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发表于 2019-8-23 12:26:13
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本帖最后由 RAMILE 于 2019-8-23 12:30 编辑
73楼答案:下面直流电位是我蒙的,标准答案见英文
交流增益是50,低频截止频率是45Hz
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ANSWER 1—For the DC analysis, start by calculating the Thevenin
equivalent of the bias network: 8.0 V and 16.67 kΩ. This sets the emitter
of Q1 at about 7.3 V. Now, consider R6. Since the voltage across it is
limited to 0.7 V, it is carrying at most about 0.15 mA. If we assume that
the contribution of Q3’s base is negligible (we’ll verify this shortly), then
that same current is flowing through R12, which gives it a voltage drop
of 1.5 V, setting the collector voltage of Q3 at about 5.8 V. This means that
R10 is carrying a total current of about 1.23 mA, which means that the
remaining current (1.08 mA) is flowing through Q3. If Q3 has a gain of
100, its base current is about 10.8 µA, which is less than 10% of the R6
current, as surmised. You could iterate through this analysis a few more
times to get more exact fgures, but that’s what circuit simulators are for.
ANSWER 2—As far as the AC analysis goes, Q1 by itself has a gain that
is set by R6 and R8 to about 21, but since Q3 has no emitter resistor,
its voltage gain is very large. Therefore, the overall gain of the circuit is
almost entirely controlled by the negative feedback (R12 and R8), which
makes the gain about 46. Each of the capacitors has a high-pass effect on
the circuit: C1 working with the impdeance of the bias network has a time
constant of 16.67 ms, which corresponds to a corner frequency of 9.5 Hz;
C2 working with R8 has a time constant of 22 ms, which corresponds to
a corner frequency of 7.2 Hz; and C5 working with R10 and Rload has a
time constant of 26.7 ms, which corresponds to a corner frequency of 6.0
Hz. The overall circuit response will be dominated by the input network,
for a cutoff frequency of about 10 Hz.
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