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Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 1 out of 13
IV. Bipolar Junction Transistors
4.0 Introduction
- Popular belief holds that the bipolar junction transistor (BJT) was developed by Schockley, Brattain, and Bardeen from Bell labs in 1948.
- This is not true, as the device invented was the point-contact transistor.
- BJTs were actually developed in the late 1951’s by Dr. Schockley.
- The transistor is a three-terminal device whose output current, voltage and/or power are controlled by its input current.
- Used primarily in communication as an amplifier to increase the strength of an ac signal.
- In digital systems it is primarily used as a switch.
4.1 Transistor Structure
- The BJT is constructed with three doped semiconductor regions separated by two pn junctions.
- The three regions are called emitter, base, and collector.
- There are two types of BJTs, either pnp (two p regions separated by one n region) and npn (two n regions separated by one p region).
- The C, B , and E symbols represent the common, emitter, and base regions, respectively.
- The base region is lightly doped and very thin compared to the heavily doped emitter and moderately doped collector regions.
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 2 out of 13
4.2 Basic Transistor Operation
- For correct operation, the two pn junctions must be correctly biased with external dc voltages.
- Operation of the pnp is similar as that of npn, but the roles of electrons and holes, bias polarities, and current directions are all reversed.
- The figure below shows the correct biasing of a BJT.
- Note the base-emitter (BE) junction is forward biased and the base-collector (BC) junction is reverse biased.
- The forward bias from base to emitter narrows the BE depletion region.
- The reverse bias from base to collector widens the BC depletion region.
- The heavily doped n- type emitter region is packed with conduction-band (free) electrons.
- The free electrons from the emitter diffuse easily through the forward biased BE junction into the p -type base region
- In the base, the electrons become minority carriers (like in a forward biased diode).
- The base region is lightly doped and very thin, so it has a limited number of holes.
- Because of that light doping, only a small percentage of all the electrons flowing through the BE junction can combine with the available holes in the base.
- These relatively few recombined electrons flow out of the base lead as valence electrons, forming the small base electron current.
- Most of the electrons flowing from the emitter into the lightly doped base region do not recombine, but diffuse into the BC depletion region.
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 4 out of 13
- 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) 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) = I (^) R RB
- Or: I (^) R RB = VBB - VBE
- Solving: I (^) B = (VBB - VBE)/RB
- Voltage at collector with respect to grounded emitter is: VCE = VCC – VR(C)
- Since drop across RC is VR(C) = I (^) C RC the voltage at the collector is also: VCE = VCC - I (^) C RC
- Where I (^) C = βDC I (^) B. Voltage across the reverse-biased collector-bias junction is VCB = VCE - VBE Example: Determine I (^) B, I (^) C , I (^) E, VBE, VCE, and VCB in the following circuit. The transistor has
β DC 150.
Vbb 5V
Rb 10kOhm
Rc 100Ohm
Vcc 10V
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 5 out of 13
Solution: We know VBE=0.7 V. Using the already known equations: I (^) B = (VBB- VBE)/RB
I B = (5 – 0.7)/10k Ω = 430 μ A
I C = βDC I B = (150)( 430 μ A) = 64.5 mA
I E = I C + I B = 64.5 mA + 430 μ A = 64.9 mA
Solving for VCE and VCB:
VCE = VCC – I C RC = 10V-(64.5mA)(100 Ω) = 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 I (^) C 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).
- I (^) C 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.
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 7 out of 13
- At the point of saturation, I (^) C = βDC I (^) B 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 = I (^) C(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 I (^) C(sat). I (^) C(sat) = (VCC – VCE(sat))/RC
Vbb 3V
Rb 10kOhm
Rc 1kOhm
Vcc gain=50 10V
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 8 out of 13
I C(sat) =(10 V – 0.2V)/10k Ω = 9.8 mA
Now let’s determine whether IB is large enough to produce I (^) C(sat).
I B = (VBB - VBE)/RB = (3 V – 0.7 V)/10k Ω = 0.23 mA
I C = β DC I B = (50)(0.23 mA) = 11.5 mA
This shows that with the specified β 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 IB, the collector current remains at its saturation value.
iv) More on βDC
- The βDC of hFE is not truly constant.
- It varies with collector current and with temperature.
- Keeping the junction temperature constant and increasing IC causes βDC to increase to a maximum.
- Further increase in I (^) C beyond this point causes βDC to decrease.
- If I (^) C is held constant and temperature varies, βDC changes directly with temperature.
- Transistor data specify βDC at specific values. Normally the βDC specified is the maximum value.
v) 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 VCEI (^) C must not exceed P (^) D(max).
