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vGS 1 VT
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EXAMPLE 910 MOSFET Self-Bias Circuit
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Determine the Q point for the MOSFET self-bias circuit of Figure 933(a) Choose RS such that vDSQ = 8 V
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Known Quantities: MOSFET drain and gate resistances; drain supply voltage; MOSFET
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parameters VT and IDSS
Find: MOSFET quiescent gate-source voltage, vGSQ ; quiescent drain current, iDQ , and RS such that the quiescent drain-source voltage, vDSQ , is 8 V
Part II
Electronics
VDD RD R1 + iG + R2 VGS GS RS
VGG RD
RG iG + VGS
iD + VDS
VDS
Figure 933(a) Self-bias circuit for Example 910
Figure 933(b) Equivalent circuit for Fig 933(a)
Schematics, Diagrams, Circuits, and Given Data: VDD = 30 V; RD = 10 k ; R1 = R2 = 12 M ; RD = 12 M ; VT = 4 V; IDSS = 72 mA Assumptions: Operation is in the active region Analysis: Let all currents be expressed in mA and all resistances in k Applying KVL
around the equivalent gate circuit of Figure 933(b) yields: VGG = vGSQ + iGQ RG + iDQ RS = vGSQ + iDQ RS where VGG = VDD /2 and RG = R1 R2 Note that iGQ = 0 because of the in nite input resistance of the MOSFET
p-Channel MOSFETs and CMOS Devices As the designation indicates, a p-channel MOSFET is characterized by p-type doping; the construction and symbol are shown in Figure 934 Note the opposite direction of the arrow to indicate that the pn junction formed by the channel and substrate is now in the opposite direction The direction of drain current is opposite; therefore vDS and vGS are now negative A more convenient reference is obtained if voltages are de ned in the direction opposite to that for the n-channel device: if one de nes vSD = vDS and vSG = vGS , then these voltages will be positive for the drain current direction indicated in Figure 934 The carriers are holes in this device, since the channel, when formed, is p-type Aside from the nature of the charge carriers and the direction of current and polarity of the voltages, the pchannel and n-channel transistors behave in conceptually the same way; however, since holes are in general less mobile (recall the discussion of carrier mobility in Section 81), p-channel MOSFETs are not used very much by themselves They do nd widespread application in complementary metal-oxide-semiconductor (CMOS) devices CMOS devices take advantage of the complementary symmetry of p- and n-channel transistors built on the same integrated circuit Focus on Measurements: MOSFET Bidirectional Analog Gate illustrates one application of CMOS technology
Gate Source p+ n Bulk (substrate) Drain p+
D iD + VDS _ G V
Figure 934 p-channel enhancement MOSFET construction and circuit symbol
9
Transistor Fundamentals
FOCUS ON MEASUREMENTS
MOSFET Bidirectional Analog Gate
The variable-resistor feature of MOSFETs in the ohmic state nds application in the analog transmission gate The circuit shown in Figure 935 depicts a circuit constructed using CMOS technology The circuit operates on the basis of a control voltage, v, that can be either low (say, 0 V), or high (v > VT ), where VT is the threshold voltage for the n-channel MOSFET and VT is the threshold voltage for the p-channel MOSFET The circuit operates in one of two modes When the gate of Q1 is connected to the high voltage and the gate of Q2 is connected to the low voltage, the path between vin and vout is a relatively small resistance, and the transmission gate conducts When the gate of Q1 is connected to the low voltage and the gate of Q2 is connected to the high voltage, the transmission gate acts like a very large resistance and is an open circuit for all practical purposes A more precise analysis follows
v vin vout vin vout
v (a) CMOS transmission gate
v (b) CMOS transmission gate circuit symbol
Figure 935 Analog transmission gate
Let v = V > VT and v = 0 Assume that the input voltage, vin , is in the range 0 vin V To determine the state of the transmission gate, we shall consider only the extreme cases vin = 0 and vin = V When vin = 0, vGS1 = v vin = V 0 = V > VT Since V is above the threshold voltage, MOSFET Q1 conducts (in the ohmic region) Further, vGS2 = v vin = 0 > VT Since the gate-source voltage is not more negative than the threshold voltage, Q2 is in cutoff and does not conduct Since one of the two possible paths between vin and vout is conducting, the transmission gate is on Now consider the other extreme, where vin = V By reversing the previous argument, we can see that Q1 is now off, since vGS1 = 0 < VT However, now Q2 is in the ohmic state, because vGS2 = v vin = 0 V < VT In this case, then, it is Q2 that provides a conducting path between the input and the output of the transmission gate, and the transmission gate is also on We have therefore concluded that when v = V and v = 0, the transmission gate conducts and provides a near-zero-resistance (typically tens of ohms) connection between the input and the output of the transmission gate, for values of the input ranging from 0 to V
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