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Bipolar Junction Transistor Amplifier

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Presentation on theme: "Bipolar Junction Transistor Amplifier"— Presentation transcript:

1 Bipolar Junction Transistor Amplifier
Chapter 3 Bipolar Junction Transistor Amplifier

3.1 Amplifier Operation 3.2 Transistor ac Model 3.3 Common-Emitter Amplifier 3.4 Common-Collector Amplifier 3.5 Common-Base Amplifier 3.6 Multistage Amplifier

3 3.1 Amplifier Operation Biasing is to establish Q-point for small ac signal from antenna, microscopes, sensors etc to amplifies, often referred as small-signal amplifier. Recall that small changes in Ib causes large changes in Ic. The small ac voltage causes Ib to increase and decrease accordingly and with this small change in Ic mimic the input only with greater amplitude. Fig 4-20a & b (stacked)

4 AC Quantities As Ic inc, Vc dec. Ic varies above and below Q-point in phase with Ib. Vce varies above and below Q-point value 1800 out of phase with Vb. Transistor always produces a phase inversion between Vb and Vc. Vce can represents rms, average, peak, or peak to peak, but rms will be assumed unless stated otherwise. vce can be any instantaneous value on the curve.

5 Boundary between cutoff and saturation is called linear region
Boundary between cutoff and saturation is called linear region. Transistor which operates in the linear region is called a linear amplifier. A voltage-divider biased transistor with a sinusoidal ac source capacitively coupled to base through C1 and load capacitively coupled to the collector through C2. C1 and C2 used to block dc and thus prevent the internal source resistance, Rs and load resistance, RL from changing the dc bias voltages at the base and collector. Only ac component reaches load because of the capacitive coupling.

6 A Graphical Picture Vb produce Ib that varies above and below Q-point on ac load line, also Ic and Vce. ac load line differ from dc load line because the effective ac collector resistance is RL in parallel with RC and is less than dc collector resistance RC.

7 Example : Let Q as the selected operating point (quiescent point). If Ib varies about 10 µA, find Vce and Ic variations

8 3.2 Transistor ac model To visualize the operation of a transistor in an amplifier circuit, it is often useful to represent the device by a model circuit. A transistor model circuit uses various internal transistor parameter to represents its operation. Transistor circuits can view by use of resistance or r parameters for better understanding.

9 r Paramaters r parameter description αac ac alpha (Ic/Ie) βac
ac beta (Ic/Ib) r’e ac emitter resistance r’b ac base resistance r’c ac collector resistance

10 Since the base resistance, r’b is small it is normally is not considered (can be replaced by a short) and since the collector resistance, r’c is fairly high and consider it as an open. The emitter and base resistance, r’e is the main parameter that is viewed. Ic = αacIe = βacIb

11 Determining r’e by a Formula
r’e is most important, and it’s value is r’e = 25 mV/IE Example : Calculate r’e, if IE = 5mA r’e = 25mV/5mA = 5 Ω Fig 6-6 r-parameter comparison Relationship of transistor symbol to r- parameter model

12 Comparison of AC Beta (ac) to DC Beta (DC)
A graph of IC versus IB is nonlinear. The two graphs best illustrate the difference between DC and ac. The two only differ slightly. Fig 6-7 IC vs IB curves

13 h parameter h parameter Description Condition hi input impedance
output short hr voltage feedback ratio input open hf forward current gain ho output admittance configuration h parameters Common-emitter hie, hre, hfe, hoe Common-base hib, hrb, hfb, hob Common-collector hic, hrc, hfc, hoc

14 Relationships of h Parameters and r Parameters
The ac current ratios, αac and βac, convert directly from h parameter as follows: αac = hfb βac = hfe r’e = hre/hoe r’c = (hre + 1)/hoe r’b = hie - (1+ hfe) hre/hoe

15 3.3 Common-Emitter Amplifier
Common-emitter amplifier exhibits high voltage and current gain. The Vout is 180º out of phase with Vin. Now lets use our dc and ac analysis methods to view this type of transistor circuit. It used voltage divider bias and coupling capacitor, C1 and C3. Fig. 6-8 ce amp Figure 6.8 : A common-emitter amplifier

