bai tap dt cao tan lan 1 bai 6.10 va 9.10
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7/28/2019 Bai Tap DT Cao Tan Lan 1 Bai 6.10 Va 9.10
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6.10 for 2 pn-diodes with abrupt junction, one of which is made of Si and the
another is made of GaAs, with NA =1017cm- 3 and ND = 2 10
14cm- 3 in both cases:
a) Find the barrier voltage
b) Find the maximun electric field and the space charge region width.
c) Plot the space charge, potential, and electric field distribution along the
diode axis
Solution:
a) The barrier voltage in both cases is found. For Si we obtain
Vdiff=Vt lnNAND
ni2
= 0.65V, and for GaAs the result is Vdiff =1.12V
b) The maximum electric field in the junction is found by setting x = 0.
Subtituting the values for Si it is seen
EO =qNAeReO
dp
where the space charge region width in the p-semiconductor is found
dp(Si) =2e VdiffNDqNA
1
NA +ND
= 4.110- 7cm
dp(GaAS) =2e VdiffNDqNA
1
NA +ND
= 5.710- 7cm
Subtuting the values into the formula for the peak electric field we find
that in Si it is equal to EO = 6.4 103(V / cm)and for GaAs SO = 7.9 10
3(V/ cm) .
c) The distributions of the space charge, potential, and electric field areidentical to the figure 6-5 c - e
9.10
A BJT is operated at f= 750MHz(and with the S-parameters as follows:
S11 = 0.56 - 78O,S12 = 8.64 122
O ,S21 = 0.05 33O,S22 = 0.66 - 42
O). Attempt to
stabilize the transistor by finding a series resistor or shunt conductance for the
input and output ports.
Solution
In this problem, the values ofS12 and S21should have been reversed
S12 = 0.05 33O,S21 = 8.64 122
O. We first investigate and find that the transistor
does not meet the requirements of unconditional stability.
In fact,
k=1 - S11 2 - S22 2 + D 2
2 S12 S21= 0.6322, D = 0.5435
Then we plot the input and output the stability circuit then we got the result
gin'=
2.0
Gin'=gin
'
50W = 0.04Sfor the output resistance
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7/28/2019 Bai Tap DT Cao Tan Lan 1 Bai 6.10 Va 9.10
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rout'
= 0.8Rout'
= 0.75 50W = 37.5W Using this resistance value, we can re-check the stability requierments by first
converting the output index into ABCD network representation followed by pre-
multiplication by converted S-parameter of the transistor. The resulting new set
of ABCD parameters is reconverted into a new parameter form and check for
stability below
The result is now : K=1.018 >1 and D = 0.5435
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