# LAB 4 : FET AMPLIFIERS

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1 LEARNING OUTCOME: LAB 4 : FET AMPLIFIERS In this lab, students design and implement single-stage FET amplifiers and explore the frequency response of the real amplifiers. Breadboard and the Analog Discovery Kit are to be used. The students will use various tools and functions from the Waveform software to perform measurement and plotting amplifier response. MATERIAL AND EQUIPMENT Material IRFD110 n-mosfet Capacitors Resistors Equipment Breadboard Analog Discovery Digilent Waveform software Multimeter IRFD110 Transistor Characteristics: - Build the circuit as shown in the schematic diagram in Figure 1 and connect the Analog Discovery instruments as indicated. Connect Scope2 probes across R1, Scope1 across the drain and the source. - Start the Digilent WaveForm software The initial settings of the generates waveforms are: AWG1: generates Vss. A triangle browses the range from (0V 5V) = (Amplitude = 2.5V, Offset=2.5V, frequency=140hz). AWG2: generates V G. There are 13 steps uniformly distributed in the range (Amplitude=150mV, Offset=3.45V, frequency=10hz). (AWG1, Scope 2+) R1, 560 IRFD110 Vs (Scope 2-, Scope 1+) V G (AWG2) GND (Scope 1-, Analog GND) Figure 1: Schematic diagram to obtain I-V characteristics of the n-channel MOSFET

2 Sample of scope windows: Figure 2: Scope sample of the n-channel MOSFET I-V characteristic 2

3 DESIGN A MOSFET AMPLIFIER: Design a FET amplifier (Figure 3) with a voltage gain AV = -2. The design is based on the I- V characteristics of the IRFD110 obtained in previous step. Design for VDD = 10 V (i.e., ±5 V supply from the Analog Discovery, ID = 1.5 ma, VDS = 2 V, and Rin > 100 kω. Assume a gm of 0.5mS in the design. Draw the operation point of your amplifier on the I-V characteristic curve of the IRFD110 n-channel MOSFET and attach it to the lab report. The following analysis is appropriate for good quality transistors where the output current ID is largely independent of the output voltage VDS (the output characteristic curves are approximately flat ). We calculate amplifier ac gain using the small signal FET transconductance gm and we assume ro can be neglected because it is very large in comparison to other circuit resistances (A=-g m.r D ). The small signal FET equivalent circuit is also shown in Figure 1. The input resistance is essentially R1 // R2 and the output impedance is essentially equal to RD if ro is very large. Small signal ac gain is calculated assuming that capacitors have negligible impedance. See Appendix 1 on selecting appropriate capacitors. +5V +5 V AWG1 (Ch1 +) (Ch 2 +) -5 V. Figure 3: FET amplifier and FET small signal model (Ch1 and Ch2 to Digilent GND to reduce noise) where gm is valid in exponential region. 3

