Using IC=βIB+ICEOI_C = \beta I_B + I_{CEO}, IC=αIE+ICBOI_C = \alpha I_E + I_{CBO} and IE=IC+IBI_E = I_C + I_B, solving gives

ICEO=(1+β) ICBO.I_{CEO} = (1 + \beta)\,I_{CBO}.

Thermal runaway …

Definition 1Unbiased transistor

A bipolar junction transistor whose terminals are not connected to any voltage source. It carries no useful current and is never used in actual practice. Since no junction is forward or reverse biased, the device stays inactive — an external DC source is what places it in the active, saturat …

Definition 2Transistor biasing

The application of a DC voltage of the correct polarity and suitable magnitude across a transistor's terminals so that it operates in the desired region for amplification, oscillation or switching. Biasing fixes the DC operating conditions — IBI_B, ICI_C, VCEV_{CE} and VBEV_{BE} — without which the tran …

Definition 3DC load line

A straight line on the output characteristics giving every possible pair of no-signal output voltage VCEV_{CE} and output current ICI_C, drawn between the fixed saturation and cutoff end points. Its usefulness is that it contains every operating point the circuit can take, since both end points a …

Definition 4Q point (operating / quiescent point)

The point where the DC load line intersects the output characteristic curve when the signal is zero; fixed by the applied DC voltages and written Q(VCEQ,ICQ)Q(V_{CEQ}, I_{CQ}). Its coordinates carry the zero-signal values ICQI_{CQ} and VCEQV_{CEQ}; for faithful amplification it is placed at the centre of the …

Definition 5Voltage divider bias

A biasing method in which R1R_1 and R2R_2 form a potential divider across VCCV_{CC} and RER_E stabilises the operating point; also called universal bias and preferred because the Q point …

Definition 6Leakage current

The small current that flows because of minority charge carriers when a transistor junction is reverse biased. In a BJT the reverse-biased collector-base junction contributes ICBOI_{CBO} — very small next to the majority-carrier collector current, but it roughly doubles for every 10 °C rise in a sil …

Definition 7Thermal runaway

The self-destruction of a transistor caused by a cumulative rise in temperature and leakage current that drives the operating point into saturation until the junctions may burn out. The loop is cumulative: a larger ICI_C heats the collector junction, raising ICBOI_{CBO}, which raises ICI_C again — which is why the bia …

Definition 8Stability factor (S)

The ratio of the change in collector current to the change in reverse saturation current at constant β\beta and VBEV_{BE}; a larger value indicates greater thermal instability. Written S=ΔIC/ΔICOS = \Delta I_C/\Delta I_{CO}, it measures how much temperature-driven leakage shifts the collector current, so a lo …

Definition 9Heat sink

A device (usually a shaped copper conductor) attached to a transistor to absorb the heat generated inside it and radiate it to the surroundings, raising the effective power ra …

Formula 10Base current and load-line equation in two-source (fixed) bias

IB=VBB−VBERBI_B = \dfrac{V_{BB} - V_{BE}}{R_B} and VCE=VCC−ICRCV_{CE} = V_{CC} - I_C R_C, from KVL on the base and collector loops of figure 2.1.2. Rearranging gives IC=(−1/RC)VCE+VCC/RCI_C = (-1/R_C)V_{CE} + V_{CC}/R_C, the $ …

Formula 11Load-line end points for fixed bias

X-axis (cutoff), set IC=0I_C = 0: VCE(cutoff)=VCCV_{CE(cutoff)} = V_{CC}. Y-axis (saturation), set VCE=0V_{CE} = 0: IC(sat)=VCC/RCI_{C(sat)} = V_{CC}/R_C. Here VCCV_{CC} is the collector supply voltage (in volts) and RCR_C the collector resistor; joining these two fixed extremes on the ICI_C and $V …

Formula 12Fixed-bias Q point

IC=βIBI_C = \beta I_B and VCE=VCC−ICRCV_{CE} = V_{CC} - I_C R_C — the coordinates of the operating point in a fixed-bias circuit (figure 2.1.7). β\beta is the current gain and IBI_B the base current set by VBBV_{BB}; e.g. with β=200\beta = 200 and IB=25 μAI_B = 25\,\mu\text{A}, IC=5 mAI_C = 5\,\text{mA} and $V_{ …

