Transistor Parameters and Characteristics

– The ratio of the dc collector current (I_C) to the dc base current (I_B) is the dc beta (b_DC).

– b_DC is called the gain of a transistor:

b_DC = I_C/I_B

– Typical values of b_DC range from less than 20 to 200 or higher.

– b_DC is usually designated as an equivalent hybrid (h) parameter:

h_FE = b_DC

– The ratio of the collector current (I_C) to the dc emitter current (I_E) is the dc alpha (a_DC). This is a less-used parameter than beta.

a_DC = I_C/I_E

– Typical values range from 0.95 to 0.99 or greater.

– a_DC is always less than 1.

– This is because I_C is always slightly less than I_E by the amount of I_B.

– From graph above we can see that there are 6 important parameters to be considered:

i) I_B: dc base current.

ii) I_E: dc emitter current.

iii) I_C: dc collector current.

iv) V_BE: dc voltage at base with respect to emitter.

v) V_CB: dc voltage at collector with respect to base.

vi) V_CE: dc voltage at collector with respect to emitter.

– V_BB forward-biases the BE junction.

– V_CC reverse-biases the BC junction.

– When the BE junction is forward biased, it is like a forward biased diode:

V_BE ? 0.7 V

– But it can be as high as 0.9 V (and is dependent on current). We will use 0.7 V from now on.

– Emitter is at ground. Thus the voltage across R_B is

V_R(B) = V_BB- V_BE

– Also

V_R(B) = I­_RR_B

– Or:

I­_RR_B = V_BB- V_BE

– Solving:

I_B = (V_BB- V_BE)/R_B

– Voltage at collector with respect to grounded emitter is:

V_CE = V_CC – V_R(C)

– Since drop across R_C is V_R(C) = I_CR_C the voltage at the collector is also:

V_CE = V_CC - I_CR_C

– Where I_C = b_DCI_B. Voltage across the reverse-biased collector-bias junction is

V_CB = V_CE - V_BE

Example:
Determine I_B, I_C, I_E, V_BE, V_CE, and V_CB in the following circuit. The transistor has b_DC 150.

Solution:

We know V_BE=0.7 V. Using the already known equations:

I_B = (V_BB- V_BE)/R_B

I_B = (5 – 0.7)/10kW = 430 mA

I_C = b_DCI_B = (150)( 430 mA) = 64.5 mA

I_E = I_C + I_B = 64.5 mA + 430 mA = 64.9 mA

Solving for V_CE and V_CB:

V_CE = V_CC – I_CR_C = 10V-(64.5mA)(100W) = 3.55 V

V_CB = V_CE – V_BE = 3.55 V – 0.7 V­ = 2.85 V

Since the collector is at higher potential than the base, the collector-base junction is reverse-biased.

– Changing the voltage supplies with variable voltage supplies in the circuit above, we can get the characteristic curves of the BJT.

– If we start at some positive V_BB and V_CC = 0 V, the BE junction and the BC junction are forward biased.

– In this case the base current is through the BE junction because of the low impedance path to ground, thus I_C is zero.

– When both junctions are forward-biased, the transistor is in the saturation region of operation.

– As V_CC is increase, V_CE gradually increases, as the I­_C increases (This is the steep slope linear region before the small-slope region).

– I_C increases as V_CC ­increase because V_CE remains less than 0.7 V due to the forward-biased base-collector junction.

– Ideally, when V_CE exceeds 0.7 V, the BC junction becomes reverse biased.

– Then, the transistor goes into the linear region of operation.

– When the BC junction is reverse-biased, I_C levels off and remains essentially constant for a given value of I_B as V_CE continues to increase.

– Actually, there is a slight increase in I_C, due to the widening of the BC collector depletion region, which results in fewer holes for recombination in the base, which causes a slight increase in b_DC.

– For the linear portion, the value of I­_C is calculated by:

I_C = b_DCI­_B

– When V_CE reaches a sufficiently large voltage, the reverse biased BC junction goes into breakdown.

– Thus, the collector current increases rapidly.

– A transistor should never be operated in this region.

– When I_B = 0, the transistor is in the cutoff region, although there is a small collector leakage current.

i) Cutoff

– As said before, when I_B = 0, transistor is in cutoff region.

