The Bipolar Junction Transistor: Switches and Amplifiers

1A switch with no moving parts

In December 1947, at Bell Laboratories in the United States, John Bardeen and Walter Brattain pressed two gold contacts onto a crystal of germanium and found that a small current into one contact controlled a larger current through the other. Within a few years William Shockley had designed a more robust version made of three layers of doped semiconductor, the bipolar junction transistor, and the three shared the 1956 Nobel Prize in Physics. A transistor does two jobs. It can act as a switch, turned fully on or fully off by a small current, which is the basis of every digital circuit. And it can act as an amplifier, copying a small varying signal as a larger one, which is how a microphone's tiny output can drive a loudspeaker.

2Structure

A bipolar junction transistor (BJT) is a single crystal with three regions, which form two p–n junctions back to back: n–p–n or p–n–p. The three regions are the emitter (heavily doped), the base (very thin, a few micrometres or less, and lightly doped) and the collector (moderately doped and the largest, because it dissipates most of the heat). "Bipolar" means that both electrons and holes take part.

Figure 4.1: npn and pnp transistors and their symbols (IEC 60617). The rest of this lesson uses the npn type; in a pnp transistor every polarity is reversed.

3How a transistor amplifies current

In normal (active) operation of an npn transistor, the base–emitter junction is forward biased (the base about 0.7 V above the emitter) and the base–collector junction is reverse biased (the collector several volts above the base).

Figure 4.2: Electron flow in an npn transistor in the active region.

1. The forward-biased base–emitter junction lets the emitter inject large numbers of electrons into the base.
2. The base is so thin and so lightly doped that only a small fraction of these electrons, typically 1% or less, meet holes and recombine there. The holes they fill are replaced through the base lead: that is the small base current $I_{B}I\_B$.
3. The remaining electrons diffuse to the base–collector junction, where its strong field sweeps them into the collector: the large collector current $I_{C}I\_C$.

By Kirchhoff's first law $I_E=I_B+I_{C}I\_E\; =\; I\_B\; +\; I\_C$. The ratio of collector current to base current is fixed by the construction, so a small change in $I_{B}I\_B$ controls a large change in $I_{C}I\_C$:

$$\beta =h_{FE}=\frac{I_C}{I_B}\text{typically }50\text{ to }500\backslash beta\; =\; h\_\{FE\}\; =\; \backslash frac\{I\_C\}\{I\_B\}\; \backslash qquad\backslash text\{typically\; \}\; 50\; \backslash text\{\; to\; \}\; 500$$

4Characteristics and the load line

With the emitter common to input and output (the common-emitter connection), the input characteristic ($I_{B}I\_B$ against $V_{BE}V\_\{BE\}$) is just that of a forward-biased diode: almost no base current until $V_{BE}\approx 0.6\mathrm{V}V\_\{BE\}\; \backslash approx\; 0.6\backslash ,\backslash mathrm\{V\}$. The output characteristics (Figure 4.3) show $I_{C}I\_C$ against $V_{CE}V\_\{CE\}$ for several fixed base currents. Above about 0.2 V, each curve is nearly flat: $I_C\approx \beta I_{B}I\_C\; \backslash approx\; \backslash beta\; I\_B$, whatever $V_{CE}V\_\{CE\}$ is. This is the active region.

Figure 4.3: Output characteristics of a transistor with $\beta \approx 150\backslash beta\; \backslash approx\; 150$, and the load line for $V_{CC}=9\mathrm{V}V\_\{CC\}\; =\; 9\backslash ,\backslash mathrm\{V\}$, $R_C=2.2\mathrm{k}\mathrm{\Omega }R\_C\; =\; 2.2\backslash ,\backslash mathrm\{k\backslash Omega\}$.

When a resistor $R_{C}R\_C$ connects the collector to a supply $V_{CC}V\_\{CC\}$, the transistor must also obey Kirchhoff's second law: $V_{CE}=V_{CC}-I_C{R}_{C}V\_\{CE\}\; =\; V\_\{CC\}\; -\; I\_CR\_C$. This straight line, the load line, runs from $V_{CE}=V_{CC}V\_\{CE\}\; =\; V\_\{CC\}$ (when $I_C=0I\_C\; =\; 0$) to $I_C=\frac{V_{CC}}{R_C}I\_C\; =\; \backslash frac\{V\_\{CC\}\}\{R\_C\}$ (when $V_{CE}=0V\_\{CE\}\; =\; 0$). The transistor's actual state is where the load line crosses the curve for its base current. Three regions matter:

• cut-off: , $I_B\approx 0I\_B\; \backslash approx\; 0$, $I_C\approx 0I\_C\; \backslash approx\; 0$, $V_{CE}\approx V_{CC}V\_\{CE\}\; \backslash approx\; V\_\{CC\}$: the switch is off;
• saturation: $I_{B}I\_B$ so large that $\beta I_B\backslash beta\; I\_B$ would exceed $\frac{V_{CC}}{R_C}\backslash frac\{V\_\{CC\}\}\{R\_C\}$; $V_{CE}V\_\{CE\}$ falls to about 0.2 V and $I_C\approx \frac{V_{CC}}{R_C}I\_C\; \backslash approx\; \backslash frac\{V\_\{CC\}\}\{R\_C\}$: the switch is on;
• active: in between, with $I_C=\beta I_{B}I\_C\; =\; \backslash beta\; I\_B$: used for amplifying.