5 APPENDIX 1 Selecting coupling capacitors: Be careful when choosing your coupling capacitors (C1 and C2). For this experiment, our largest non-polarized capacitors may be used. Polarized capacitors tend to have higher capacitance values, usually 5 µf, and they are always marked with either a + or a (or both) next to one of their terminals. They may also be marked with a band to indicate the negative end (same convention as a diode). Remember that the potential of the + terminal should be always higher than the terminal when connected in a circuit. Otherwise, it will induce the leak current between the two terminals and eventually damage the capacitor. In the signal path of a circuit such as C1 and C2, this condition may not be met in all cases since the connected circuits are unknown. Therefore you should avoid polarized capacitors in the signal path. Coupling capacitors must be chosen so that they have a small impedance at the frequency of interest compared with the input impedance of the circuit to which they re connected. This is to ensure that little voltage will be dropped or lost across the capacitor itself after all, an amplifier is supposed to amplify voltages, not attenuate them. A good rule of thumb is that Zcoupling C should be no more than approximately 10% of the input impedance of the amplifier (for the input coupling capacitor), or the input impedance of whatever circuit the amplifier drives (for the output coupling capacitor). For the FET amplifier you just constructed, the input impedance is supposed to be > 100 kω. Therefore the impedance of C1 at the lowest frequency the amplifier is expected to see should be no more than approximately 10 kω. If this lowest frequency is expected to be 100 Hz, then C1 > 0.16 µf. For this experiment, select the appropriate coupling capacitors for C1 at lowest frequency of 50 Hz. Similarly, the amplifier drives a load of 1 kω (Figure 2). Following the same argument the impedance of C2 at the lowest expected frequency should be no more than approximately 100 Ω. If this lowest frequency is 100 Hz, then C2 > 16 µf. If the largest non-polarized capacitors available are 2 µf, then C2 would have to be made up of eight 2 µf capacitors in parallel. Alternately, a polarized capacitor could be used with appropriate care given to the polarity of the capacitor. APPENDIX 2 Frequency response of the FET amplifier: The typical Frequency Response of an amplifier is presented in a form of a graph that shows output amplitude (or, more often, voltage gain) plotted versus log frequency. Typical plot of the voltage gain is shown in Figure 3. The gain is null at zero frequency, then rises as frequency increases, level off for further increases in frequency, and then begins to drop again at high frequencies. The frequency response of an amplifier can be divided into three frequency regions. 5

6 Figure 3: Diagram of voltage gain versus frequency for an amplifier. The frequency response begins with the lower frequency region designated between 0 Hz and lower cutoff frequency. At lower cutoff frequency, fl,the gain is equal to Amid. Amid is a constant midband gain obtained from the midband frequency region. The third, the upper frequency region covers frequency between upper cutoff frequency and above. Similarly, at upper cutoff frequency, fh, the gain is equal to Amid. After the upper cutoff frequency, the gain decreases with frequency increases and dies off eventually. The Lower Frequency Response: Since the impedance of coupling capacitors increases as frequency decreases, the voltage gain of a FET amplifier decreases as frequency decreases. At very low frequencies, the capacitive reactance of the coupling capacitors may become large enough to drop some of the input voltage or output voltage. Also, the source-bypass capacitor, the capacitor in parallel with the resistor from source to ground (source-resistor), may become large enough so that it no longer shorts the source-resistor to ground. Approximately, the following equations can be used to determine the lower cutoff frequency of the amplifier, where the voltage gain drops 3 db from its midband value (=0.707 times the midband Amid): (1) f1 = 1/ ( 2πrinC1 ) where: f1 = lower cutoff frequency due to C1, C1 = input coupling capacitance, rin = input resistance of the amplifier. (2) f2 = 1/ ( 2πrout C2 ) where: f2 = lower cutoff frequency due to C2, C2 = output coupling capacitance, rout = output resistance of the amplifier. Provided that f1 and f2, are not close in value, the actual lower cutoff frequency is approximately equal to the largest of the two. The Upper Frequency Response: Transistors have inherent shunt capacitances between each pair of terminals. At high frequencies, these capacitances effectively short the ac signal voltage. 6

7 The design using my available resistors, capacitors: - R1=220K, R2=300K, C1=C2=100nF, Rd=3.9K, Rs=1.6K - Vdd to Vss=9.96V, Vg=0.712V, Vs=-2.664V, Vd=-0.636V - Input: 100mV-peak, 1KHz, note that polarities of Channel 2 are swapped to have the same phase with input to calculate amplifier gain. Scope: - XY (C1:C2) shows the slope which is the gain of the amplifier) - M1: 20* Lg ( ( Max ( C2, 0.095) ) / ( Max ( C1, 0.095) ) ) = gain in db - M2: ( Max ( C2, 0.095) ) / ( Max ( C1, 0.095) ) is the gain 7

8 Appendix 3 IRFD110 Specifications: 8

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