Formula 13Divider voltage and collector current in voltage divider bias

V2=VCCR2R1+R2V_2 = \dfrac{V_{CC}R_2}{R_1 + R_2}; since IE≈ICI_E \approx I_C, IC=V2−VBEREI_C = \dfrac{V_2 - V_{BE}}{R_E}. V2V_2 is the divider voltage across R2R_2, VBEV_{BE} the base-emitter drop and RER_E the emitter resistor; because β\beta is absent, ICI_C b …

Formula 14Collector-to-emitter voltage and load-line end points for voltage divider bias

VCE=VCC−IC(RC+RE)V_{CE} = V_{CC} - I_C(R_C + R_E). End points: VCE=VCCV_{CE} = V_{CC} at IC=0I_C = 0, and IC=VCC/(RC+RE)I_C = V_{CC}/(R_C + R_E) at VCE=0V_{CE} = 0. Obtained from KVL on the collector loop, with RC+RER_C + R_E acting as the total DC load; joining the two end points gives the DC …

Formula 15Q-point coordinates for voltage divider bias

ICQ=V2−VBEREI_{CQ} = \dfrac{V_2 - V_{BE}}{R_E} and VCEQ=VCC−ICQ(RC+RE)V_{CEQ} = V_{CC} - I_{CQ}(R_C + R_E), with V2=VCCR2/(R1+R2)V_2 = V_{CC}R_2/(R_1 + R_2). Because β\beta is absent, the Q point is nearly independent of β\beta. Both come from KVL — the base loop gives ICQI_{CQ}, the collector loop gives VCEQV_{CEQ} — and tog …

Formula 16Total collector current and leakage-current relation

IC(TOTAL)=IC(MAJORITY)+ICBO(MINORITY)I_{C(TOTAL)} = I_{C(MAJORITY)} + I_{CBO(MINORITY)}. Using IC=βIB+ICEOI_C = \beta I_B + I_{CEO}, IC=αIE+ICBOI_C = \alpha I_E + I_{CBO} and IE=IC+IBI_E = I_C + I_B, the two leakage currents relate as …

Formula 17Stability factor

S=ΔICΔICOS = \dfrac{\Delta I_C}{\Delta I_{CO}} at constant β\beta and VBEV_{BE}; the higher SS, the more thermally unstable the circuit. ΔIC\Delta I_C is the change in collector current and ΔICO\Delta I_{CO} the change in reverse saturation current; SS quantifies how stron …

Figure 18Output characteristics of a transistor in CE mode with the DC load line joining the saturation and cutoff points across the active region.
Fig. 18 — Output characteristics of a transistor in CE mode with the DC load line joining the saturation and cutoff points across the active region.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Shows a family of ICI_C–VCEV_{CE} output curves for increasing base currents IB0I_{B0} up to IB4I_{B4}, with the DC load line running from the saturation point on the ICI_C axis down to the cutoff point on the VCEV_{CE} axis and the active region in between. Illustrates how the lo …

Figure 19Two-source common-emitter biasing circuit used to derive the DC load line.
Fig. 19 — Two-source common-emitter biasing circuit used to derive the DC load line.

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An npn transistor with base source VBBV_{BB} through RBR_B and collector source VCCV_{CC} through RCR_C, marking IBI_B, ICI_C, VBEV_{BE} and VCEV_{CE}. This is the circuit whose KVL equations give th …

Figure 20The DC load line plotted between its saturation and cutoff end points.
Fig. 20 — The DC load line plotted between its saturation and cutoff end points.

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A straight line, labelled DC LOAD, from IC(sat)I_{C(sat)} on the ICI_C axis to VCE(cutoff)=VCCV_{CE(cutoff)} = V_{CC} on the VCEV_{CE} axis (figure 2.1.3). Notice that the line's slope is −1/RC-1/R_C: every no-signal pair (VCE,IC)(V_{CE}, I_C) the circuit can take lies on this li …

Figure 21Worked-illustration circuit for finding the load-line end points, with a 2 kΩ collector resistor, 10 kΩ base resistor, 12 V collector supply and 4 V base supply.
Fig. 21 — Worked-illustration circuit for finding the load-line end points, with a 2 kΩ collector resistor, 10 kΩ base resistor, 12 V collector supply and 4 V base supply.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

