ITT Technical Institute. ET215 Devices 1. Unit 7 Chapter 4, Sections
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1 ITT Technical Institute ET215 Devices 1 Unit 7 Chapter 4, Sections
2 Chapter 4 Section 4.1 Structure of Field-Effect Transistors Recall that the BJT is a current-controlling device; the field-effect transistor (FET) is a voltage-controlling device. Objectives: Describe the basic classifications for a fieldeffect transistor (FET) 1. Discuss principal differences between FETs and BJTs
3 The FET Family Field-effect transistors (FETs) are a class of semiconductors that operate on an entirely different principal than bipolar transistors. In an FET, a narrow conducting channel (made up of either n-type or p-type material) is connected to two leads called the source and the drain. Conduction in the channel is controlled by a voltage applied to a third lead called a gate. In an FET (JFET) the gate forms a pn junction with the channel.
4 The FET Family Another type of FET, called a MOSFET (Metal Oxide Semiconductor) uses an insulated gate to control conduction in the channel. This type dominates the transistors used in IC circuits due to physical size. They are biased differently than BJTs, contain very high input resistance and low electrical noise, inherently more linear.
5 QUIZ BONUS QUESTIONS The FET Family What are the three terminals of a FET called? Source, drain and gate What is another name for an insulated-gate FET? MOSFET Why are MOSFETs the dominant type of transistor used in ICs? Easier to manufacture without resistors or diodes, and are produced smaller What are some important differences between BJTs and FETs? biased is different, very high input resistance & low electrical noise, inherently more linear
6 Chapter 4 Section 4.2 FET Characteristics n-channel the carrier are electrons p-channel the carrier are holes With no external voltage, the channel can conduct in either direction
7 FET Characteristics The width of the channel, and therefore it s ability to conduct current, is controlled by the gate voltage. With no gate voltage, the channel conducts the maximum current. When reverse bias is applied to the gate, the channel width narrows, and the conductivity drops.
8 FET Characteristics The channel width, and thus the channel resistance, is controlled by varying the gate voltage, thereby controlling the amount of drain current, I D. The white areas represent the depletion region created by the reverse bias. It is wider towards the drain end of the channel because the reverse-bias voltage between the gate and the drain is greater than that between the gate and the source.
9 JFET Symbols Schematic symbols for both n-channel and p-channel JFETs: Notice that the arrow on the gate points in on the n-channel and out on the p-channel
10 FET Drain Characteristic Curve Drain Characteristic: It is a plot of drain current, I D, vs. the drain-tosource voltage, V DS. Ohmic region: V DS and I D are related by Ohm s Law Constant-current region: linear region to curve Pinch-Off Voltage: equivalent to the knee of the BJT Breakdown Voltage: exceeding this voltage can damage
11 FET Drain Characteristic Curve V GS Controls I D : Let s connect a bias voltage Notice by setting V GG and increasing V DD, we can plot a characteristic curve
12 FET Cutoff Voltage, V GS(off) : It is the value that makes I D approximately zero. The JFET must be operated between V GS = 0 and V GS(cut). For this range of gate to source voltages, I D will vary from a maximum to a minimum of almost zero.
13 FET JFET at Cutoff: For an n-channel device, this is a positive voltage at which the drain current becomes constant. Cutoff can also be measured on the drain characteristic and represents the negative gate-to-source voltage that reduces the drain current to zero.
14 FET EXAMPLE: For the n-channel JFET shown: V GS(off) = -4V and I DSS = 12mA. Determine the minimum value of V DD required to put the device in the constant-current region of operation. V DS = V P = 4 V I D = I DSS = 12 ma V RD = (12mA)(560Ω) = 6.7 V V DD = V DS + V RD = 4 V V = 10.7 V
15 JFET Transconductance Curve Transfer curve is a useful way of looking at any circuit; look at the output for a given input. You can think of a voltage input being transferred to the output as a current. A Transconductance curve is a plot of the transfer characteristics (I D vs. V GS ) of a FET.
16 JFET Transconductance Curve The Transconductance curve (g m ) is directly related to the drain characteristic as shown in fig b, and can be found by dividing a small change in current by a small change in gate-to-source voltage: g m = ΔI D / ΔV GS or g m = I D / V GS The Transconductance curve is not a straight line, indicating that the relationship between the output current and the input voltage is nonlinear.
