KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 1 7 DC BIASING FETS
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1 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 7 DC BIASING FETS Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 013
2 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 4.1 Introduction We know that the biasing levels for a silicon transistor configuration can be obtained using the approximate characteristic equations V BE = 0.7 V, I C = βi B, and I C = I E. The link between input and output variables is provided by β, which is assumed to be fixed in magnitude for the analysis to be performed. The fact that beta is a constant establishes a linear relationship between I C and I B. For the fieldeffect transistor, the relationship between input and output quantities is nonlinear due to the squared term in Shockley s equation. Linear relationships result in straight lines when plotted on a graph of one variable versus the other, whereas nonlinear functions result in curves as obtained for the transfer characteristics of a JFET. The nonlinear relationship between and can complicate the mathematical approach to the dc analysis of FET configurations. A graphical approach may limit solutions to tenthsplace accuracy, but it is a quicker method for most FET amplifiers. Since the graphical approach is in general the most popular, the analysis of this chapter will have graphical solutions rather than mathematical solutions. Another distinct difference between the analysis of BJT and FET transistors is that: The controlling variable for a BJT transistor is a current level, whereas for the FET a voltage is the controlling variable. In both cases, however, the controlled variable on the output side is a current level that also defines the important voltage levels of the output circuit.
3 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 3 The general relationships that can be applied to the dc analysis of all FET amplifiers are = 0 A = For JFETs and depletiontype MOSFETs and MESFETs, Shockley s equation is applied to relate the input and output quantities: = SS 1 For enhancementtype MOSFETs and MESFETs, the following equation is applicable: ID = k VGS VT It is particularly important to realize that all of the equations above are for the field effect transistor only The first few sections of this chapter are limited to JFETs and the graphical approach to analysis. The depletiontype MOSFET will then be examined with its increased range of operating points, followed by the enhancementtype MOSFET. Finally, problems of a design nature are investigated
4 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Fixedbias Configuration (JFET) C C C 1 AC output C 1 AC output AC input R G AC input R G V GG V GG The network will be isolated from indicated AC signals DC Equivalent
5 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 5 = V GG = SS 1 = SS 1 V GG = 0 C It is interesting to note that because the drain current is controlled by. The level of will determine the magnitude of, which is an important parameter. R G V GG = = V GG
6 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 6 Example 7.1: (Fixedbias Configuration) = 16 V V GG = V SS = 10 ma = 8 V = 1 k R G = 1 M R G Determine,, = V = = SS = 5,65 ma 1 V D = = = 4.75 V = SS 1 V GG V GG
7 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Selfbias Configuration (JFET) The selfbias configuration eliminates the need for two dc supplies. The controlling gate to source voltage is now determined by the voltage across a resistor. C C 1 AC output AC input R G
8 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 8 For the dc analysis, the capacitors can again be replaced by open circuits and the resistor R G replaced by a shortcircuit equivalent since = 0 A. = V S = = = SS 1 = SS 1 = 0 = = ( ) =
9 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 9 Example 7.: (Selfbias Configuration) = 0 V SS = 8 ma = 6 V = 3.3 k R G = 1 M = 1 k Determine,, = = SS 1 =,6 ma = =.6 V = = 8.8 V Intersection of two equations is the Q point
10 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 10 Example 7.3: (Selfbias Configuration) For the previous problem, Determine,, given = 100 k and = 10 k For = 100 k = 6.4 ma = 0.64 V For = 10 k = 0.46 ma = 4.6 V In particular, note how lower levels of bring the load line of the network closer to the axis, whereas increasing levels of bring the load line closer to the axis.
11 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Voltagedivider Configuration (JFET) AC input C 1 R 1 R C C 3 AC output The voltagedivider bias arrangement applied to BJT transistor amplifiers is also applied to FET amplifiers. The basic construction is exactly the same, but the dc analysis of each is quite different. = 0 A for FET amplifiers, but the magnitude of I B for commonemitter BJT amplifiers can affect the dc levels of current and voltage in both the input and output circuits. Recall that I B provides the link between input and output circuits for the BJT voltagedivider configuration, whereas does the same for the FET configuration.
12 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 For the dc analysis, the capacitors can again be replaced by open circuits and the resistor R G replaced by a shortcircuit equivalent since = 0 A. = 0 V G = R R 1 R = = ( ) V G = V S = = R 1 = V G = SS 1 = SS 1 (V G ) R = V G
13 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 13 Example 7.4: (Voltage Divider Configuration) = 16 V SS = 8 ma = 4 V R 1 =.1 M R = 70 k =.4 k = 1.5 k R 1 R Determine,, V G = R = 1.8 V R 1 R = V G = SS =.4 ma = 1.8 V = 6.64 V 1 Intersection of two equations is the Q point
14 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Common Gate Configuration (JFET) AC input C 1 C AC output For this configuration the gate terminal is grounded and the input signal typically applied to the source terminal and the output signal obtained at the drain terminal V SS
15 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 15 V SS = 0 V SS = 0 = V SS = V SS = SS 1 = SS 1 (V SS ) = V SS = V SS ( ) = V SS
16 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 16 Example 7.5: (Common gate Configuration) = 1 V V SS = 0 V SS = 1 ma = 6 V = 1.5 k = 680 Determine,, = V SS = = SS = 3.8 ma =.6 V = 3.7 V 1 Intersection of two equations is the Q point V SS
17 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Depletiontype MOSFETS The similarities in appearance between the transfer curves of JFETs and depletiontype MOSFETs permit a similar analysis of each in the dc domain. The primary difference between the two is the fact that depletiontype MOSFETs permit operating points with positive values of and levels of that exceed SS. In fact, for all the configurations discussed thus far, the analysis is the same if the JFET is replaced by a depletiontype MOSFET. The only undefined part of the analysis is how to plot Shockley s equation for positive values of. How far into the region of positive values of and values of greater than SS does the transfer curve have to extend? For most situations, this required range will be fairly well defined by the MOSFET parameters and the resulting bias line of the network. A few examples will reveal the effect of the change in device on the resulting analysis.
18 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 18 Example 7.6: (Voltage Divider Configuration) = 18 V SS = 6 ma = 3 V R 1 = 110 M R = 10 M = 1.8 k = 750 For the n channel depletiontype MOSFET Determine,, R 1 R V G = R = 1.5 V R 1 R = V G = SS = 3.1 ma = 0.8 V = 10.1 V 1 Intersection of two equations is the Q point
19 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 19 Example 7.6 Repeat previous example with = 150. = 7.6 ma = 0.35 V = ( ) = 3.18 V
20 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 0 Example 7.7: (Selfbias Configuration) = 0 V SS = 8 ma = 8 V = 6. k R G = 1 M =.4 k For the n channel depletiontype MOSFET Determine,, = = SS 1 = 1,7 ma = = 4.3 V = = 9.46 V Intersection of two equations is the Q point
21 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 Example 7.8: = 0 V SS = 10 ma = 4 V = 1.5 k Determine,, = 0 V = SS 1 = SS = 10 ma V D = = = 5 V
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