A two-source CE circuit with numeric component values (RC=2 kΩR_C = 2\,\text{k}\Omega, RB=10 kΩR_B = 10\,\text{k}\Omega, VCC=12 VV_{CC} = 12\,\text{V}, VBB=4 VV_{BB} = 4\,\text{V}) used to work out VCE(cutoff)=12 VV_{CE(cutoff)} = 12\,\text{V} and $I_ …

Figure 22Load line for the illustration circuit, joining 6 mA on the current axis to 12 V on the voltage axis.
Fig. 22 — Load line for the illustration circuit, joining 6 mA on the current axis to 12 V on the voltage axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The straight load line for figure 2.1.4, from (0, 6 mA)(0,\ 6\,\text{mA}) to (12 V, 0)(12\,\text{V},\ 0) (figure 2.1.5). The intercepts follow from the component values: IC(sat)=VCC/RC=12 V/2 kΩ=6 mAI_{C(sat)} = V_{CC}/R_C = 12\,\text{V}/2\,\text{k}\Omega = 6\,\text{mA}, and $V_{CE(cutoff)} = V_{CC} = 12,\text …

Figure 23First comparison circuit for load-line slope, with a 1 kΩ collector resistor and a 10 V supply.
Fig. 23 — First comparison circuit for load-line slope, with a 1 kΩ collector resistor and a 10 V supply.

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A CE biasing circuit whose 1 kΩ collector resistor gives a saturation current of 10 mA10\,\text{mA}; used to compare load-line slopes (figure 2.1.6a). Compare it with the 5 kΩ circuit of figure 2.1.6b: the smaller collector resistor gives the larger saturation current …

Figure 24Second comparison circuit for load-line slope, with a 5 kΩ collector resistor and a 10 V supply.
Fig. 24 — Second comparison circuit for load-line slope, with a 5 kΩ collector resistor and a 10 V supply.

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The same form as figure 2.1.6a but with a 5 kΩ collector resistor, giving a saturation current of only 2 mA2\,\text{mA} (figure 2.1.6b). With the same 10 V supply, the larger RCR_C lowers IC(sat)=VCC/RCI_{C(sat)} = V_{CC}/R_C from 10 mA to 2 mA, so its load line is shallo …

Figure 25Two DC load lines compared on one graph, showing how a larger collector resistor lowers the saturation current.
Fig. 25 — Two DC load lines compared on one graph, showing how a larger collector resistor lowers the saturation current.

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Both load lines share the 10 V cutoff point; the 1 kΩ circuit reaches 10 mA and the 5 kΩ circuit only 2 mA on the ICI_C axis (figure 2.1.6c). Notice both lines meet the voltage axis at the same cutoff point VCE=VCCV_{CE} = V_{CC}; only the slope −1/RC-1/R_C differs, so a larger collector re …

Figure 26Fixed-bias circuit with a variable base supply, a 1 kΩ collector resistor, a 10 kΩ base resistor and \(\beta = 200\), used to locate the Q point.
Fig. 26 — Fixed-bias circuit with a variable base supply, a 1 kΩ collector resistor, a 10 kΩ base resistor and \(\beta = 200\), used to locate the Q point.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Varying VBBV_{BB} sets the base current so that different Q points can be calculated along the load line (figure 2.1.7). For IB=25 μAI_B = 25\,\mu\text{A} it gives IC=βIB=5 mAI_C = \beta I_B = 5\,\text{mA} and VCE=5 VV_{CE} = 5\,\text{V} — the mid-line point Q1Q_1; larger or smaller base cur …

Figure 27Output characteristics with the load line and three operating points Q1, Q2 and Q3 at different positions.
Fig. 27 — Output characteristics with the load line and three operating points Q1, Q2 and Q3 at different positions.

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Q1 sits at the middle of the load line, Q2 near saturation and Q3 near cutoff, showing how the choice of base current moves the operating point (figure 2.1.8). Q1Q_1 (VCE=5 VV_{CE} = 5\,\text{V}, IC=5 mAI_C = 5\,\text{mA}) suits faithful amplification; Q2Q_2 and Q3Q_3 show why a drifting bias risks distortion as the signal swings into saturation or cutoff. …

Figure 28Voltage divider (universal) bias circuit using an npn transistor with divider resistors R1 and R2 and an emitter resistor R_E.
Fig. 28 — Voltage divider (universal) bias circuit using an npn transistor with divider resistors R1 and R2 and an emitter resistor R_E.