Example: The transistor shown in the figure below has the following maximum ratings: PD(max)=800 mW, VCE(max) = 15 V, and I (^) C(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 I (^) B, so that you can determine I (^) C.
Vbb 5V
Rb 22kOhm
Rc 1kOhm
Vcc_Variable gain=
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 10 out of 13
- Consider the following circuit.
- An ac voltage, Vin , is superimposed on the dc bias voltage VBB.
- DC bias voltage VCC is connected to the collector through the collector resistance, RC.
- The ac input voltage produces an ac base current, which results in a much larger ac collector current.
- The ac collector current produces an ac voltage across RC , thus producing an amplified, but inverted, reproduction of the ac input voltage in the active region.
- The forward biased base-emitter junction present low resistance to the ac wave.
- This internal ac emitter resistance is designated r’ (^) e. I (^) e ≈ I (^) c = Vb / r’ (^) e
- The ac collector voltage, Vc = I (^) cRC.
- Since I (^) e ≈ I (^) c, the ac collector voltage is Vc ≈ I (^) eRC.
- Vb can be considered the transistor ac input voltage where Vb = Vin – I (^) b RB.
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 11 out of 13
- Vc can be considered the transistor ac output voltage.
- The ratio of Vc to Vb is the ac voltage gain, Av , of the transistor circuit. Av = Vc/Vb
- Substituting I (^) eRC for Vc and I (^) e r’ (^) e for Vb yields Av = Vc/Vb ≈ (I (^) eRC )/(I (^) e r’ (^) e ) = RC / r’ (^) e
- Thus, amplification depends on the ratio of RC and r’ (^) e.
- RC is always considerably larger in value than r’ (^) e , thus the output voltage is larger than the input voltage.
Example: Determine the voltage gain and the ac output voltage for the following circuit if
r’ e = 50 Ω.
Solution: The voltage gain is
Av ≈ RC /r’ e = 1 k Ω /50 Ω = 20
Thus the output voltage is Vout = AvVb = (20)(100 mV) = 2 Vrms
4.5 The Transistor as a Switch
- One major application of a transistor is as an amplifier.
- The other major application is switching applications.
- In this case, it is operated alternately in cutoff and saturation.
- Analyze the following graph.
Vbb
Rb
1kOhmRc
Vcc
Vin
100 mV Vout
Chapter 4 – Bipolar Junctions Transistors (BJTs) Page 13 out of 13 the amplitude of the square wave input voltage necessary to make sure that the transistor saturates. Use double the minimum value of base current as a safety
margin to ensure saturation. VCC = 9 V, VCE(sat) = 0.3 V, RC = 270 Ω , RB =3.3 k Ω ,
and β DC = 50.
Solution: When the square wave is at 0 V, the transistor is in cutoff and, since there is no collector current, the LED does not emit light. When the square wave goes to its high level, the transistor saturates. This forward-biases the LED, and the resulting collector current through the LED causes is to emit light.
I C(sat) = (VCC – VCE(sat))/RC = (9 V – 0.3 V)/270 Ω = 32.2 mA
I B(min) = I C(sat)/ β DC = 32.2 mA/50 = 644 μ A
To ensure saturation, use twice the value of IB(min), that is, 1.29 mA. Then
I B = VR(B)/RB = (Vin – VBE)/R B = (Vin – 0.7)/3.3k Ω
Solving for the voltage amplitude of the square wave input, Vin , we get:
Vin – 0.7 = 2 I B(min)RB = (1.29 mA)(3.3 k k Ω)
Vin = (1.29 mA)(3.3 k k Ω ) + 0.7 V = 4.96 V
Homework 4 (All from chapter 4 problems) 4: 13, 15, 17, 19, 23, 25.
Reading: Chapter 5
Chapter Summary
- BJTs have three regions: base, collector, and emitter.
- BJTs have two pn junctions: base-emitter, and base-collector.
- Current in a BJT consists of both free electrons and holes, thus the term bipolar.
- Base is lightly doped compared to emitter and collector.
- There are two types of BJTs: npn and pnp.
- To operate as amplifier, the BE junction must be forward-biased and BC junction must be reverse-biased ( forward-reverse bias).
- I (^) B is very small compared to I (^) C and I (^) E.
- The dc current gain is βDC = I (^) C /I (^) B.
- When a BJT is forward-reverse biased, the voltage gain depends on the internal emitter resistance and the external collector resistance.
- A transistor can be operated as a switch in cutoff and saturation.
- In cutoff, both pn junctions are reverse-biased. Thus no IC. Therefore, there is an open between collector and emitter.
- In saturation, both pn junctions are forward-biased and the collector current is maximum. Behaves like a closed switch between collector and emitter.
