If an emitter current is changed by $4 mA ,$ the collector current changes by $3.5 mA$. The value of $\beta$ will be

Question

What is the output of a NAND gate when both inputs are HIGH?

Answer 1

To solve this problem, we need to understand the relationship between the emitter current $(I_E)$, collector current $(I_C)$, and the current gain $(\beta)$ of a transistor. The given data states that the emitter current changes by $4 \, \text{mA}$ and the collector current changes by $3.5 \, \text{mA}$.

In a transistor, the current relationships are given by the following equations:

$I_E = I_B + I_C$, where $I_B$ is the base current.
$\beta = \frac{I_C}{I_B}$, which is the DC current gain of the transistor.
Also, $\beta = \frac{\Delta I_C}{\Delta I_B}$

Considering the change in currents, we can express the base current change $(\Delta I_B)$ as:

$\Delta I_B = \Delta I_E - \Delta I_C$

Substitute the given changes in emitter ($\Delta I_E = 4 \, \text{mA}$) and collector currents ($\Delta I_C = 3.5 \, \text{mA}$):

$\Delta I_B = 4 \, \text{mA} - 3.5 \, \text{mA} = 0.5 \, \text{mA}$

Now, calculate $\beta$:

$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3.5 \, \text{mA}}{0.5 \, \text{mA}} = 7$

Thus, the correct value of $\beta$ is 7.

Therefore, the correct option is 7.

Answer 2