Question

In the circuit shown below, there is an open port at the emitter. Suppose Vs 5.000 V). We take Rs 1.000 k2. We then connect a resistor RL between the emitter and ground. Assume β-140, VBE-0.6 V a. Write VE,E as a function of V, RG, R b. Calculate VE, for R-1.00 k,R 2.00 k2 15v 5 V s V E

Answer 1

$\dpi{100} V_{E}$ is the voltage between emitter and ground. And $\dpi{100} V_{E}$ can be calculated as

$\dpi{100} V_{E}= V_{s} - V_{BE}$

So, From Ohm's law

$\dpi{100} I_{E} = \frac{V_{E}}{R_{s}/ \beta + R_{E}} = \frac{V_{s} - V_{BE}}{R_{s}/ \beta + R_{L}}$

(b) Given, $\dpi{100} V_{S} = 5 \, V$

$\dpi{100} R_{S} = 1 \, k \, ohm$

  $\dpi{100} V_{BE} = 0.6 \, V$

So, $\dpi{100} V_{E}= V_{s} - V_{BE} = 5 - 0.6 = 4.4 \, V$

For $\dpi{100} R_{L} = 1 \, k \, ohm$

  $\dpi{100} I_{E} = \frac{V_{E}}{R_{S}/ \beta + R_{L}} = \frac{4.4}{\frac{1 \times 10^{3}}{140} + 1 \times 10^{3}} = 4.368 \times 10^{-3} \, \, A= 4.368 \, \, mA$

For $\dpi{100} R_{L} = 2 \, k \, ohm$

  $\dpi{100} I_{E} = \frac{4.4}{\frac{1 \times 10^{3}}{140} + 2 \times 10^{3}} = 2.19 \times 10^{-3} \, \, A= 2.19 \, \, mA$

For any doubt please comment.