Transistor Characteristics and Parameters
– The ratio of the dc collector current (IC) to the dc base current (IB) is the dc beta (bDC).
– bDC is called the gain of a transistor:
bDC = IC/IB
– Typical values of bDC range from less than 20 to 200 or higher.
– bDC is usually designated as an equivalent hybrid (h) parameter:
hFE = bDC
– The ratio of the collector current (IC) to the dc emitter current (IE) is the dc alpha (aDC). This is a less-used parameter than beta.
aDC = IC/IE
– Typical values range from 0.95 to 0.99 or greater.
– aDC is always less than 1.
– This is because IC is always slightly less than IE by the amount of IB.
– From graph above we can see that there are 6 important parameters to be considered:
i) IB: dc base current.
ii) IE: dc emitter current.
iii) IC: dc collector current.
iv) VBE: dc voltage at base with respect to emitter.
v) VCB: dc voltage at collector with respect to base.
vi) VCE: dc voltage at collector with respect to emitter.
– VBB forward-biases the BE junction.
– VCC reverse-biases the BC junction.
– When the BE junction is forward biased, it is like a forward biased diode:
VBE ? 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 RB is
VR(B) = VBB- VBE
– Also
VR(B) = IRRB
– Or:
IRRB = VBB- VBE
– Solving:
IB = (VBB- VBE)/RB
– Voltage at collector with respect to grounded emitter is:
VCE = VCC – VR(C)
– Since drop across RC is VR(C) = ICRC the voltage at the collector is also:
VCE = VCC - ICRC
– Where IC = bDCIB. Voltage across the reverse-biased collector-bias junction is
VCB = VCE - VBE
Example:
Determine IB, IC, IE, VBE, VCE, and VCB in the following circuit. The transistor has bDC 150.
Solution:
We know VBE=0.7 V. Using the already known equations:
IB = (VBB- VBE)/RB
IB = (5 – 0.7)/10kW = 430 mA
IC = bDCIB = (150)( 430 mA) = 64.5 mA
IE = IC + IB = 64.5 mA + 430 mA = 64.9 mA
Solving for VCE and VCB:
VCE = VCC – ICRC = 10V-(64.5mA)(100W) = 3.55 V
VCB = VCE – VBE = 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 VBB and VCC = 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 IC is zero.
– When both junctions are forward-biased, the transistor is in the saturation region of operation.
– As VCC is increase, VCE gradually increases, as the IC increases (This is the steep slope linear region before the small-slope region).
– IC increases as VCC increase because VCE remains less than 0.7 V due to the forward-biased base-collector junction.
– Ideally, when VCE 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, IC levels off and remains essentially constant for a given value of IB as VCE continues to increase.
– Actually, there is a slight increase in IC, 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 bDC.
– For the linear portion, the value of IC is calculated by:
IC = bDCIB
– When VCE 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 IB = 0, the transistor is in the cutoff region, although there is a small collector leakage current.
i) Cutoff
– As said before, when IB = 0, transistor is in cutoff region.
– There is a small collector leakage current, ICEO.
– Normally it is neglected so that VCE = VCC.
– 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, IC also increase (IC – bDCIB) and VCE decreases as a result of more drop across the collector resistor (VCE = VCC – ICRC).
– When VCE reaches its saturation value, VCE(sat), the BC junction becomes forward-biased and IC can increase no further even with a continued increase in IB.
– At the point of saturation, IC = bDCIB is no longer valid.
– VCE(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 (IC = 0 and VCE = VCC).
– Top of load line is at saturation (IC = IC(sat) and VCE = VCE(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 VCE(sat) = 0.2 V.
First determine IC(sat).
IC(sat) = (VCC – VCE(sat))/RC
IC(sat) =(10 V – 0.2V)/10kW = 9.8 mA
Now let’s determine whether IB is large enough to produce IC(sat).
IB = (VBB - VBE)/RB = (3 V – 0.7 V)/10kW = 0.23 mA
IC = bDCIB = (50)(0.23 mA) = 11.5 mA
This shows that with the specified bDC, this base current is capable of producing an IC greater than IC(sat). Thus, the transistor is saturated, and the collector current value of 11.5 mA is never reached. If you further increase IB, the collector current remains at its saturation value.
i) More on bDC
– The bDC of hFE is not truly constant.
– It varies with collector current and with temperature.
– Keeping the junction temperature constant and increasing IC causes bDC to increase to a maximum.
– Further increase in IC beyond this point causes bDC to decrease.
– If IC is held constant and temperature varies, bDC changes directly with temperature.
– Transistor data specify bDC at specific values. Normally the bDC 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 VCEIC must not exceed PD(max).
Example:
The transistor shown in the figure below has the following maximum ratings: PD(max)=800 mW, VCE(max) = 15 V, and IC(max) = 100 mA. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first?
Solution:
First, find IB, so that you can determine IC.
IB = (VBB – VBE)/RB = (5 V – 0.7 V)/22 kW = 195 mA
IC = bDCIB = (100)(195 mA) = 19.5 mA
IC is much less than IC(max) and will not change with VCC. It is determined only by IB and bDC.
The voltage drop across RC is
VR(C) =ICRC = (19.5 mA)(1 kW) = 19.5 V
Now we can determine the value of VCC when VCE = VCE(max) = 15 V.
VR(C) = VCC - VCE
So,
VCC(max) = VCE(max) + VR(C) = 15 V + 19.5V = 34.5 V
VCC can be increased to 34.5 V, under the existing conditions, before VCE(max) is exceeded. However, at this point it is not known whether or not PD(max) has been exceeded:
PD = VCE(max)IC = (15 V)(19.5 mA) = 293 mW
Since PD(max) is 800 mW, it is not exceeded when VCC = 34.5 V. So, VCE(max) = 15 V is the limiting rating in this case. If the base current is removed, causing the transistor to turn off, VCE(max) will be exceeded first because the entire supply voltage, VCC, will be dropped across the transistor.
