Transistor Characteristics

Background

The transistor ranks as one of the greatest inventions of 20th century technology. It finds application in virtually all electronic devices from radios to computers. Integrated circuits typically contain millions of transistors, formed on a single tiny chip of silicon. Two of the basic uses of a transistor, which will be explored in this experiment, are as an amplifier and as a switch.

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Figure 1: a) Schematics of the pnp transistor. b) Circuit symbol (note the emitter, base and collector symbols) c) The common emitter circuit.

The php transistor, shown in Fig. 1a contains three distinct regions, a p-type ”emitter”, an n-type “base” and a p-type “collector”, which together form two pn junctions. In a typical amplifier circuit, voltages are supplied so that the emitter-base junction is forward-biased and the collector-base junction is reverse-biased. This means that V_{CE} > V_{BE}. Fig. 1c illustrates a “common emitter” circuit, so called because the emitter is common to the input circuit on the left and the output circuit on the right.

Consider first the forward-biased emitter-base junction. The doping of the emitter is made much heavier than that of the base so that positive holes from the emitter form almost all of the current, I_E, from emitter to base. The base, being lightly doped, does not have many electrons available for recombination with these holes to form neutral atoms. It is also very narrow (< 1 \mu m) making it easy for a large fraction, \alpha, of the holes to diffuse across to the collector-base junction where the junction voltage accelerates them into the collector region to form the collector current, I_C.

Thus,

(1)I_C = \alpha I_E

The remaining fraction, (1-\alpha), of holes leave the base through the external connection to form the base current, I_B, where

(2)I_B = (1-\alpha)I_E

The “current gain”, \beta, of the transistor is defined by

(3)\beta = I_C/I_B = \frac{\alpha}{(1-\alpha)}

For typical transistors, 0.9 \leq \alpha \leq 0.995 , giving values of 10 \leq \beta \leq 200. Thus we have a “current amplifier”, in that a small change in I_B will cause a large change in I_C. The “voltage gain”, A_V, is the ratio of the voltage drop, I_C R_C, across the output resistor, R_C, to the voltage, V_{BB}, of the input source:

(4)A_V = \frac{I_C R_C}{V_{BB}}

Applying the loop theorem to the input circuit in Fig 1c, and assuming I_E \approx I_C = \beta I_B, one obtains

(5)A_V = \frac{\beta R_C}{R_B + \beta r_B}

where r_B is the resistance of the emitter-base junction. By a suitable choice of resistors, an appreciable voltage amplification may be obtained.

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Figure 2: The AC amplifier circuit.

Fig. 2 shows how the transistor may be used as an AC amplifier to amplify a small signal from a signal generator (e.g. the output of the DAC, a digital to analog converter, of a CD or MP3 player). The two batteries in the circuit behave like large capacitors with impedances 1/\omega C \approx 0. Once again the voltage gain is given by equation (5). However, this is a simplified situation. In reality, the transistor junctions possess capacitance, and the corresponding reactances are frequency dependent. Thus we can expect Av to be a function of frequency.

Experiment

Preliminary Studies

Set up the circuit shown in Fig. 1c using the power supply outputs for the voltages V_{BB} and V_{CC}. Note the symbols e, b and c denoting the transistor connections. Use a 3000\Omega resistor for R_B and a 220\Omega resistor for R_C. Turn the supply outputs to zero then turn on the unit. Set one of the digital meters to the 20 V-DC range and connect it to measure V_{CC} (+ lead to ground on the transistor board). Adjust V_{CC} to approximately 15 V. Reconnect the meter to measure V_{CE}. This should also read 15 V, indicating I_C = 0. Connect the second meter to measure V_{BE} (also with the + lead to ground). Slowly increase V_{BB} up to 2 V and note V_{CE} decreasing, indicating an increasing I_C. Your amplifier is now working.

DC operation

Set V_{BB} to 0.75 V. Reconnect the meter to measure V_{BE} and calculate

(6)I_B = \frac{V_{BB}-V_{BE}}{R_B}.

Now reconnect the meter to measure V_{CC}. (The first meter should still be measuring V_{CE}.) Adjust V_{CC} to 1, 3, 5, .... 15 V and calculate corresponding values of

(7)I_C = \frac{V_{CC}-V_{CE}}{R_C}.

Repeat with V_{BB} = 0.90 and 1.05 V. Plot on a single graph I_C (y-axis) vs V_{CE}; (x-axis) for each value of I_B. The “operating region” of the transistor is where the curves level off. Determine \beta for the middle curve at V_{CE} = 3 V. Is \beta generally constant?

Transistor as a switch

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Figure 2: The transistor as a switch.

Reconnect the meters to measure V_{CC} and V_{BB}, set V_{CC} to 6 V and decrease V_{BB} to zero. Replace R_C with the light bulb. Slowly increase V_{BB} until the bulb is at maximum intensity. Connect the two-way switch as shown in Fig. 3. Note the effect of operating the switch. Measure V_{BE} to determine the very small current I_B that you are turning on and off to control the much larger current (~ 1A) through the light bulb. ,

AC Amplifier

Reconnect the circuit of Fig 1c with the meters to measure V_{CC} and V_{CE}. Increase V_{CC} to 15 V and turn V_{BB} down to zero. Increase V_{BB} until V_{CE} = 7.5 V. Connect the signal generator in series with V_{BB} as shown in Fig. 2 and adjust it to 100Hz. Connect the oscilloscope also as in Fig. 2 WITH THE BLACK LEADS TO GROUND ON THE TRANSISTOR BOARD FOR BOTH CONNECTIONS. Observe the input and output waveforms. Note that adjusting V_{BB} causes distortion of the output waveform. Can you explain this? From the ratio of peak-to-peak voltages, determine A_V for frequencies of 100 Hz, 1 kHz, 10 kHz and 20kHz. Would this amplifier be good as an audio-amplifier? Replace R_C with a 120\Omega resistor and note the effect on A_V. Is this an expected result?

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