The junction transistor

Three layers, two junctions

Narrated video. Transcript:

The junction transistor is no longer in the current NCERT syllabus, but it is still asked in competitive exams, and it is in every phone and computer you use. So let us see how it works.

It was invented in 1947 at Bell Labs by Bardeen, Brattain and Shockley. A transistor is a sandwich of three doped regions, so it has two p-n junctions back to back.

In an n-p-n transistor, a thin p-type layer sits between two n-type regions. In a p-n-p transistor, a thin n-type layer sits between two p-type regions.

The three regions have names. The emitter is heavily doped, because it must supply a lot of charge carriers. It is of moderate size.

The base, in the middle, is very thin and lightly doped. The collector is moderately doped and the largest of the three, because it collects most of the carriers and must get rid of the heat.

In the circuit symbol, the arrow is on the emitter. It shows the direction of conventional current: pointing out for n-p-n, and pointing in for p-n-p.

To use the transistor as an amplifier, we forward bias the emitter-base junction and reverse bias the base-collector junction. This is called the active state.

Watch the electrons in an n-p-n transistor. The forward bias pushes a huge number of them from the emitter into the base. The base is so thin and has so few holes that only a few, about 1 to 5 percent, recombine there. These make the small base current I B.

The rest are swept across into the collector by the reverse-biased junction. They make the large collector current I C. So the emitter current splits: I E equals I B plus I C.

For example, if I E is 10 milliampere and I B is 0.2 milliampere, then I C is 9.8 milliampere. Both junctions reverse biased gives cut-off, with almost no current. Both forward biased gives saturation. The active state lies in between.

Characteristics in the CE configuration

Narrated video. Transcript:

A transistor has three terminals, so one terminal is shared between the input and the output. Most often it is the emitter. This is the common emitter, or CE, configuration.

The input is between base and emitter: V B E and I B. The output is between collector and emitter: V C E and I C. We vary each and measure the currents.

First, the input characteristic: I B against V B E, with V C E held fixed, say at 10 volt. Like a forward-biased diode, almost no current flows until about 0.7 volt for silicon. Then I B rises steeply.

Its slope gives the input resistance, r i equals ΔV B E over ΔI B, at constant V C E. If V B E goes from 0.70 to 0.75 volt and I B goes from 40 to 60 microampere, r i equals 0.05 volt over 20 microampere, which is 2.5 kilo-ohm.

Next, the output characteristics: I C against V C E, one curve for each fixed value of I B. Each curve rises quickly, then becomes almost flat. In the flat part, I C is controlled by I B, not by V C E.

The output resistance is r o equals ΔV C E over ΔI C, at constant I B. On the 30 microampere curve, raising V C E from 6 to 10 volt raises I C only from 3.00 to 3.05 milliampere. So r o equals 4 volt over 0.05 milliampere, which is 80 kilo-ohm.

Now hold V C E at 10 volt and step I B from 20 to 30 microampere. I C rises from 2.0 to 3.0 milliampere. The a.c. current gain is β a c equals ΔI C over ΔI B, so 1 milliampere over 10 microampere: β a c is 100.

The d.c. current gain is just the ratio, β d c equals I C over I B. At 30 microampere and 3.0 milliampere, that is also 100. In the flat region the two are nearly equal.

So a tiny base current, tens of microampere, controls a collector current a hundred times larger. That is the whole secret of the transistor.

A switch and an amplifier

Narrated video. Transcript:

Here is the CE circuit. The input V i drives the base through R B. The collector is fed from V C C through R C, and the output V o is taken between collector and emitter.

Two loop equations describe it. On the input side, V i equals I B R B plus V B E. On the output side, V C C equals I C R C plus V C E, so V o equals V C C minus I C R C.

Now slowly raise V i and plot V o. While V i is below about 0.7 volt, there is no base current, so no collector current. V o equals V C C. This is cut-off.

Above 0.7 volt, I B flows and I C equals β I B. As V i rises, I C rises and V o falls along a straight line. This is the active region.

Raise V i further and I C cannot grow any more: it is limited to about V C C over R C. V o drops to nearly zero. This is saturation.

As a switch, we use only the two ends. Low input gives high output: switch OFF. High input gives low output: switch ON. This is exactly a NOT gate.

As an amplifier, we fix the operating point Q in the middle of the active region, then add a small a.c. signal v i to the input. The base current wobbles by ΔI B, and the collector current wobbles by β a c times as much.

The output changes by ΔV C E equals minus R C ΔI C. So the voltage gain is A v equals minus β a c R C over r, where r is R B plus r i. The minus sign means the output is 180 degrees out of phase with the input.

Take β a c equal to 100, R C equal to 4 kilo-ohm and r equal to 10 kilo-ohm. A v equals minus 100 times 4 over 10, which is minus 40. A 20 millivolt input becomes a 0.8 volt output, upside down.

The power gain is the current gain times the voltage gain: 100 times 40, which is 4000. The extra energy is not free: it comes from the battery V C C.

Try it: the CE transistor

Interactive simulation. Things to try:

• Slide V_i from 0 to 5 V and watch Q move along the load line: cut-off, then active, then saturation. The transfer graph traces V_o against V_i.
• In the active region, check that I_C / I_B equals β and that I_E = I_B + I_C. In saturation, I_C / I_B drops below β.
• Tick “Small a.c. input”. The output v_CE is a bigger, upside-down copy of the input: the gain is about −β R_C / R_B.
• Raise the a.c. peak until the output clips, or move Q towards cut-off or saturation. A good amplifier keeps Q in the middle of the load line.

Audio amplifier

In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. The current gain β is 100 and the base resistance is 1 kΩ. Find the input signal voltage and the base current.

1. v_i = 10 mV, i_B = 10 μA
2. v_i = 20 mV, i_B = 10 μA
3. v_i = 1 V, i_B = 1 mA
4. v_i = 0.1 V, i_B = 100 μA

Switching it on

A silicon transistor switch has V_CC = 5 V, R_C = 1 kΩ, R_B = 100 kΩ and β = 250. Take V_BE = 0.7 V and V_CE ≈ 0 in saturation. What is the smallest input V_i that drives it into saturation (switch ON)?

1. 0.7 V
2. 2.0 V
3. 2.7 V
4. 5.7 V

Summary

• Currents: I_E = I_B + I_C
• a.c. current gain (CE): β_ac = ΔI_C / ΔI_B
• Output loop: V_o = V_CC − I_C R_C
• Voltage gain: A_v = −β_ac R_C / r

• Extra topic, beyond the current NCERT syllabus. A junction transistor (n-p-n or p-n-p) has a heavily doped emitter, a thin, lightly doped base and a moderately doped, large collector.
• In the active state the emitter-base junction is forward biased and the collector-base junction reverse biased; most carriers cross the thin base, so I_C ≫ I_B.
• Cut-off (both junctions reverse biased) and saturation (both forward biased) let the transistor act as a switch: low input gives high output, like a NOT gate.
• From the CE characteristics: r_i = ΔV_BE/ΔI_B (V_CE fixed), r_o = ΔV_CE/ΔI_C (I_B fixed), β_dc = I_C/I_B. In the amplifier r = R_B + r_i, and the power gain is A_p = β_ac × A_v.
• With Q in the middle of the active region it is an amplifier; the CE output is 180° out of phase with the input, and the extra power comes from V_CC.