# Decreasing the resistance in the input circuit of a transistor and its effect on the output current and voltage

With reference to this circuit:

Question: When the resistance $R$ is decreased, how does the lamp's brightness and the voltmeter's reading change?

My answer: The input resistance of a transistor circuit is given as $r = (\frac{\Delta V_{BE}}{\Delta I_{B}})$. So when R decreases, $r$ decreases, and hence $\Delta I_B$ increases (as they are inversely proportional from the equation). From the output characteristics, an increase in $I_B$ leads to an increase in $I_C$ and therefore the current in the lamp increases; leading to it getting brighter. As $I_C$ gets higher, voltage in the output circuit increases and the voltmeter has a higher reading as compared to its initial value.

Even though my notion that the brightness and the voltmeter increases is right, my reasoning is quite wrong. How does the circuit become "more forward biased" when $R$ is decreased? And why is it that $I_B$ decreases while $I_C$ increases. Why aren't both of them increasing? And if possible, could you say why my concepts are wrong?

1) (r=(ΔVBEΔIB)) is the formula for dynamic resistance when the input is A.C. Here you should apply the KVL. (Vbb – IB Rb = 0.7). Thus, decreasing the value of R will increase (Ib). And if the transistor is in active region, Ic will be proportional to Ib and hence the bulb will glow brighter.
2) If you keep on decreasing the value of R, transistor will enter into saturation. I am guessing the book means to say saturation by using the term more forward biased. Because in saturation, (VCE) = 0.2V.