Question

Consider the circuit shown in the figure below.

$\text{The npn transistor has}\beta =25,V_{CE\text{(sat)}}=0.5\text{V and}{V}_{BEON)}=0.7V\text{. If the maximum output voltage that can appear is 4.5 V, then the range of input voltage for which}\text{the transistor will work in active region is}$

• A. $0.7V\le V_{\text{in}}\le 4.5V$
• B. $1.86V\le V_{\text{in}}\le 4.255V$
• C. $4.5V\le V_{in}\le 5V$
• D. $3.5V\le V_{\text{in}}\le 5V$

Answer 1

The correct option is B $1.86V\le V_{\text{in}}\le 4.255V$


$V_{\text{out}}\text{maximum}=4.5V$

$V_C=V_{CE}=4.5V$

$\therefore I_C=\frac{5-V_C}{1k\Omega }=\frac{5-4.5}{1}\times {10}^{-3}=0.5\text{mA}$

So, the base current is,

$I_B=\frac{I_C}{\beta }=\frac{0.5}{25}=20\mu A$

$\text{Now,}{V}_B=V_{BE\left(ON\right)}=0.7V$

$\therefore I_2=\frac{V_B-(-5)}{100}\times {10}^{-3}=57\mu A$

$\therefore I_2=20+57=77\mu A$

$\therefore V_{in}=V_B+I_1\times 15\times {10}^3$

$=0.7+77\times {10}^{-6}\times 15\times {10}^3$

$=1.855V\approx 1.86V$

For saturation region,

$V_C=V_E+V_{CE(\text{(sat)}}=0.5V$

$\therefore I_C=\frac{5-0.5}{1}\times {10}^{-3}=4.5\text{mA}$

$I_B=\frac{I_C}{\beta }=\frac{4.5}{25}=180\mu A$

$I_2=57\mu A$

$\therefore I_1=(180+57)\times {10}^{-6}$

$I_1=237\mu A$

$V_{\text{in}}=0.7+I_1\times 15\times {10}^3$

$=0.7+15\times {10}^3\times 237\times {10}^{-6}$

$=4.255$

$\therefore 1.86V<V_{\text{in}}<4.255V$