Bipolar Junction Transistor (BJT)

What a BJT is

A junction transistor is a semiconductor device having two junctions and three terminals. The current in it is carried by both electrons and holes, which is why it is called a bipolar junction transistor. It has three doped regions forming a structure with two back to back p-n junctions.

Two types exist:

• n-p-n transistor: a p-type semiconductor (base) layer separates two layers of n-type semiconductor (emitter and collector). It is made by growing a thin layer of p-type semiconductor between two relatively thick layers of n-type semiconductor.

• p-n-p transistor: an n-type semiconductor (base) layer separates two layers of p-type semiconductor (emitter and collector), made by growing a thin n-type layer between two relatively thick p-type layers.

Symbols, described in words

Both symbols show a vertical bar (the base region) with the base lead coming in horizontally from the left, and two slanting leads leaving the bar on the right — the emitter (lower) and the collector (upper). The emitter lead carries an arrow, and the arrow shows the direction of the conventional current.

• n-p-n: the arrow points away from the base, i.e. outwards along the emitter lead.

• p-n-p: the arrow points from the emitter towards the base, i.e. inwards.

The three regions

Why the base is thin and lightly doped: if it were thick or heavily doped, most of the carriers injected from the emitter would recombine with the majority carriers in the base and form base current instead of reaching the collector. A thin, lightly doped base lets about $95\%$ of the emitter carriers diffuse across into the collector, leaving only about $5\%$ as base current — this is what makes transistor action possible.

Why the three regions are doped differently: heavy emitter doping provides plenty of carriers; light base doping and small base thickness minimise recombination in the base; intermediate collector doping with a large area allows the collector to collect nearly all the carriers and to dissipate the heat generated.

Depletion regions, currents, resistances

• Depletion regions are formed at the emitter-base junction and at the base-collector junction.

• The three currents are the emitter current $I_E$, the base current $I_B$ and the collector current $I_C$.

• The emitter-base junction has low resistance while the base-collector junction has high resistance.

Two-diode analogy

A transistor can be thought of as two junction diodes connected back to back — for an n-p-n, two diodes with their p-sides (anodes) joined at the base; for a p-n-p, two diodes with their n-sides (cathodes) joined at the base. This is only an analogy for remembering the structure; a real transistor cannot be made by wiring two diodes together, because the base must be thin enough for carriers to cross it.

Biasing

There are two p-n junctions in a transistor, the emitter-base (E-B) junction and the collector-base (C-B) junction, and they can be biased in different ways. In the most common method of biasing, used when the transistor is to work as an amplifier, the emitter-base junction is forward biased and the collector-base junction is reverse biased. This helps the easy flow of the majority carriers supplied by the emitter through the transistor.

Working of an n-p-n transistor

• Electrons are the majority carriers in the emitter of an n-p-n transistor, so the emitter current $I_E$ is due to electrons.

• The current through the E-B junction is large because that junction is forward biased. The current through the B-C junction is also large although that junction is reverse biased — the point of the whole explanation is why.

• At the instant the forward bias is applied to the E-B junction, the electrons in the emitter region have not yet entered the base region.

• When the biasing voltage $V_{BE}$ becomes greater than the barrier potential ($0.6$ to $0.7\ \text{V}$ for the commonly used silicon transistors), many electrons enter the base region and form the emitter current $I_E$.

• These electrons can flow in two directions. They can flow through the base circuit and constitute the base current $I_B$, or they can flow through the collector circuit and contribute to the collector current $I_C$.

• The base current is small — about $5\%$ of $I_E$ — because the base is thin and lightly doped compared with the emitter.

• Electrons injected from the emitter into the base diffuse into the collector-base depletion region because the base region is thin. When they enter that depletion region they are pushed into the collector region by the electric field there. The collector current $I_C$ then flows through the external circuit. $I_C$ is about $95\%$ of $I_E$.

• A p-n-p transistor works in exactly the same way, except that the majority carriers are now holes.

From this working,

$$I_E = I_B + I_C$$

and since $I_B$ is very small, $I_C \approx I_E$.

Transistor configurations

The three possible configurations of a transistor in a circuit are common emitter (CE), where the emitter is common to both input and output; common base (CB), where the base is common to both; and common collector (CC), where the collector is common to both.

