求助一道电路题。英文。

只看该作者 · 2012-10-5 22:02

本帖最后由 jxd1207 于 2012-10-5 22:03 编辑

2. The shunt-series feedback amplifier is shown in the following figure. All
transistors have β=100 and VBE=0.7 V. Neglect ro.
(a) Perform a dc analysis to find the dc emitter currents in Q1 and Q2 and hence
determine their small-signal parameters.
(b) Give the A circuit (basic amplifier with input and output loading effect) and
determine A, Ri and Ro. Note Ro is the resistance determined by breaking the emitter
loop of Q2 and measuring the resistance between the terminals thus created.
(c) Find the β circuit and determine the value of β.
(d) Find Aβ, 1+Aβ, Af, Rif, and Rof. Note that Rof represents the resistance that appears
in the emitter of Q2 as a result of the feedback.

只看该作者 · 2012-10-5 22:03

还有一道。。。
这两道题。我都没有什么思路。求助大家了。
3. The feedback transconductance amplifier (series-series feedback amplifier) is
shown below. All three MOSFETs are biased and sized to operate at gm = 4 mA/V.
You may neglect their ro’s.
(a) Considering the feedback amplifier as a transcoductance amplifier with output
current Io, find the value of RF that results in a closed-loop transcoductance of
approximately 100 mA/V. (Hint: assume Aβ>>1, thus Af≒1/β then you can find RF).
(b) Sketch the A circuit and find the value of A≣Io/Vi.
(c) Find 1+Aβ and Af≣Is/Vo.
(d) Assuming ro
= 20 kΩ, find the output resistance Rout1. Since the current sampled by
the feedback network is exactly equal to the output current, you can use the feedback
formula.

2万 · 2012-10-5 22:19

我知道要做什么但是忘了怎么做了
2（a）（b）还行。。。

只看该作者 · 2012-10-5 22:21

3# Siderlee 好吧。能不能教我一下2 題的 b

只看该作者 · 2012-10-5 22:57

求助啊！！！！

1万 · 2012-10-5 23:16

2. The shunt-series feedback amplifier is shown in the following figure. All
transistors have β=100 and VBE=0.7 V. Neglect ro.
(a) Perform a dc analysis to find the dc emitter currents in Q1 and Q2 and hence
determine their small-signal parameters.
解：
设Q1基极对地电压为Vb，基极电流为Ib。则有：
（12 - Vb）/ 100 = Vb /15 + Ib ---------- (1);
(Vb - 0.7) / 0.87 = 101* Ib ---------------(2);
联力求解（1）、（2）式，得：Ib = 8.574(uA);
所以Q1的发射极电流Ie = 101 * Ib = 0.866 （mA）；
下面容易求得Q2的发射极电流为： Ie2 = 0.802（mA）；