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With reference to this circuit: enter image description here

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.

Answer given: enter image description here

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?

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I will divide the answer into two parts:

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.

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  • \$\begingroup\$ I wrote the subscripts notation in word and copied here. Please help me edit properly. \$\endgroup\$ – explorer Dec 12 '15 at 14:07
  • \$\begingroup\$ Ah that makes sense. \$\endgroup\$ – Indo Ubt Dec 12 '15 at 14:37
  • \$\begingroup\$ Then is the answer given wrong? For it states that IB decreases and IC increases. IB must increase, right? And how could I increase/decrease the value of Vbe from 0.7? Will it change when I add an AC voltage in the input circuit? \$\endgroup\$ – Indo Ubt Dec 12 '15 at 14:40
  • \$\begingroup\$ yes, the "answer given" is wrong. The "answer given" should have the words "decreasing the base current" replaced with "increasing base current". \$\endgroup\$ – Marla Dec 12 '15 at 15:05
  • \$\begingroup\$ Yes, as @Marla said, the answer is wrong. Also, in saturation Ib is more than (b)Ic [beta*Ic]. You can change Vb (voltage that you measure between the base terminal and ground) but not Vbe (voltage between base terminal and the emitter). It will always be 0.7. \$\endgroup\$ – explorer Dec 12 '15 at 15:09

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