Electronic Devices. Floyd. Chapter 9. Ninth Edition. Electronic Devices, 9th edition Thomas L. Floyd

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1 Electronic Devices Ninth Edition Floyd Chapter 9

2 The Common-Source Amplifier In a CS amplifier, the input signal is applied to the gate and the output signal is taken from the drain. The amplifier has higher input resistance and lower gain than the equivalent CE amplifier. V in C 1 R G +V DD R D C 3 V out R S C 2 R L The voltage gain is given by the equation A v = g m R d.

3 The Common-Source Amplifier Recall that conductance is the reciprocal of resistance and admittance is the reciprocal of impedance. Data sheets typically specify the forward transfer admittance, y fs rather than transconductance, g m. The definition of y fs is y fs I = V D G DYNAMIC CHARACTERISTICS Forward Transfer Admittance (V DS = 15 Vdc, V GS = 0) 2N5457 2N5458 Symbol Min Typ Max Unit Y fs µ mhos An alternate gain expression for a CS amplifier is A v = y fs R d.

4 The Common-Source Amplifier You can estimate what the transfer characteristic looks like from values on the specification sheet, but keep in mind that large variations are common with JFETs. For example, the range of specified values for a 2N5458 is shown. V GS (V) I D (ma) OFF CHARACTERISTICS Gate-Source Cutoff Voltage (V DS = 15 Vdc, i D = 10 nadc) ON CHARACTERISTICS Zero Gate-Source Drain Current (V DS = 15 Vdc, V GS = 0) 2N5457 2N5458 2N5457 2N5458 Symbol Min Typ Max Unit V GS(off) I DSS Vdc Symbol Min Typ Max Unit madc

5 The Common-Source Amplifier To analyze the CS amplifier. you need to start with dc values. It is useful to estimate I D based on typical values; specific circuits will vary from this estimate. For a typical 2N5458, what is the drain current? V DD +12 V R D 2.7 kω From the specification sheet, the typical I DSS = 6.0 ma and V GS(off) = 4 V. These values can be plotted along with the load line to obtain a graphical solution. C µ F V in 100 mv R G 10 MΩ See the following slide 2N5458 R S 470 Ω C 2 10 µ F V out

6 The Common-Source Amplifier (continued) A graphical solution is illustrated. On the transconductance curve, plot the load line for the source resistor. Load line for 470 Ω resistor Then read the current and voltage at the Q-point. I D = 2.8 ma and V GS = 1.3 V V GS (V) I D (ma) 6 Q 2.8 ma V

7 The Common-Source Amplifier (continued) Alternatively, you can obtain I D using Equation 9-2: I D IDR = IDSS 1 V S GS(off) 2 The solution to this quadratic equation is simplified using a calculator that can handle quadratic equations. After entering the equation, enter the known values, but leave I D open. For the typical values for the 2N5458, (I DSS = 6 ma and V GS(off) = 4 V) with a source resistance of 470 Ω, we find 2.75 ma. ID=IDSS (1 ( ID RS/VG... ID= IDSS=.006 RS= 470 VGSOFF= 4.0 bound=( 1E99,1E99) GRAPH RANGE ZOOM TRACE SOLVE press enter absolute value F5

8 g The Common-Source Amplifier 2I Assume I DSS is 6.0 ma, V GS(off) is 4 V, and V GS = 1.3 V as found previously. What is the expected gain? ( ) ma DSS m0 = = = V 4 V GS(off) g m V GS = gm0 1 V GS(off) 1.3 V = 3.0 ms V 2.02 ms 3.0 ms A v = g m R d = (2.02 ms)(2.7 kω) = 5.45 C µ F V in 100 mv R G 10 MΩ V DD +12 V R D 2.7 kω 2N5458 R S 470 Ω Output is inverted C 2 10 µ F V out

9 The Common-Source Amplifier The gain is reduced when a load is connected to the amplifier because the total ac drain resistance (R d ) is reduced. How does the addition of the 10 kω load affect the gain? R d RR D L = R + R D L ( 2.7 kω)( 10 kω) = 2.7 kω+ 10 kω = 2.13 kω A v = g m R d = (2.02 ms)(2.13 kω) = 4.29 C µ F V in 100 mv R G 10 MΩ V DD +12 V R D 2.7 kω 2N5458 R S 470 Ω V out C 2 10 µ F R L 10 kω

10 The D-MOSFET In operation, the D-MOSFET has the unique property in that it can be operated with zero bias, allowing the signal to swing above and below ground. This means that it can operate in either D-mode or E-mode. I D +V DD Enhancement Q C 1 R D C 2 V out Depletion I d R L V GS 0 +V GS V in R G V gs

11 The E-MOSFET The E-MOSFET is a normally off device. The n-channel device is biased on by making the gate positive with respect to the source. A voltagedivider biased E-MOSFET amplifier is shown. I D Enhancement +V DD R D R 1 C 3 V out I DQ Q C 1 I d V C in R 2 R 2 S R L 0 V GS(th) V gs V GS V GSQ

12 The E-MOSFET The E-MOSFET amplifier in Example 9-8 is illustrated in Multisim using a 2N7000 MOSFET.

13 The Common-Drain Amplifier In a CD amplifier, the input signal is applied to the gate and the output signal is taken from the source. There is no drain resistor, because it is common to the input and output signals. V in C 1 R G +V DD R S C 2 V out R L The voltage gain is given by the equation A v gmrs = 1 + g R m s The voltage gain is always < 1, but the power gain is not.

