Lecture 16 MOSFET (cont d) Sunday 3/1/017 MOSFET 1-1
Outline Continue Enhancement-type MOSFET Characteristics C Biasing Circuits and Examples ntroduction to BJT-FET Combination Circuits Combination of BJT and FET devices in a circuit MOSFET 1-
Enhancement-Type MOSFET (Quick Review) MOSFET is also known as insulated gate FET (GFET) MOSFET 1-3
Enhancement-Type MOSFET Symbol n-channel E-Type MOSFET p-channel E-Type MOSFET MOSFET 1-4
Enhancement-Type MOSFET Characteristic k V VT for V VT k V V ( on) ( on) T MOSFET 1-5
Enhancement-Type MOSFET Characteristic k V V 8 ( on) ( on) T 10m 0.78 ma/v V 0.78m MOSFET 1-6
Enhancement-Type MOSFET Transfer Curve Plotting all the V (on) from the characteristic curve, the transfer curve can be obtained: V 0.78m MOSFET 1-7
Enhancement-Type MOSFET mportant Relationships k G 0 k S V V T (on) V V (on) T for V V T MOSFET 1-8
Enhancement-Type MOSFET C Biasing Fixed-bias, self-bias and many more bias configuration can be applied to enhancement-type MOSFET Two most popular MOSFET biasing configurations Feedback-bias configuration Voltage-divider bias configuration MOSFET 1-9
Feedback-Bias Configuration As the situation G = 0 still applied, the resistor R G will be ignored resulting in the drain and gate terminal to have the same voltage (V G = V ) MOSFET 1-10
Example (1) etermine V Q and Q for the E-Type MOSFET shown in the figure MOSFET 1-11
Example (1) Solution For the enhancement-type MOSFET s equation, the value of k have to be obtained first: k V V 8 3 (on) (on) the equation for the device: V V 1 G T 0.4m V 6m 3 for V 3 Since G equals zero, then 0.4mA/V nserting the V equation into the device equation: V V V 1 0 1 G S MOSFET 1-1
Example (1) Solution Substituting the equation for the MOSFET 960 9.64 19.44m 0 0.4m V 3 0.4m 1 3 0.4m 9 0.4m 81 36 4 19.44m 8.64 960 Solving the equation, we get: b b 4ac 9.64 a ( 9.64) 4(960)(19.44m) (960) 7.5mA and.79 ma Enhancement-type MOSFET doesn t have limitation for saturation current ( SS ), the true value of is the smaller one (since V should be greater than V T ).79 ma and V 1 6.4 V Q MOSFET 1-13
Example (1) Solution For graphical approach, several plot points have to be obtained first: 0.4m V V 3 for V 3 3 V 0 ma 4 V 0.4 ma 5 V 0.96 ma 6 V.16 ma 7 V 3.84 ma 8 V 6 ma For the bias line, only two plot points are required: V V 1 1 V 0 ma 0 V 6 ma MOSFET 1-14
Example (1) Solution Plots all the device transfer curve and device representation points: V T MOSFET 1-15
Voltage-ivider Bias Configuration Basically, the configuration is the same as in depletiontype MOSFET, JFET or BJT except the change of device to the enhancementtype MOSFET All the calculation would be the same except for the transfer curve of enhancement-type MOSFET is different from those depletion-type MOSFET and JFET MOSFET 1-16
Example () etermine Q and V Q MOSFET 1-17
Example () Solution etermining V G : For V S : So, for V : k 18M V G 40* 18V 18M M 3 V S 0.810 3 V V V 180.810 G V V 105 ( on) ( on) T S 3m 0.1mA/V nserting the circuit representation equation into the device equation: 0.110 3 V 5 for V 5 3 3 0.110 5 0.110 18 0.8 5 V 3 80.69 3.56 0.8 10 0 MOSFET 1-18
Example () Solution Solving the equation, we get: V b b 4ac 3.56 a 37.4 ma and 6.7 ma We take the smaller value: 18 0.8 ( 3.56) 4(80.69)(0.8m) (80.69) 6.7 ma 3 For 6.7 ma, V 18 0.810 (6.7 m ) 1.49 V The Q-point for MOSFET is defined by V Q Q 6.7mA 1.49V MOSFET 1-19
Example () Solution For graphical approach, several plot points have to be obtained first: 0.1m V 5 for V 5 For the bias line, only two plot points are required: V 18 0. 8k V 5 V 0 ma 10 V 3 ma 15 V 1 ma 0 V 7 ma 5 V 48 ma 30 V 75 ma V 18 V 0 ma 0 V 1.95 ma MOSFET 1-0