17 FET EXAMPLE: For the curve shown, determine the transconductance at I D = 1.0 ma. select a small change in I D g m = ΔI D / ΔV GS = (1.25 ma 0.75 ma) (-1.1 V (-1.8 V) ) = ms
18 JFET Input Resistance & Capacitance A pn junction has a very high resistance when it is reverse-biased. A JFET operates with its gate-source reversebiased; therefore, the input resistance at the gate is very high. The input resistance can be determined by: R IN = V GS / I GSS
19 FET EXAMPLE: The data sheet for an MPF3821 n-channel JFET shows a maximum I GSS of -0.1 na at 25 C for V GS = -30 V and a maximum I GSS of -100 na at 150 C for V GS = -30 V. Determine the minimum input resistance at 25 C.. R IN = V GS / I GSS = 30 V / 0.1 na = 300 GΩ
20 Chapter 4 Section 4.3 JFET Biasing Self-Biasing a JFET An n-channel JFET can be reverse-biased by applying a negative V GS voltage. Notice (in figure) that the gate voltage V G is 0 by resistor R G being connected to ground. We then make the source positive with respect to the gate which produces a reverse-bias condition.
21 JFET Biasing Self-Biasing a JFET The drain voltage with respect to ground is determined as follows: V D = V DD I D R D The drain-to-source voltage is: V DS = V DD I D (R D + R S )
22 FET EXAMPLE: Find V DS and V GS in figure. For the particular JFET in this circuit, the internal parameter values such as gm, V GS(off ), and I DSS are such that a drain current (I D ) of approximately 5.0 ma is produced. Another JFET, even of the same type, may not produce the same results when connected in this circuit due to variations in parameter values. V S = I D R S = (5.0 ma)(220ω) = 1.10 V V D = V DD - I D R D = 10 V ( (5.0 ma)(1kω) ) = 5.00 V V DS = V D V S = 5.00 V 1.10 V = 3.90V V GS = V G V S = 0 V 1.10 V = -1.10V
23 JFET Biasing Graphical Methods: a) Draw a straight line from the origin to the point where V GS(off) (-4V) intersects I DSS (2.5 ma). The reciprocal of the slope of the line represents a choice for R S. R S = V GS(off) / I DSS = 4 V / 2.5 ma = 1.6kΩ
24 EXAMPLE: FET A 2N5457 general-purpose JFET has the following specifications: I DSS(min) = 1 ma, I DSS(max) = 5 ma, V GS(off)(min) = -0.5 V, V GS(off)(max) = -6 V. Select a self-bias resistor for this JFET. R S = V GS(off)(min) / I DSS(min) = 0.5 V / 1.0 ma = 500 Ω R S = V GS(off)(max) / I DSS(max) = 6 V / 5.0 ma = 1.2k Ω A good choice is 820Ω; a standard value between extremes.
25 JFET Biasing Voltage Divider Bias: Bias can be more stable with a voltage divider on the gate circuit. Gate voltage is found by applying the voltage-divider rule to R 1 and R 2 V G = (R 2 / (R 1 + R 2 ) ) V DD
26 FET Assume you are troubleshooting the 2N5458 JFET shown. You do not know the transconductance of the transistor, but you need to find out of the circuit is working properly. a) Estimate the expected V G and V S Start with V G : V G = (R 2 / (R 1 + R 2 ) ) V DD = ( 330 / (1 MΩ + 330kΩ)) 12 V = 2.98 V you know immediately that the circuit is operating correctly because the source voltage must be more positive than this. You estimate the source voltage should be about 4 V.
27 Assume you are troubleshooting the 2N5458 JFET shown. You do not know the transconductance of the transistor, but you need to find out of the circuit is working properly. b. Assume you measure the source voltage and found it was 5.4 V. Is the circuit operating as expected? Based on the measurement, what is the drain voltage? The measured value of 5.4V may indicate a problem. It is larger than the expected 4V and is nearly half of V DD, possibly a drain-to-source short in the transistor, causing V D to also be 5.4V.
28 Assume you are troubleshooting the 2N5458 JFET shown. You do not know the transconductance of the transistor, but you need to find out of the circuit is working properly. c) Assume you replace the transistor. The measured source voltage for the new transistor is 4.0 V. From this measurement, what is the expected drain voltage? Apply Ohm s Law to the new transistor to find drain current. I D = 4.0 V / 1.8kΩ = 2.2mA Subtracting V RD from V DD gives V D. V D = V DD I D R D = 12V (2.2mA)(2.2k) = 7.16 V
29 JFET Biasing Current Sources: Review of Current Sources
30 JFET Biasing Current Sources: This form of bias is widely used in ICs but requires an additional transistor. One transistor acts as a current source to force I D to stay constant, creating a very stable form of bias. Example b is better due to the BJT having the base grounded and a constant V BE voltage of 0.7V at R E.
31 FET EXAMPLE: Shown is a current-source biasing circuit. What is I D? I E = V RE / R E = 14.3 V / 15kΩ = 0.95 ma This current is applied to the FET; therefore, I D = 0.95 ma
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