Transistor Static Characteristics

A number of current-voltage plots are available for the analysis and design of a transistor circuit. The static characteristic curves give information on the value of current flowing into or out of one terminal, for either a given current flowing into or out of another terminal, or a given voltage applied between two terminals. Four sets of characteristics can be plotted for each configuration: (a) the input characteristic, (b) the transfer characteristic, (c) the output charac­teristic and (d) the mutual characteristic. In this book, however, the characteristics for the common-base and the common-collector circuits will not be discussed. Two versions of set (c) are available.

The method of determining the static characteristics of a transistor is to connect the transistor into a suitable circuit and then to vary the appropriate currents and/or voltages in a number of discrete steps, noting the corresponding values of the other currents at each step. Fig. 10. shows a suitable circuit for the measurement of the static characteristics of an n-p-n transistor in the common-emitter configuration.

The collector and base currents are shown as flowing into the transistor and are therefore, by definition, positive; the emitter current is shown flowing out of the transistor and must be taken as negative. If the static characteristics of a p-n-p transistor were to be measured, the polarities of the two batteries would need to be reversed.

(a) Common-emitter Input Characteristic

The input characteristic shows the way in which the base current varies with change in the base-emitter voltage, the collector-emitter voltage remaining constant. The input characteristic is measured by main­taining the collector-emitter voltage constant at a convenient value and increasing the base-emitter voltage in a number of discrete steps, noting the base current at each step. This procedure is then repeated for a different but constant value of collector-emitter voltage V_CE, since any change in this voltage has an effect on the input charac­teristic. A typical input characteristic is shown in Fig. 11. The input resistance, h_ie , of the transistor for a given base-emitter voltage V_BE is given by the reciprocal of the slope of the curve at that point. For example, consider the input resistance of the transistor at the point

V_BE = 0.7 V.

h_ie = δV_BE/δI_B = V_be/I_b (V_CE constant) (1.13)

= 0.02/(22 X 10^-6) = 909 Ω

The corresponding p-n-p characteristics would have negative values of I_B, V_BE and V_CE.

Since the input characteristic is non-linear, the a.c. input resistance of a transistor will vary with the base current. For example,

when I_B = 2 μA h_ie – 6.7 kΩ

Some manufacturers’ data sheets give a graph that shows how h_ie varies with the collector current.

The d.c. input resistance h_IE is given by the ratio V_BE/I_B. At the point of measurement

h_IE = 0.7/(25xl0^-6) = 28 kΩ

It is evident that a considerable difference may exist between the values of h_IE and h_ie. If the point of measurement is reduced below about V_BE = 0.68 V, the slope of the input characteristic is much smaller and hence the a.c. input resistance h_IE will be larger so that any difference between h_IE and h_ie will be much less. For base currents of the order of a few tens of microamps , h_ie is typically about 3 kΩ.

(b) Common-emitter Transfer Characteristic

The transfer characteristic shows how the collector current changes with changes in the base current, the collector-emitter voltage being held at a constant value. For this measurement the collector-emitter voltage is kept constant and the base current is increased in a number of discrete steps and at each step the collector current is noted. Finally a plot is made of collector current against base current. Since the transfer characteristic is not independent of the value of the collector-emitter voltage, the procedure can be repeated for a number of different collector-emitter voltages to give a family of curves; Fig. 12. shows a typical n-p-n transistor transfer characteristic.

The slope of the transfer characteristic gives the short-circuit current gain h_fe of the transistor.

h_fe = δI_C/δI_B = I_C/I_b = (l X 10^-3)/(4Xl0^-6) = 250

(c) Common-emitter Output Characteristic

The output characteristic illustrates the changes that occur in collector current with changes in collector-emitter voltage, for constant values of base current. Alternatively, the collector current can be plotted against collector-emitter voltage for constant values of base-emitter voltage. The base current, or the base-emitter voltage, is set to a convenient value and is then maintained constant and the collector-emitter voltage is increased from zero in a number of discrete steps, the collector current being noted at each step. The collector-emitter voltage is then restored to zero and the base current, or the base-emitter voltage, is increased to another convenient value and the procedure repeated. In this way a family of curves (see Figs. 13 and 14) can be obtained. For the corresponding p-n-p characteristic the polarities of I_c, I_b and v_ce should be changed to negative.

The slope δI_C/δV_BE of the output characteristic is the output admittance, h_oe, of the transistor.

The output resistance of the transistor is equal to the reciprocal of the slope of the output characteristic. From Fig. 13 the output resistance at the point

V_CE = 6 V and I_B = 30 μA is

R_out, = 1/ h_oe = δV_ce/δI_C = V_ce/I_c (I_B constant) (1.14)

= 2/0.2xl0^-3) = 10 000 Ω

 

The output admittance h_oe, and hence the output resistance, varies somewhat with the d.c. collector current. Manufacturers* data sheets may include a graph showing h_oe plotted against I_c.

When a characteristic is non-linear its slope will vary according to the point of measurement and therefore the point of measurement should always be quoted. It is usual, unless specified otherwise, to measure the slope in the middle of the characteristic. For the greatest accuracy the increments taken either side of the chosen point should be as small as possible although this has not been done in this chapter in order to clarify the diagrams.

The output characteristics of Fig. 13 can also be used to determine the short-circuit current gain h_fe of the transistor, since, for a given value of collector-emitter voltage V_CE, the change in collector current δI_B produced by a change in base current δI_B can be obtained by projecting from the appropriate curves. Thus, for V_CE = 4 V a change in the base current from 20 μA to 30 μA will produce a change in the collector current from 2.1 to 2.9 mA. The current gain h_fe is therefore equal to

[(2.9 – 2.1)x 10^-3]/[(30-20)xl0^-6] = 80

The current gain h_fe of a transistor is not a constant quantity but varies with the d.c. collector current. Fig. 15 shows a typical graph of h_FE plotted against I_c. Manufacture’s data generally quote the collector current at which the maximum value of h_fe is obtained, e.g. for me BC 108 the typical h_fe is 180 at I_c = 2 mA. (Both h_fe and h_FE vary.)

It will be seen that Fig. 13 shows a collector current flowing even when the base current is zero. This current is the common-emitter leakage current, symbol I_CEO, which is related to the common-base leakage current I_CBO according to the expression

I_ceo = I_CBO(1+h_FE) (1.15)

Example 6

The collector leakage current I_CBO of a transistor is 10 nA. If the d.e. current gain of the transistor is h_fb = 0.995, calculate (a) the collector leakage current when the transistor is connected with common emitter, and (b) the d.c. collector current if a d.c. base current of 10 μA is supplied to the base.

Solution

h_FE = 0.995/(1 -0.995) = 199

(a) I_ceo = 10 X 10^-9 X 200 = 2 μA (Ans.)

(b) I_c = h_FEI_B + I_CEO = 199 x 10 x 10^-6 + 2 x l0^-6

= 1-992 mA (Ans.)

(Clearly the contribution of the collector leakage current is negligibly small.)

The collector leakage current I_CBO of a common-base transistor is extremely temperature-sensitive and is approximately doubled for every 12°C rise in temperature for silicon transistors and every 8°C rise for germanium transistors. However, the leakage current of a silicon transistor at a given temperature is much less than the leakage current of an equivalent germanium transistor at the same temperature.

Typically, I_cbo at 20°C may be about 10 μA for a germanium transistor but only about 10 nA for a silicon transistor.

(d) Common-emitter Mutual Characteristic

The mutual characteristics of a common-emitter connected transistor show the changes in collector current that occur with changes in the base-emitter voltage, with the collector-emitter voltage held constant. Fig. 16 shows a typical mutual characteristic. The slope of the mutual characteristic is the mutual conductance g_m of the transistor.

Thus, when V_BE changes from 0.6 V to 0.65 V the resulting change in I_c is from 1.5 raA to 11.5 mA and so

g_m = δI_c/δV_BE = (10 x 10^-3)/0.05 = 200 mS

The mutual conductance can also be determined from the output characteristics of Fig. 14 using a similar method to that employed to obtain h_fe from Fig. 13. When V_CE = 8 V, a change in V_BE from 610 mV to 630 mV causes I_c to vary from 3.3 mA to 6.7 mA. Hence

g_m = δI_c/δV_BE = (3.4 X 10^-3)/(20 X 10^-3) = 170 mS

Also g_m = δI_c/δV_BE (1.16)

= (δI_C/δI_B) X (δ1_B/δV_BE)

= h_fe/h_ie (3.17)

The mutual conductance g_m depends only on the d.c. collector current I_c, according to the equation

g_m = I_C/26mS (3.18)

where I_c is in mA.

Thus for the transistor just considered I_c = (3.3 + 6.7)/2 = 5 mA, and g_m = 5/26 – 192 mS.

From equation (3.7), the voltage gain of a transistor is given by A_v = A_iR_L/R_in. This can be written as A_v = h_feR_L/h_ie and, using equation (3.16)

A_v = g_mR_L (3-19)

Example 7

An 18 mV peak signal is applied to the base of a common-emitter connected transistor. The resulting change in the collector current is 2 mA. Calculate (a) the mutual conductance of the transistor, (b) the current gain of the transistor, (c) the d.c. collector current, (d) the value of collector load resistance that will give a voltage gain of 220 and (e) the output voltage. The transistor has an input resistance of 1200 Ω.

Solution

(a) g_m = (2 X 10^-3)/(18 x 10^-3) = 111 mS (Ans.)

(b) h_fe = g_mh_ie = 111 x 10^-3 x 1200 = 133 (Ans.)

(c) I_c = 26g_m – 26 X 111 X 10^-3 = 2.9mA (Ans.)

(d) 220 = 111 x 10^–3 R_L, therefore R_L = 1982 ≈ 2000 Ω (Ans.)

(e) V_out = 220 x 18 x 10^-3 = 3.96V (Ans.)