Question

10 V 4k 4k $V_{O1}$ $V_{O2}$ + 8k $V_{id}/2$ o $Q_1$ $Q_2$ o -$V_{id}/2$ 4k $Q_3$ 6k 2.65k - 10 V Q.1: Assume $\beta = \infty$ (Ignore base current) \begin{itemize} \item Find all the DC currents, voltages, and $I_{DC}$ \item Draw small-signal and find $|A_{vd}| = \frac{|V_{o1} - V_{o2}|}{|V_{id}|}$ \item Find $V_{CM}^{Max}$, $V_{CM}^{Min}$ \end{itemize}

          10 V
4k
4k
$V_{O1}$ 
$V_{O2}$
+ 8k
$V_{id}/2$ o
$Q_1$
$Q_2$
o -$V_{id}/2$
4k
$Q_3$
6k
2.65k
- 10 V
Q.1: Assume $\beta = \infty$ (Ignore base current)
\begin{itemize}
    \item Find all the DC currents, voltages, and $I_{DC}$
    \item Draw small-signal and find $|A_{vd}| = \frac{|V_{o1} - V_{o2}|}{|V_{id}|}$ 
    \item Find $V_{CM}^{Max}$, $V_{CM}^{Min}$
\end{itemize}

10 V
4k
4k
VO1
VO2
+ 8k
Vid/2 o
Q1
Q2
o -Vid/2
4k
Q3
6k
2.65k
- 10 V
Q.1: Assume β = ∞ (Ignore base current)

* Find all the DC currents, voltages, and IDC

* Draw small-signal and find |Avd| = (|Vo1 - Vo2|)/(|Vid|)

* Find VCM^Max, VCM^Min

Added by Concepci-N W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suman K

Step 1

Since we are ignoring the base current, we can assume that the base-emitter junction is forward biased and the base-collector junction is reverse biased. This means that the transistor is in active mode. Show more…

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