What is Cut off and Saturation Points
Fig. 8.49 (i) shows CE transistor circuit while Fig. 8.49 (ii) shows the output characteristcs along with the d.c. load line.
(i) Cut off
The point where the load line intersects the IB = 0 curve is known as cut off. At this point, IB = 0 and only small collector current (i.e. collector leakage current ICEO) exists. At cut off, the base-emitter junction no longer remains forward biased and normal transistor action is lost. The collector-emitter voltage is nearly equal to VCC i.e. VCE (cut off) = VCC
The point where the load line intersects the IB = IB(sat) curve is called saturation. At this point, the base current is maximum and so is the collector current. At saturation, collector base junction no longer remains reverse biased and normal transistor action is lost.
If base current is greater than IB(sat), then collector current cannot increase because collector-base
junction is no longer reverse-biased.
(iii) Active region
The region between cut off and saturation is known as active region. In the active region, collector-base junction remains reverse biased while base-emitter junction remains forward biased. Consequently, the transistor will function normally in this region.
Note. We provide biasing to the transistor to ensure that it operates in the active region. The reader may find the detailed discussion on transistor biasing in the next articles.
Summary. A transistor has two pn junctions i.e., it is like two diodes. The junction between base
and emitter may be called emitter diode. The junction between base and collector may be called
collector diode. We have seen above that transistor can act in one of the three states:
cut-off, saturated and active. The state of a transistor is entirely determined by the states of the emitter diode and collector diode [See Fig. 8.50]. The relations between the diode states and the transistor states are :
CUT-OFF : Emitter diode and collector diode are OFF.
ACTIVE : Emitter diode is ON and collector diode is OFF.
SATURATED : Emitter diode and collector diode are ON.
In the active state, collector current [See Fig 8.51 (i)] is β times the base current (i.e. IC = βIB). If the transistor is cut-off, there is no base current, so there is no collector or emitter current. That is collector emitter pathway is open [See Fig. 8.51(ii)]. In saturation, the collector and emitter are, in effect, shorted together. That is the transistor behaves as though a switch has been closed between the collector and emitter [See Fig. 8.51 (iii)].
Note. When the transistor is in the active state, IC = βIB. Therefore, a transistor acts as an amplifier when operating in the active state. Amplification means linear amplification. In fact, small signal amplifiers are the most common linear devices.
Example 8.31. Find IC(sat) and VCE(cut off) for the circuit shown in Fig. 8.52 (i).
Solution. As we decrease RB, base current and hence collector current increases. The increased
collector current causes a greater voltage drop across RC ; this decreases the collector-emitter voltage. Eventually at some value of RB, VCE decreases to Vknee. At this point, collector-base junction is no longer reverse biased and transistor action is lost. Consequently, further increase in collector current is not possible. The transistor conducts maximum collector current ; we say the transistor is saturated.
As we increase RB, base current and hence collector current decreases. This decreases the voltage drop across RC. This increases the collector-emitter voltage. Eventually, when IB = 0, the emitter base junction is no longer forward biased and transistor action is lost. Consequently, further increase in VCE is not possible. In fact, VCE now equals to VCC.
VCE(cut-off) = VCC = 20 V
Figure 8.52 (ii) shows the saturation and cut off points. Incidentally, they are end points of the d.c. load line.
Note. The exact value of VCE(cut-off) = VCC − ICEO RC. Since the collector leakage current ICEO is very small, we can neglect ICEO RC as compared to VCC.
Example 8.32. Determine the values of VCE (off) and IC (sat) for the circuit shown in Fig. 8.53.
Example 8.33. Determine whether or not the transistor in Fig. 8.54 is in stauration. Assume Vknee = 0.2V
This shows that with specified β, this base current (= 0.23 mA) is capable of producing IC greater
than IC (sat). Therefore, the transistor is saturated. In fact, the collector current value of 11. 5 mA is
never reached. If the base current value corresponding to IC (sat) is increased, the collector current
remains at the saturated value (= 9.8 mA).
Example 8.34. Is the transistor in Fig. 8.55 operating in saturated state ?
IC = βIB= (100)(100 µA) = 10 mA
VCE = VCC – ICRC= 10V – (10 mA)(970Ω) = 0.3V
Let us relate the values found to the transistor shown in Fig. 8.56. As you can see, the value of VBE is 0.95V and the value of VCE = 0.3V. This leaves VCB of 0.65V (Note that VCE = VCB + VBE). In this case, collector – base junction (i.e., collector diode) is forward biased as is the emitter-base junction (i.e., emitter diode). Therefore, the transistor
is operating in the saturation region.
Note. When the transistor is in the saturated state, the base current and collector current are independent of each other. The base current is still (and always is) found only from the base circuit. The collector current is found apporximately by closing the imaginary switch
between the collector and the emitter in the collector circuit.
Example 8.35. For the circuit in Fig. 8.57, find the base supply voltage (VBB) that just puts the transistor into saturation. Assume β = 200.
Solution. When transistor first goes into saturation, we can assume that the collector shorts to the emitter (i.e. VCE = 0) but the collector current is still β times the base current.
Therefore, for VBB ≥ 1.95, the transistor will be in saturation.
Example. 8.36. Determine the state of the transistor in Fig. 8.58 for the following values of
collector resistor : (i) RC= 2 kΩ (ii) RC= 4 kΩ (iii) RC= 8 kΩ
Solution. Since IE does not depend on the value of the collector resistor RC, the emitter current
(IE) is the same for all three parts.