16.4.1 The Common Emitter (CE) configuration

The CE configuration is studied in detail because it is the most commonly used. In the common emitter, or grounded emitter, configuration the input signal is applied between the base and the emitter, and the output is obtained between the collector and the emitter.

Since the electrical relationship between $I_B$, $I_C$ and $I_E$ is fixed by the physical construction of the transistor, any small change in the base current $I_B$ produces a much larger change in the collector current $I_C$. A small current in the base thus controls the current in the emitter-collector circuit. Typical values of $\beta_{DC}$ lie between $20$ and $200$ for general purpose transistors; if a transistor has $\beta_{DC} = 100$, then one electron flows from the base terminal for every $100$ electrons flowing between the emitter and collector terminals.

The CE amplifier produces the highest current gain and power gain of all three configurations, mainly because the input impedance is low (the input is a forward biased p-n junction) while the output impedance is high (the output is taken from a reverse biased p-n junction).

The current gains

Since the current flowing out of the transistor must equal the current flowing in,

$$I_E = I_C + I_B$$

The current gain, called the current amplification factor, is defined as

$$\beta_{DC} = \dfrac{I_C}{I_B}$$

and the ratio of the collector current to the emitter current is defined as

$$\alpha_{DC} = \dfrac{I_C}{I_E}$$

Combining these,

$$I_C = \alpha I_E = \beta I_B$$

Relation between $\alpha$ and $\beta$ — the derivation the board wants

Start from $I_E = I_B + I_C$ and divide throughout by $I_C$:

$$\dfrac{I_E}{I_C} = \dfrac{I_B}{I_C} + 1$$

$$\therefore\ \dfrac{1}{\alpha_{dc}} = \dfrac{1}{\beta_{dc}} + 1$$

$$\therefore\ \alpha_{DC} = \dfrac{\beta_{dc}}{1 + \beta_{dc}}$$

$$\therefore\ \beta_{DC} = \dfrac{\alpha_{dc}}{1 - \alpha_{dc}}$$

16.4.2 The CE characteristics

Circuit in words: an n-p-n transistor with its emitter earthed. In the input (base) circuit, a variable DC supply drives the base through a resistance, with a voltmeter reading $V_{BE}$ across the base-emitter and a microammeter reading $I_B$ in the base lead. In the output (collector) circuit, a second variable DC supply drives the collector through the load resistance $R_L$, with a voltmeter reading $V_{CE}$ across the collector-emitter and a milliammeter reading $I_C$ in the collector lead. The input is applied as $V_{in}$ between base and emitter and the output $V_{out}$ is taken between collector and emitter.

Input characteristic

The variation of the base current $I_B$ with the base-emitter voltage $V_{BE}$ is called the input characteristic. While studying the dependence of $I_B$ on $V_{BE}$, the collector-emitter voltage $V_{CE}$ is kept fixed.

Described in words: the curve starts at the origin and stays almost flat along the $V_{BE}$ axis — the current is very small until the barrier potential is overcome. Beyond the barrier potential the curve turns sharply upwards and $I_B$ rises rapidly with $V_{BE}$, exactly like the characteristic of a forward biased diode.

The input dynamic resistance $r_i$ of a transistor is the ratio of the change in the base-emitter voltage to the resulting change in the base current, at constant collector-emitter voltage:

$$r_i = \left( \dfrac{\Delta V_{BE}}{\Delta I_B} \right)_{V_{CE}}$$

Output characteristic

The variation of the collector current $I_C$ with the collector-emitter voltage $V_{CE}$ is called the output characteristic of a transistor. The base current $I_B$ is kept constant while this is done.

Described in words: for each fixed value of $I_B$ a separate curve is drawn. Each curve rises very steeply from the origin for small $V_{CE}$ and then flattens out, running almost parallel to the $V_{CE}$ axis. The board's two statements about this graph are: the collector current $I_C$ is independent of $V_{CE}$ as long as the collector-emitter junction is reverse biased, and the collector current $I_C$ is large for large values of the base current $I_B$ when $V_{CE}$ is constant. So the family of curves rises in steps, one above another, as $I_B$ is increased.

The output dynamic resistance $r_o$ of a transistor is the ratio of the change in the collector-emitter voltage to the change in the collector current, at constant base current:

$$r_o = \left( \dfrac{\Delta V_{CE}}{\Delta I_C} \right)_{I_B}$$

Because the flat part of the curve has a very small slope, $r_o$ is large.