14 The Cascode Amplifier The cascode connection is a combination of CS and CG amplifiers. This forms a good high-frequency amplifier. The input and output signals at 10 MHz are shown for this circuit on the following slide

15 The Cascode Amplifier The input signal for the cascode amplifier is shown in red; the output is blue. What is the gain? The peak of the input is 24.7 mv. The peak of the output is 2.33 V. A V = 94.3

16 The Class-D Amplifier MOSFETs are useful as class-d amplifiers, which are very efficient because they operate as switching amplifiers. They use pulse width modulation, a process in which the input signal is converted to a series of pulses. The pulse width varies proportionally to the amplitude of the input signal. Pulse-width modulation is easy to set up in Multisim. The following slide shows the circuit. A sine wave is compared to a faster triangle wave of the about the same amplitude using a comparator (a 741 op-amp can be used at low frequencies).

17 The Class-D Amplifier A circuit that you can use in lab or in Multisim to observe pulse width modulation in action. The scope display is shown on the following slide Op-amp set up as a comparator

18 The Class-D Amplifier The signal is the yellow sine wave and is compared repeatedly to the triangle (cyan). The result of the comparison is the output (magenta).

19 The Class-D Amplifier The modulated signal is amplified by class-b complementary MOSFET transistors. The output is filtered by a low-pass filter to recover the original signal and remove the higher modulation frequency. Modulated input +V DD Q 1 Low-pass filter PWM is also useful in control applications such as motor controllers. MOSFETs are widely used in these applications because of fast switching time and low onstate resistance. Q 2 V DD R L

20 The Analog Switch MOSFETs are also used as analog switches to connect or disconnect an analog signal. Analog switches are available in IC form for example the CD4066 is a quad analog switch that used parallel n- and p-channel MOSFETs. The configuration shown allows signals to be passed in either direction. Advantages of MOSFETs are that they have relatively low on-state resistance and they can be used at high frequencies, such as found in video applications. IN/OUT Control Simplified internal construction of a bidirectional IC analog switch. OUT/IN

21 Selected Key Terms Commonsource source is the (ac) grounded A FET amplifier configuration in which the terminal. Common-drain A FET amplifier configuration in which the drain is the (ac) grounded terminal. The common-drain amplifier. Source-follower A nonlinear amplifier in which the transistors Class-D are operated as switches. amplifier

22 Selected Key Terms Pulse-width A process in which a signal is converted to a modulation series of pulses with widths that vary proportionally to the signal amplitude. Analog switch CMOS A device that switches an analog signal on and off. Complementary MOS.

23 Quiz 1. Compared to a common-emitter amplifier, a commonsource amplifier generally will have a. higher gain and higher input resistance b. higher gain and lower input resistance c. lower gain and higher input resistance d. lower gain and lower input resistance

24 Quiz 2. The abbreviation y fs means a. forward transfer admittance b. forward on-state resistance c. reverse transfer susceptance d. reverse on-state conductance

25 Quiz 3. The plot shown is a graphical solution for a self-biased FET amplifier. The red line represents the a. gate resistor b. source resistor c. drain resistor d. none of the above I D (ma) 6 V GS (V) 4 0

26 Quiz 4. The resistance represented by the red line is a. 150 Ω b. 240 Ω I D (ma) c. 470 Ω d. 666 Ω 6 V GS (V) 4 0

27 Quiz gmrs 5. The gain equation Av = is used to calculate the 1 gain of + g m R s a. a CS amplifier b. a CD amplifier c. a CG amplifier d. any of the above

28 Quiz 6. A FET that can be biased with zero bias is a a. an n-channel JFET b. a D-MOSFET c. an E-MOSFET d. all of the above

29 Quiz 7. The cascode amplifier shown uses a. A CS and a CD stage b. Two CS stages c. Two CD stages d. none of the above

30 Quiz 8. The principle circuit used in creating a pulse width modulator is a a. peak detector b. clipper c. comparator d. low-pass filter

31 Quiz 9. The circuit is an amplifier for a pulse width modulated signal. The load has the demodulated signal. The yellow box represents a a. peak detector b. clipper c. comparator d. low-pass filter Modulated input +V DD Q 1 Q 2 R L V DD

32 Quiz 10. When the control signal is active, the output of an analog switch should look like a. the input signal b. a square wave c. a modulated pulse d. a dc level

33 Quiz Answers: 1. c 6. b 2. a 7. d 3. b 8. c 4. d 9. d 5. b 10. a

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