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R1R_1 and R2R_2 set the base voltage across R2R_2 while RER_E stabilises the operating point; VCEV_{CE} and VBEV_{BE} are marked. This is the circuit analysed for the β\beta-i …

Figure 29DC load line for voltage divider bias, between the end points V_CC and V_CC/(R_C + R_E).
Fig. 29 — DC load line for voltage divider bias, between the end points V_CC and V_CC/(R_C + R_E).

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A straight DC load line from IC=VCC/(RC+RE)I_C = V_{CC}/(R_C + R_E) on the current axis to VCE=VCCV_{CE} = V_{CC} on the voltage axis (figure 2.1.10). Note that the saturation intercept uses the total DC load RC+RER_C + R_E, not RCR_C alone, so this line is shallower than a fixed-bia …

Figure 30Voltage divider bias circuit with numeric values — a 12 V supply, 10 kΩ and 1 kΩ divider resistors, a 2 kΩ collector resistor and a 1 kΩ emitter resistor — used in the worked problems.
Fig. 30 — Voltage divider bias circuit with numeric values — a 12 V supply, 10 kΩ and 1 kΩ divider resistors, a 2 kΩ collector resistor and a 1 kΩ emitter resistor — used in the worked problems.

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The circuit whose load-line end points and Q point are estimated in the chapter's worked problems (figure 2.1.11). Trace the divider action: V2=VCCR2/(R1+R2)V_2 = V_{CC}R_2/(R_1 + R_2) sets the base voltage, RER_E fixes IC≈(V2−VBE)/REI_C \approx (V_2 - V_{BE})/R_E, and RC+RER_C + R_E …

Figure 31Load line for the voltage divider problem, joining 4 mA on the current axis to 12 V on the voltage axis.
Fig. 31 — Load line for the voltage divider problem, joining 4 mA on the current axis to 12 V on the voltage axis.

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Load line for figure 2.1.11, from (0, 4 mA)(0,\ 4\,\text{mA}) to (12 V, 0)(12\,\text{V},\ 0) (figure 2.1.12). The saturation intercept uses the total DC load: IC=VCC/(RC+RE)=12 V/3 kΩ=4 mAI_C = V_{CC}/(R_C + R_E) = 12\,\text{V}/3\,\text{k}\Omega = 4\,\text{mA}, while the cutoff intercept …

Figure 32Load line for the second voltage divider problem, joining 2 mA on the current axis to 12 V on the voltage axis.
Fig. 32 — Load line for the second voltage divider problem, joining 2 mA on the current axis to 12 V on the voltage axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Load line for the problem with a 5 kΩ collector resistor and a 1 kΩ emitter resistor, from (0, 2 mA)(0,\ 2\,\text{mA}) to (12 V, 0)(12\,\text{V},\ 0) (figure 2.1.13). Here the total DC load is 5+1=6 kΩ5 + 1 = 6\,\text{k}\Omega, so IC(sat)=12 V/6 kΩ=2 mAI_{C(sat)} = 12\,\text{V}/6\,\text{k}\Omega = 2\,\text{mA}; the cutoff intercept re …

Figure 33Collector-to-base leakage current path with the emitter open and the collector-base junction reverse biased.
Fig. 33 — Collector-to-base leakage current path with the emitter open and the collector-base junction reverse biased.

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With IE=0I_E = 0 (emitter open), the collector still carries IC=ICBOI_C = I_{CBO} through the reverse-biased collector-base junction, illustrating the collector-to-base leakage current (figure 2.1.14). Note the current path from collector to base while the open emitter carries nothing; this ICBOI_{CBO} gro …

Figure 34Collector-to-emitter leakage current path with the base open and the collector reverse biased.
Fig. 34 — Collector-to-emitter leakage current path with the base open and the collector reverse biased.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

With IB=0I_B = 0 (base open), the collector carries IC=ICEOI_C = I_{CEO}, illustrating the collector-to-emitter leakage current (figure 2.1.15). Compare with figure 2.1.14: here the leakage crosses both junctions to the emitter, and since ICEO=(1+β)ICBOI_{CEO} = (1 + \beta)I_{CBO} it i …