16 DC Analysis DC component of the circuit “sees” only the part of the circuit that is within the boundaries of C1, C2, and C3 as the dc will not pass through these components and consider open. RINbase = βDCRE = 150 x 560Ω = 84KΩ Since RINbase is 10 x R2, so neglect. VB = (R2/(R1+R2)) x VCC = (6.8KΩ/28.8KΩ) x 12V = 2.83V VE = VB – VBE = 2.83V – 0.7V = 2.13V IE = VE/RE = 2.13V/560Ω = 3.80mA = IC VC = VCC – ICRC = 12V – 3.8mAx1KΩ = 8.2V VCE = VC – VE = 8.2V – 2.13V = 6.07V Fig 6-9 dc eq. ce amp

17 AC equivalent circuit for the amplifier in Figure 6.8
AC Analysis Basically replaces the capacitors C1, C2 and C3 with shorts because their value are selected so Xc is negligible at signal frequency and can be considered to be 0 Ω. dc source being replace with ground, assume, Vs has internal resistance, rint= 0Ω, so V ac across ac source = 0. So Vcc are also effectively shorts to ground for ac analysis. Fig 6-10&6-11 ac eq. ce circuit AC equivalent circuit for the amplifier in Figure 6.8

18 RC and R1 have one end connected to ac ground, because in actual circuit they are connected to VCC which is in fact ac ground. It a common-emitter amplifier because, C2 keeps emitter at ac ground which is the common point in circuit. If internal resistance (re), of ac source is 0Ω, then all source voltage appears at base terminal. If ac source has a nonzero internal resistance, 3 factor determine base voltage, source resistance, Rs, bias resistance R1||R2, and input resistance, Rinbase. Total input resistance, Rin(tot) = R1||R2||Rin(base) Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs If Rs << Rin(tot), then Vb = Vs where Vb is input voltage, Vin to amplifier.

19 Input Resistance and Output Resistance
Input resistance looking at base, Rin(base) = Vin/Iin = Vb/Ib Base voltage, Vb = Ier’e and since Ie = Ic, then Ib = Ie/βac So, Rin(base) = Ier’e/Ie/β ac = β ac r’e Total input resistance, Rin(tot) = R1||R2||Rin(base) Output resistance, looking at collector, Rout = RC || r’c, but since r’c >> RC So, Rout = RC Fig 6-10&6-11 ac eq. ce circuit

20 Example : If ac input with Vs = 10mV rms, source resistance of 300Ω and IE of 3.8mA, find the signal voltage at base. r’e = 25mV/IE = 25mV/3.8mA = 6.58Ω Rin(base) = βr’e = 160 x 6.58Ω = 1.05KΩ Rin(tot) = R1||R2||Rin(base) = 873 Ω Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs = 7.44mV

21 Voltage Gain Ac voltage gain for common-emitter amplifier is developed using the model circuit below Voltage gain is; Av = Vout/Vin = Vc/Vb Since, VC = α IeRC = IeRC and Vb = =Ier’e So Av = IeRC/Ier’e Av = RC/r’e Fig 6-14

22 Attenuation = Vs/Vb = (Rs + Rin(total)) / Rin(total)
Attenuation is the reduction in signal voltage as it passes through a circuit and correspond to a gain of less than 1. Attenuation from the ac supply internal resistance and input resistance must be determined since it affects the overall gain. Attenuation = Vs/Vb = (Rs + Rin(total)) / Rin(total) The overall voltage gain of the amplifier, A’v is the voltage gain from base to collector, Av times the reciprocal attenuation. A’v = (Vb/Vs)Av = Vc/Vs Fig 6-19

23 Example: Suppose we have a source, Vs of 10mV
Example: Suppose we have a source, Vs of 10mV. Due to resistances, the voltage felt at base (amplifier input),Vb is 5mV. Determine the overall gain. Attenuation factor = 10mV/5mV = 2 The overall gain, A’v = Av x reciprocal of attenuation = 20 x (2)- = 10

24 Effect of the Emitter Bypass Capasitor on Voltage Gain
Emitter bypass capacitor, C2 provides an effective short to ac signal around RE, thus keep emitter at ac ground. With bypass capacitor, the gain of a given amplifier is maximum and equal to RC/r’e. The value of bypass capacitor must be large enough so that the reactance over the frequency range of the amplifier is very small (ideally 0Ω) compared to RE. The XC(bypass) should be 10 Xc << RE at minimum frequency for which the amplifier must operate. Fig 6-19

25 Example : Select min C2 value if amp freq is 2kHZ to 10kHz
Since RE = 560Ω = 10XC, so XC = 56Ω So C2 at 2kHz = 1/(2πfXC) = 1/(2(2 kHz)( 56Ω)) = 1.42µF This is a minimum value for the bypass capacitor for this circuit. A large value can be use, although cost and physical size usually impose limitations. Fig 6-8 ce amp

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