Collector voltage, VC = VCC – IC RC
= 10V – 2 mA×2 kΩ = 10V – 4V = 6V
Since VC (= 6V) is greater than VE (= 2V), the transistor is active. Therefore, our assumption that transistor is active is correct.
(ii) When RC = 4 kΩ. Suppose the transistor is active.
∴ IC = 2mA and IB = 0.02 mA … as found above Collector voltage,VC = VCC – IC RC
= 10V – 2 mA × 4 kΩ = 10V – 8V = 2V
Since VC = VE , the transistor is just at the edge of saturation. We know that at the edge of saturation, the relation between the transistor currents is the same as in the active state. Both answers are correct.
(iii) When RC = 8 kΩ. Suppose the transistor is active.
∴ IC = 2mA ; IB = 0.02 mA … as found earlier. Collector voltage, VC = VCC – IC RC
= 10V – 2 mA × 8 kΩ = 10V – 16V = – 6V
Since VC < VE , the transistor is saturated and our assumption is not correct.
Example 8.37. In the circuit shown in Fig. 8.59, VBB is set equal to the following values :
(i) VBB = 0.5V (ii) VBB = 1.5V (iii) VBB = 3V
Determine the state of the transistor for each value of the base supply voltage VBB.
Solution. The state of the transistor also depends on the base supply voltage VBB.
What is Power Rating of Transistor
The maximum power that a transistor can handle without destruction is known as power rating of
the transistor. When a transistor is in operation, almost all the power is dissipated at the reverse biased*collector-base junction. The power rating (or maximum power dissipation) is given by :
PD (max) = Collector current × Collector-base voltage
= IC× VCB
∴ PD (max) = IC× VCE
[ VCE = VCB + VBE. Since VBE is very small, VCB j VCE]
While connecting transistor in a circuit, it should be ensured that its power rating is not exceeded
otherwise the transistor may be destroyed due to excessive heat. For example, suppose the power
rating (or maximum power dissipation) of a transistor is 300 mW. If the collector current is 30 mA,
then maximum VCE allowed is given by ;
PD (max) = IC × VCE (max)
or 300 mW = 30 mA × VCE (max)
or VCE (max) = 300 mW/30 mA = 10V
This means that for IC = 30 mA, the maximum VCE allowed is 10V. If VCE exceeds this value, the
transistor will be destroyed due to excessive heat. Maximum power dissipation curve. For **power transistors, it is sometimes necessary to draw maximum power dissipation curve on the output characteristics. To draw this curve, we should know the power rating (i.e. maximum power dissipation) of the transistor. Suppose the power rating
of a transistor is 30 mW.
PD (max) = VCE × IC
or 30 mW = VCE × IC
Using convenient VCE values, the corresponding collector currents are calculated for the maximum power dissipation. For example, for VCE = 10V,
This locates the point A (10V, 3 mA) on the output characteristics. Similarly, many points such as
B, C, D etc. can be located on the output characteristics. Now draw a curve through the above points to obtain the maximum power dissipation curve as shown in Fig. 8.60. In order that transistor may not be destroyed, the transistor voltage and current (i.e. VCE and IC) conditions must at all times be maintained in the portion of the characteristics below the maximum
power dissipation curve.
Example 8.38. The maximum power dissipation of a transistor is 100mW. If VCE = 20V, what is
the maximum collector current that can be allowed without destruction of the transistor?
PD (max) = VCE × IC (max)
or 100 mW = 20 V × IC (max)
∴ IC (max) =100 mW/20 V = 5 mA
Thus for VCE = 20V, the maximum collector current allowed is 5 mA. If collector current exceeds t his value, the transistormay be burnt due to excessive heat.
Note. Suppose the collector current becomes 7mA. The power produced will be 20 V × 7 mA = 140 mW. The transistor can only dissipate 100 mW. The remaining 40 mW will raise the temperature of
the transistor and eventually it will be burnt due to excessive heat.
Thus for VCE = 20V, the maximum collector current allowed is 5 mA. If collector current exceeds this value, the transistor may be burnt due to excessive heat.
Note. Suppose the collector current becomes 7mA. The power produced will be 20 V × 7 mA = 140 mW. The transistor can only dissipate 100 mW. The remaining 40 mW will raise the temperature of the transistor and eventually it will be burnt due to excessive heat.
Example 8.39. For the circuit shown in Fig. 8.61, find the transistor power dissipation. Assume that β = 200.
Example 8.40. For the circuit shown in Fig. 8.62, find thepower dissipated in the transistor. Assume β = 100.
Solution. The transistor is usually used with a resistor RC connected between the collector and its power supply VCC as shown is Fig. 8.62. The collector resistor RC serves two purposes. Firstly, it allows us to control the voltage VC at the collector.
Secondly, it protects the transistor from excessive collector current IC and, therefore, from excessive power dissipation.
Example 8.41. The transistor in Fig. 8.63 has the following maximum ratings :
PD (max) = 800 mW ; VCE (max) = 15V ; IC (max) = 100 mA
Determine the maximum value to which VCC can be adjusted without exceeding any rating.
Which rating would be exceeded first ?
Since PD (max) = 800 mW, it is not exceeded when VCC = 34.5V.
If base current is removed causing the transistor to turn off, VCE (max) will be exceeded because
the entire supply voltage VCC will be dropped across the transistor.
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Reference: Principles Of Electronics By V K Mehta And Rohit Mehta