– There is a small collector leakage current, I_­CEO.

– Normally it is neglected so that V_CE = V_CC.

– In cutoff, both the base-emitter and the base-collector junctions are reverse-biased.

ii) Saturation

– When BE junction becomes forward biased and the base current is increased, I_C also increase (I­_C – b_DCI_B) and V_CE decreases as a result of more drop across the collector resistor (V_CE = V_CC – I_CR_C).

– When V_CE reaches its saturation value, V_CE(sat), the BC junction becomes forward-biased and I_­C can increase no further even with a continued increase in I_B.

– At the point of saturation, I_C = b_DCI_B is no longer valid.

– V_CE(sat) for a transistor occurs somewhere below the knee of the collector curves.

– It is usually only a few tenths of a volt for silicon transistors.

iii) DC load line

– Cutoff and saturation can be illustrated by the use of a load line.

– Bottom of load line is at ideal cutoff (I_C = 0 and V_CE = V_CC).

– Top of load line is at saturation (I_C = I_C(sat) and V_CE = V_CE(sat))

– In between cutoff and saturation along the load line is the active region.

– More to come later.

Example

Determine whether or not the transistor in circuit below is in saturation. Assume V_CE(sat) = 0.2 V.

First determine I_C(sat).

I_C(sat) = (V_CC – V_CE(sat))/R_C

I_C(sat) =(10 V – 0.2V)/10kW = 9.8 mA

Now let’s determine whether I_B is large enough to produce I_C(sat).

I_B = (V_BB - V_BE)/R_B = (3 V – 0.7 V)/10kW = 0.23 mA

I_C = b_DCI_B = (50)(0.23 mA) = 11.5 mA

This shows that with the specified b_DC, this base current is capable of producing an I_C greater than I_C(sat). Thus, the transistor is saturated, and the collector current value of 11.5 mA is never reached. If you further increase I_­B, the collector current remains at its saturation value.

i) More on b_DC

– The b_DC of h_FE is not truly constant.

– It varies with collector current and with temperature.

– Keeping the junction temperature constant and increasing I_C causes b_DC to increase to a maximum.

– Further increase in I_C beyond this point causes b_DC to decrease.

– If I_C is held constant and temperature varies, b_DC changes directly with temperature.

– Transistor data specify b_DC at specific values. Normally the b_DC specified is the maximum value.

ii) Maximum transistor ratings

– Maximum ratings are given for collector-to-base voltage, collector-to-emitter voltage, emitter-to-base voltage, collector current, and power dissipation.

– The product V_CEI_C must not exceed P_D(max).

Example:

The transistor shown in the figure below has the following maximum ratings: P_D(max)=800 mW, V_CE(max) = 15 V, and I_C(max) = 100 mA. Determine the maximum value to which V_CC can be adjusted without exceeding a rating. Which rating would be exceeded first?

Solution:

First, find I_B, so that you can determine I_C.

I_­B = (V_BB – V_BE)/R_B = (5 V – 0.7 V)/22 kW = 195 mA

I_C = b_DCI_B = (100)(195 mA) = 19.5 mA

I_C is much less than I_C(max) and will not change with V_CC. It is determined only by I_B and b_DC.

The voltage drop across R_C is

V­_R(C) =I_CR_C = (19.5 mA)(1 kW) = 19.5 V

Now we can determine the value of V_CC when V_CE = V_CE(max) = 15 V.

V_R(C) = V_CC - V_CE

So,

V_CC(max) = V_CE(max) + V_R(C) = 15 V + 19.5V = 34.5 V

V_CC can be increased to 34.5 V, under the existing conditions, before V_CE(max) is exceeded. However, at this point it is not known whether or not P_D(max) has been exceeded:

P_D = V_CE(max)I_C = (15 V)(19.5 mA) = 293 mW

Since P_D(max) is 800 mW, it is not exceeded when V_CC = 34.5 V. So, V_CE(max) = 15 V is the limiting rating in this case. If the base current is removed, causing the transistor to turn off, V_CE(max) will be exceeded first because the entire supply voltage, V_CC, will be dropped across the transistor.
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