5The transistor as a switch

Used as a switch, the transistor is driven hard between cut-off and saturation, never left in between, so that little power is wasted in it: when off, the current is zero; when on, the p.d. across it is only about 0.2 V. A small current from a sensor or from a logic chip can then switch a lamp, a motor or a relay that needs a much larger current. Because $\beta \backslash beta$ varies from one transistor to the next and with temperature, designers provide two to five times the minimum base current, to be sure of saturation.

Figure 4.4: A light-activated switch (night light). The LDR's symbol has arrows pointing in, for incoming light.

A light-dependent resistor (LDR) has a resistance that falls in light. In Figure 4.4, in daylight the LDR's resistance is low, so it takes only a small share of the 9 V and the base is below 0.6 V: the transistor is off. At dusk the LDR's resistance rises, the base p.d. rises above 0.7 V, and the transistor turns on and lights the LED. Swapping the LDR and $R_{1}R\_1$ gives a switch that turns on in light; using a thermistor gives a temperature alarm.

6The transistor as an amplifier

Figure 4.5: A simple common-emitter amplifier. The coupling capacitors pass the a.c. signal but block d.c., so the bias is not disturbed.

To amplify, the transistor is first biased into the middle of its active region: the resistor $R_{B}R\_B$ supplies a steady base current that sets the quiescent point Q, usually with $V_{CE}\approx \frac{V_{CC}}{2}V\_\{CE\}\; \backslash approx\; \backslash frac\{V\_\{CC\}\}\{2\}$ so that the output can swing equally up and down (Figure 4.3). A small signal is then added to the base. An increase $\mathrm{\Delta }{I}_B\backslash Delta\; I\_B$ in the base current increases the collector current by $\mathrm{\Delta }{I}_C=\beta \mathrm{\Delta }{I}_B\backslash Delta\; I\_C\; =\; \backslash beta\backslash ,\backslash Delta\; I\_B$, which increases the p.d. across $R_{C}R\_C$ and so decreases $V_{CE}V\_\{CE\}$:

$$\mathrm{\Delta }{V}_{\text{out}}=-R_C\mathrm{\Delta }{I}_C=-\beta R_C\mathrm{\Delta }{I}_B,A_v=\frac{\mathrm{\Delta }{V}_{\text{out}}}{\mathrm{\Delta }{V}_{\text{in}}}\backslash Delta\; V\_\{\backslash text\{out\}\}\; =\; -R\_C\backslash ,\backslash Delta\; I\_C\; =\; -\backslash beta\; R\_C\backslash ,\backslash Delta\; I\_B,\; \backslash qquad\; A\_v\; =\; \backslash frac\{\backslash Delta\; V\_\{\backslash text\{out\}\}\}\{\backslash Delta\; V\_\{\backslash text\{in\}\}\}$$

The minus sign means that the output is inverted: 180° out of phase with the input. If the signal is too large, or Q is badly placed, the transistor reaches saturation or cut-off on the peaks and the output is clipped: distortion. Simple bias from one resistor depends on $\beta \backslash beta$; practical amplifiers use a potential divider and an emitter resistor to make Q independent of $\beta \backslash beta$.

7Experiment 4.1: measuring β

Connect an npn transistor (for example a BC547) with a microammeter in the base lead, a 10 kΩ potentiometer and a 100 kΩ resistor to set the base current, a 1 kΩ resistor and a milliammeter in the collector lead, and a 9 V supply. Increase $I_{B}I\_B$ in steps and record $I_{C}I\_C$.

Illustrative readings. The graph of $I_{C}I\_C$ against $I_{B}I\_B$ is a straight line through the origin: gradient $=\frac{(7.7-1.6)\times {10}^{-3}}{(50-10)\times {10}^{-6}}=153=\; \backslash frac\{(7.7\; -\; 1.6)\; \backslash times\; 10^\{-3\}\}\{(50\; -\; 10)\; \backslash times\; 10^\{-6\}\}\; =\; 153$, so $\beta \approx 150\backslash beta\; \backslash approx\; 150$. At larger $I_{B}I\_B$ the line levels off as the transistor saturates: here $I_{C}I\_C$ cannot exceed $\frac{9\mathrm{V}}{1\mathrm{k}\mathrm{\Omega }}=9\mathrm{m}\mathrm{A}\backslash frac\{9\backslash ,\backslash mathrm\{V\}\}\{1\backslash ,\backslash mathrm\{k\backslash Omega\}\}\; =\; 9\backslash ,\backslash mathrm\{mA\}$.