16.4.3 Transistor as an amplifier

An amplifier is a device used for increasing the amplitude of an alternating signal (voltage, current or power). The standard circuit is an n-p-n transistor in common emitter configuration.

Circuit in words: the base is biased by a DC supply $V_{BB}$ through a base resistance $R_B$, and the small sinusoidal input signal $v_i$ is superimposed on this DC bias. The collector is supplied by $V_{CC}$ through the load resistance $R_L$. The output is taken across $R_L$, through a capacitor connected in the output circuit to block the DC component so that only the amplified AC signal appears at the output.

Working

With no signal applied, Kirchhoff's law for the output loop gives

$$V_{CC} = V_{CE} + I_C R_L$$

and for the input loop

$$V_{BB} = V_{BE} + I_B R_B$$

When the AC signal voltage $v_i$ is applied at the input, the emitter-base voltage changes, so the emitter current changes and hence the collector current changes. The base current $I_B$ and the collector current $I_C$ carry the sinusoidal variation superimposed on their DC values.

Because $V_{CC}$ is fixed, a change in $I_C$ forces a change in $V_{CE}$:

$$\Delta V_{CE} = - R_L\, \Delta I_C$$

This change in the collector voltage is the amplified output of the input variation, and the negative sign shows that it is in antiphase with the input.

Gains

The AC current gain is defined as

$$\beta_{AC} = \dfrac{i_c}{i_b} = \dfrac{\Delta I_C}{\Delta I_B}$$

and it is almost the same as $\beta_{DC}$ at normal operating voltages. The output voltage is

$$V_o = \Delta V_{CE} = - \beta_{AC} R_L\, \Delta I_B$$

The voltage gain is

$$A_v = \dfrac{v_o}{v_i} = \dfrac{\Delta V_{CE}}{r_i\, \Delta I_B} = - \dfrac{\beta_{AC} R_L}{r_i}$$

The negative sign indicates that the output voltage and the input voltage are out of phase (by $180^\circ$). Since there is also a current gain $\beta_{AC}$ in the common emitter configuration, the power gain is

$$A_p = \beta_{AC} \times A_v$$

taking the magnitude of $A_v$. A transistor can give a power gain because $\beta_{AC} \gg 1$.

An amplifier does not generate power. The additional power in the output comes from the DC supply $V_{CC}$; the transistor only controls how much of that power reaches the load, following the small input signal.

Method of biasing used for operating a transistor as an amplifier: emitter-base junction forward biased and collector-base junction reverse biased.

Standard results

• $I_E = I_B + I_C$; $I_B \approx 5\%$ of $I_E$; $I_C \approx 95\%$ of $I_E$; $I_C \approx I_E$. The largest current in a BJT is the emitter current.

• $\alpha_{DC} = \dfrac{I_C}{I_E}$, $\beta_{DC} = \dfrac{I_C}{I_B}$, $I_C = \alpha I_E = \beta I_B$.

• $\alpha_{DC} = \dfrac{\beta_{DC}}{1 + \beta_{DC}}$, $\beta_{DC} = \dfrac{\alpha_{DC}}{1 - \alpha_{DC}}$, typical $\beta_{DC}$ between $20$ and $200$.

• $r_i = \Delta V_{BE}/\Delta I_B$ at constant $V_{CE}$; $r_o = \Delta V_{CE}/\Delta I_C$ at constant $I_B$.

• $V_{CC} = V_{CE} + I_C R_L$; $V_{BB} = V_{BE} + I_B R_B$; $A_v = -\beta_{AC} R_L / r_i$; $A_p = \beta_{AC} A_v$.

• Barrier potential of a silicon transistor: $0.6$ to $0.7\ \text{V}$.

Worked numericals from the exercises

• Common base DC current gain $0.967$, emitter current $10\ \text{mA}$. $I_C = \alpha_{DC} I_E = 0.967 \times 10 = 9.67\ \text{mA}$, so $I_B = I_E - I_C = 10 - 9.67 = 0.33\ \text{mA}$.

• Emitter current $10\ \text{mA}$, collector current $9.8\ \text{mA}$. $I_B = I_E - I_C = 10 - 9.8 = 0.2\ \text{mA}$.

• Emitter current $6.28\ \text{mA}$, collector current $6.20\ \text{mA}$. $\alpha_{DC} = I_C/I_E = 6.20/6.28 = 0.987$.