Example () Solution Plots all the device transfer curve and device representation points: MOSFET 1-1
p-channel Enhancement-Type MOSFET t is the complement to n-channel enhancement-type MOSFET All the current flow will be in the opposite direction Although the current direction is reverse, however the current equation are still the same (just like in JFET and depletion-type MOSFET) Construction Transfer Curve Characteristics MOSFET 1-
Comparisons between MOSFETs and BJTs High input impedance MOSFETs Pros Minimal drive power, no C current required at gate Simple drive circuits evices can be easily paralleled Max. operating temp. ~ 00 o C, less temp. sensitive Very low switching losses High switching speed High on-resistance Low transconductance Cons Low input impedance BJTs Cons Large drive power, continuous C current required at base Complex drive circuits as large +ve and ve currents are involved evices cannot be easily paralleled Max. operating temp. ~ 150 o C, more sensitive to temp Medium to high switching losses (depends on trade-off with conduction losses) Lower switching speed Low on-resistance Pros High transconductance FET Small AC Signal Model 1-3
BJT-FET Combination Circuits Combination of BJT and FET device in a circuit nnovative circuits that take some advantages of FETs, such as the high-input-impedance and low input power operation, and some merits of BJTs, such as high output current-driving capability How to analyze such circuits Firstly, recognize both of the devices and their current flows To make the calculation simple and easier to view, transform the circuit into the equivalent form to avoid complexity List down all the important relationships that involve for both of the devices Start with approaching the device that is closer to the ground (bottom device) MOSFET 1-4
Example (3) etermine V and V C MOSFET 1-5
Example (3) Solution We know that for the JFET device, G = 0 making the resistor R G = 1 MΩ useless and can be remove from the circuit By analyzing the circuit, we notice that the configuration is a voltage-divider bias for both the JFET and BJT device ue to involvement of BJT, we have to check βr E 10R to use the approximate analysis As for βr E = (180)(1.6k) = 88k and 10R = 10(4k) = 40k, situation βr E 10R is satisfied and we can use approximate analysis for this configuration Obtaining the E TH : E TH 16* 4 48 3.6V MOSFET 1-6
Example (3) Solution Transforming the circuit into its equivalent form: E TH = 3.6 V MOSFET 1-7
Example (3) Solution By approaching BJT (bottom device) first, we know V BE = 0.7 active operating mode From earlier calculation, we got E TH = V B = 3.6 V Obtaining V E : V R 1 R E E E B E 181 1.6 89.6 B B Obtaining B from V BE = 0.7: E TH = 3.6 V VBE VB VE 0.7 0.7 3.6 89.6k 10.08A B B MOSFET 1-8
Example (3) Solution From the circuit, B is not really important but C is very important because C = S = As for that, obtain C : C B ( 180)(10.08) 1.81mA Knowing the value of, V can be obtained: E TH = 3.6 V V 16.710 3 11.11V MOSFET 1-9
Example (3) Solution From the configuration, we notice that V S = V C By obtaining V for the JFET, the value of V S can be achieved: V V G V S 3.6V S 3.6V C V 1 V 1.81m 1m1 V 7.9V C SS P 3.6V 6 C E TH = 3.6 V MOSFET 1-30
Conclusion: FET Advantages FETs provide: Excellent voltage gain High input impedance Low-power consumption Good frequency range MOSFET 1-31
Lecture Summary Covered material Continue Enhancement-type MOSFET Characteristics Biasing Circuits and Examples ntroduction to BJT-FET Combination Circuits Material to be covered next lecture Small AC Signal Analysis for FETs MOSFET 1-3