ITT Technical Institute ET215 Devices 1 Chapter 4.6 4.7
Chapter 4 Section 4.6 FET Linear Amplifiers Transconductance of FETs The output drain current is controlled by the input signal voltage. As we earlier learned: g m = I d / V gs and has the units of Siemens (S), the reciprocal of resistance and g m is analogous to r e for BJTs r s = 1 / g m
Common-Source (CS) Amplifiers: JFET Shown is a common-source amplifier with a self-biasing n-channel JFET R G acts to keep the gate approx 0 V and prevents ac noise
Common-Source (CS) Amplifiers: D-MOSFET Shown is a common-source amplifier with a zero-biasing n-channel D-MOSFET R G acts to keep the gate approx 0 V and with a source voltage of 0 V dc; V GS = 0 V. The signal causes V GS to swing above and below the 0 V value, producing a swing in I d.
Common-Source (CS) Amplifiers: E-MOSFET Shown is a common-source amplifier with a voltage-divider biased E-MOSFET As with the JFET and D_MOSFET, the signal voltage produces a swing in V GS above and below Q point value; thus causing a swing in I d.
Voltage Gain: Voltage gain, A v, of an amplifier always equals V out / V in In the case of the CS Amplifier A v = V out / V in = I d R d / V gs A v = -g m R d or A v = -R d / r s (negative sign indicates it is an inverting amplifier)
Input Resistance: When the transistor's internal resistance is ignored, the input resistance seen by the signal source is determined only by the bias resistor(s).
EXAMPLE: FET Linear Amplifiers (a) What is the total output voltage of the amplifier when the g m is 1500 µs, I D is 2.0 ma, and V GS is 3 v? (b) What is the input resistance seen by the signal source? First, find the dc output voltage V D = V DD I D R D = 15 V (2mA)(3.3kΩ) = 8.4 V
(CS) EXAMPLE (continued): (a) What is the total output voltage of the amplifier when the g m is 1500 µs, I D is 2.0 ma, and V GS is 3 v? (b) What is the input resistance seen by the signal source? Next, find the gain A v = -g m R D = -(1500 µs)(3.3kω) = - 5 V
(CS) EXAMPLE (continued): (a) What is the total output voltage of the amplifier when the g m is 1500 µs, I D is 2.0 ma, and V GS is 3 v? (b) What is the input resistance seen by the signal source? The output voltage is the gain times the input voltage V out = A v V in = (-5 V)(100mV) = - 0.5 V rms The total output voltage is an ac signal with a p-p of 0.5 V rms x 2.828 = 1.4V, riding on a dc level of 8.4 V.
(CS) EXAMPLE (continued): (a) What is the total output voltage of the amplifier when the g m is 1500 µs, I D is 2.0 ma, and V GS is 3 v? (b) What is the input resistance seen by the signal source? The input resistance is R in R G 10 MΩ
Common-Drain (CD) Amplifiers: Shown is a common-drain amplifier with a self-biasing JFET NOTE: there is no drain resistor
Common-Drain (CD) Amplifier Voltage Gain: As with all amplifiers, the gain, A v, always equals V out / V in A v = V out / V in A v = V in (R S / (r s + R S )) A v = (g m R S ) / (1 + g m R S )
Common-Drain (CD) Amplifier Input Resistance: for self-biasing, the input resistance is R in R G for voltage-divider bias, the input resistance is R in R 1 ǁ R 2
(CD) EXAMPLE: Determine the minimum and maximum voltage gain of the amplifier shown based on the data sheet information.
EXAMPLE: Determine the minimum and maximum voltage gain of the amplifier shown based on the data sheet information. On the data sheet, g m is shown as y fs with a range of 2000 µs to 8000 µs (µmhos). The max value of r s is r s = 1 / g m = 1 / 2000 µs = 500 Ω The min value of r s is r s = 1 / g m = 1 / 8000 µs = 125 Ω
(CD) EXAMPLE (continued): Determine the minimum and maximum voltage gain of the amplifier shown based on the data sheet information. Finding the gain A v A v(min) = R S / (r s + R s ) = 10 k / (500Ω + 10 kω) = 0.95 A v(max) = R S / (r s + R s ) = 10 k / (125Ω + 10 kω) = 0.99
Common-Drain (CD) Amplifiers: Shown is a common-drain amplifier with a Current-Source Biasing JFET NOTE: Improvements include: (a) higher input resistance, lower distortion, and the ability to dc couple the signal at both the input and the output without coupling capacitors. The two FETs and two resistors should match.
(CD) EXAMPLE: Determine the drain current, I D, and the source voltage, V S, of Q 1 for the amplifier shown. Assume the FETs are matched and each has a transconductance curve as shown. On the curve, draw a line representing the 1.0 kω bias resistor for the current source (Q 2 ). The crossing point indicates that I D = 1.8 ma which causes the source of Q to be 1.8 V
Common-Gate (CG) Amplifiers: Limited use by itself, but can be used in the second stage of a FET differential amplifier. Principal advantage is high input resistance in linear applications. A V = g m R d R in = 1 / g m
Chapter 4 Section 4.7 FET Switching Circuits Types of Solid-State Switches: Analog switch: takes advantage of the FET s low internal drain-source resistance when it is on and its high drain-source resistance when it is off to either pass or block and ac signal Digital switch: frequently designed for power switching where current and voltages may be large
FET Switching Circuits Analog switch : Switched on and off by the control voltage on the gate The diode on the gate and the resistor across the source-gate prevent the gate from becoming positive by the ac input To turn switch on: V GS is made zero volts To turn switch off: V GS is made more negative than the lowest value of the source input. r DS(on) = - (V GS(off) / 2I DSS )
Switching EXAMPLE: Compute r DS(on) for a JFET with the transconductance curve shown. Reading the graph, V GS(off) = -4V and I DSS = 5 ma. r DS(on) = - (V GS(off) / 2I DSS ) = - ( 4V/ 2(5.0mA) ) = 400 Ω
FET Switching Circuits MOSFET Analog switch: Widely used as switches as they offer both positive and negative control. Note: there are no high current JFETs available for this application. Can be used as Solid-State Relays
Digital switch : FET Switching Circuits
FET Switching Circuits IC Digital switch: A switch to control direction of motor: