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QUESTION In the circuit, if current gain βdc of transistor amplifier circuit is 200, then what will be the voltage drop across resistor R under quiescent conditions? Assume quiescent base current is 1 mA.

enter image description here

ANSWER okay now if Ib=1mA. Ic=(Beta)*Ib Ic=200mA which means voltage across the resistor 4.7 k-ohm will be (200/1000)*4.7*1000(since resistance was in k-ohm). well this much voltage drop is not possible. i know my analysis is going wrong, but how? i am sure i am missing something but what is the fault in my method and what is the correct approach for this question?

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    \$\begingroup\$ If the base-current is 1mA then the voltage drop is I*R. \$\endgroup\$
    – sstobbe
    Jun 5 '17 at 0:47
  • \$\begingroup\$ It's saturated, then. \$\endgroup\$
    – jonk
    Jun 5 '17 at 0:48
  • \$\begingroup\$ At Ib = 1uA, Ve = 0.2 volts, Vc ~~ +9 volts. ............. At Ib = 10uA, Ve = 2v, Vc = 12 - 2* 4.7 = 12 - 9.4 = 2.6volts, and already the device is on edge of saturation what with Ve + Vbe > Vc. [ 2v + 0.6 > 2.6v ] \$\endgroup\$ Jun 5 '17 at 4:23
  • \$\begingroup\$ Correction: at Ib = 1uA, Vc = 11.1 volts. \$\endgroup\$ Jun 5 '17 at 4:33
  • \$\begingroup\$ It it possible the question actually says the quiescent collector current is 1 mA? \$\endgroup\$
    – Evan
    Jun 5 '17 at 5:10
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It would be nice if you'd use the schematic editor that's available here, when you are forming (or editing) your question. I'll use it right now:

schematic

simulate this circuit – Schematic created using CircuitLab

It's pretty clear that the BJT is saturated, as shown on the right. From the right diagram, we have this nodal analysis:

$$\frac{V_E}{R_E}+\frac{V_E+V_{CE}}{R_C}=\frac{V_{CC}}{R_C}+1\:\textrm{mA}$$

The solution of which is:

$$V_E=\left(V_{CC}-V_{CE} + R_C\cdot 1\:\textrm{mA}\right)\cdot\frac{R_E}{R_E+R_C}$$

We also know that \$V_B=V_E+V_{BE}\$. One might quibble about the exact value of \$V_{CE}\$ and \$V_{BE}\$, but in this case it's not worth quibbling over. For now, set \$V_{CE}=0\:\textrm{V}\$ and \$V_{BE}=700\:\textrm{mV}\$. (You can always apply different values, later.) This works out to:

$$\begin{align*} V_E&=\left(12\:\textrm{V}-0\:\textrm{V} + 4.7\:\textrm{k}\Omega\cdot 1\:\textrm{mA}\right)\cdot\frac{1\:\textrm{k}\Omega}{1\:\textrm{k}\Omega+4.7\:\textrm{k}\Omega}\\\\ &\approx 2.93\:\textrm{V}\\\\ V_B&=V_E+V_{BE}\\\\ &\approx 3.63\:\textrm{V} \end{align*}$$

From that, it's clear that:

$$V_{R_B}=V_{CC}-V_B=12\:\textrm{V}-3.63\:\textrm{V}\approx 8.37\:\textrm{V}$$

Which isn't the answer you say it should be.

As a double-check, let's compute a few things:

$$\begin{align*} I_C&=\frac{V_{CC}-V_E-V_{CE}}{R_C}\\\\ &=\frac{12\:\textrm{V}-2.93\:\textrm{V}-0\:\textrm{V}}{4.7\:\textrm{k}\Omega}\\\\ &\approx 1.93\:\textrm{mA}\\\\ I_E&=I_B+I_C\\\\& \approx 2.93\:\textrm{mA}\\\\ V_E&=R_E\cdot I_E\\\\ &=1\:\textrm{k}\Omega\cdot 2.93\:\textrm{mA}\\\\ &\approx 2.93\:\textrm{V} \end{align*}$$

That last value confirms the original calculation made at the outset, which demonstrates that it all works out.


You can play around with the inputs of \$V_{CE}\$ and \$V_{BE}\$ and I think you'll see it doesn't really change that much. Either you didn't provide full and complete and accurate information here, or else there is a problem with the answer you mentioned in one of your comments here.

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  • \$\begingroup\$ well, you are right the answer given is wrong then i guess. \$\endgroup\$
    – Saad Anwar
    Jun 5 '17 at 3:20
  • \$\begingroup\$ @SaadAnwar Hopefully, you can get some attention applied to it and save some others the trouble you faced. I hope these results make sense to you and to others there. \$\endgroup\$
    – jonk
    Jun 5 '17 at 3:30
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Your answer is unphysical for another reason -- it implies the transistor would go below its saturation voltage!

This means the transistor is saturated, and the collector current is not necessarily = beta * base current. (It can be substantially less.)

However we are told the base current is 1mA, so the drop across the resistor R is R/1000. Seems like kind of a trick question.

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  • \$\begingroup\$ if it under saturation the please give the answer of this question i.e. voltage drop accross R. well, the answer given is 10.3v. but what is the step to reach to that answer? \$\endgroup\$
    – Saad Anwar
    Jun 5 '17 at 0:53
  • \$\begingroup\$ @SaadAnwar You can't reach that answer. If that were true, then \$V_E=1\:\textrm{V}\$ (assuming \$V_{BE}=700\:\textrm{mV}\$.) But that would mean that there cannot be any collector current since then \$I_E=1\:\textrm{mA}=I_B\$ and \$\therefore I_C=0\$. But under saturation (which also must be true), there would be collector current. So the answer cannot be as you say. \$\endgroup\$
    – jonk
    Jun 5 '17 at 1:27
  • \$\begingroup\$ @jonk this is the same thing which i thought after seeing the answer.so, i guess the answer given is wrong then. \$\endgroup\$
    – Saad Anwar
    Jun 5 '17 at 3:18
  • \$\begingroup\$ Or the question is wrong. It seems strange to ask for a voltage drop across the resistor which they give you the current through, since it makes the rest of the circuit irrelevant. \$\endgroup\$
    – Luke Wren
    Jun 28 '17 at 15:17

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