The transistor circuit in shown in the schematic I have included:


simulate this circuit – Schematic created using CircuitLab

I have tried solving it like this:

First I calculated the potential at node A by applying voltage divider law:

V(A)=(680/680+4.7)*20. The answer is 19.86V. Now, applying KVL in input loop:

19.86=Ib*R2+0.7V. And this gives the value of Ib to be 0.028mA.
But my text book has the value of Ib=15.45 microA. Why am I getting this much difference in the base current?(Which ultimately causes lots of difference in collector current if gain is high). This is how the textbook has solved it:enter image description here

I dont know where I am getting it wrong. Would be of great help if someone corrected me. Thanks in advance...

  • \$\begingroup\$ Can you explain to us why do you think that the voltage divider equation will work in this case? Because the voltage divider equation assumes the that we have only one current flow in the circuit (two resistors in series), which is not the case with the transistor circuit. We have one current (Ib+Ic) entering the Va node and two currents Ib and Ic that leaves the Va node. Also, why did you omit the transistor Vbe voltage? \$\endgroup\$
    – G36
    Commented Aug 31, 2018 at 13:53
  • \$\begingroup\$ Can't I assume the current entering node A i.e. (Ib+Ic) to be single current,say I? And this I is divided into Ib and Ic, just like in case of current divider.....And I applied voltage divider rule at A because when two resistors are placed in parallel, the drop across one is equal to other. So if any resistor is in series with this parallel combination, I still can apply voltage divider rule. Correct me if am wrong please. \$\endgroup\$
    – G-aura-V
    Commented Aug 31, 2018 at 17:28
  • 1
    \$\begingroup\$ @GauravBhattarai you do not have two resistors in parallel. Actually, you don't have any two components in parallel. \$\endgroup\$ Commented Aug 31, 2018 at 18:05
  • \$\begingroup\$ @Kevin Cruse would you kindly explain further? \$\endgroup\$
    – G-aura-V
    Commented Sep 1, 2018 at 1:31

3 Answers 3


19.86=Ib*R2+0.7V. And this gives the value of Ib to be 0.028mA. But my text book has the value of Ib=15.45 microA.

If only it was a simple potential divider.

Why am I getting this much difference in the base current?

Because, as you start drawing base current the collector current rises and drags down point A to a lower value. It's not just a simple potential divider in that respect.

You could iteratively solve this by recalculating the collector voltage to use in the "next step" - clearly it would be lower than 19.86 V and this would lead to a lower base current and that in turn would lead to a lower collector current.

Iterate a few cycles and see what you get. I might suggest you use a spreadsheet.

  • \$\begingroup\$ Would you help me with some articles regarding the wrong ways of using potential dividers ?...I really cant find any of those in websites. \$\endgroup\$
    – G-aura-V
    Commented Aug 31, 2018 at 17:05

I'll start with a traditional approach: nodal analysis. I'll end with a simple voltage divider approach that may surprise you, a little. Both approaches result in the same answers. But it can show you how to "see" this as a simple voltage divider between two resistors, if you make one adjustment to account for the BJT.


Staying with the NPN, as shown, we can develop the nodal equation for the collector voltage, \$V_\text{C}\$ as:

$$\frac{V_\text{C}}{R_\text{B}} + \frac{V_\text{C}}{R_\text{C}} + I_\text{C} = \frac{V_\text{CC}}{R_\text{C}} + \frac{V_\text{B}}{R_\text{B}}$$

Since \$I_\text{C} = \beta\cdot I_\text{B}\$ and since \$I_\text{B}=\frac{V_\text{C}-V_\text{B}}{R_\text{B}}\$, we can transform the above equation to:

$$\begin{align*} \frac{V_\text{C}}{R_\text{B}} + \frac{V_\text{C}}{R_\text{C}} + \beta\cdot I_\text{B} &= \frac{V_\text{CC}}{R_\text{C}} + \frac{V_\text{B}}{R_\text{B}}\\\\ \frac{V_\text{C}}{R_\text{B}} + \frac{V_\text{C}}{R_\text{C}} + \beta\cdot \frac{V_\text{C}-V_\text{B}}{R_\text{B}} &= \frac{V_\text{CC}}{R_\text{C}} + \frac{V_\text{B}}{R_\text{B}}\\\\ V_\text{C}\left(\frac{1}{R_\text{C}} + \frac{\beta+1}{R_\text{B}}\right) &= \frac{V_\text{CC}}{R_\text{C}} + V_\text{B}\cdot\frac{\beta+1}{R_\text{B}} \end{align*}$$

At this point, we could just solve for \$V_\text{C}\$. (We know in advance that \$V_\text{B}=V_\text{BE}\$, but let's worry about that detail later.)


Something else looks interesting in the above equation. What if we substituted in a new variable, \$R^{'}_\text{E}=\frac{R_\text{B}}{\beta+1}\$? (This is the equivalent of moving \$R_\text{B}\$ through the base-emitter of the BJT and into the emitter leg.)

Then we'd have:

$$\begin{align*} V_\text{C}\left(\frac{1}{R_\text{C}} + \frac{1}{R^{'}_\text{E}}\right) &= \frac{V_\text{CC}}{R_\text{C}} + \frac{V_\text{B}}{R^{'}_\text{E}}\\\\ V_\text{C} &= \left(\frac{V_\text{CC}}{R_\text{C}} + \frac{V_\text{B}}{R^{'}_\text{E}}\right)\cdot\left(R_\text{C}\:\mid\mid\:R^{'}_\text{E}\right)\\\\ &=\frac{V_\text{CC}\cdot R^{'}_\text{E}+V_\text{B}\cdot R_\text{C}}{R_\text{C}+R^{'}_\text{E}}\\\\ \text{applying the fact that }V_\text{B}&=V_\text{BE},\\\\ V_\text{C}&=\frac{V_\text{CC}\cdot R^{'}_\text{E}+V_\text{BE}\cdot R_\text{C}}{R_\text{C}+R^{'}_\text{E}} \end{align*}$$

This is exactly the same equation you'd get if you had a voltage divider of two resistors sitting between a voltage of \$V_\text{CC}\$ and of \$V_\text{B}=V_\text{BE}\$!

By moving \$R_\text{B}\$ from a base-perspective, through the BJT, to the emitter-perspective, we've transformed the problem into a simple voltage divider problem using the original collector resistor and an equivalent emitter resistor sitting between \$V_\text{CC}\$ and one base-emitter junction voltage above ground.


simulate this circuit – Schematic created using CircuitLab

The reason this works so conveniently is because the collector resistor and the transformed emitter resistor both carry the same current. (The collector resistor carries the collector current plus the base current and the transformed emitter resistor carries the emitter current, which is the same value, of course.) So now the resistor divider assumption (current is the same in both resistors) once again is valid, allowing that approach to be successfully applied.


If you apply the above equations, you will get results quite similar to what your book says. I've avoided re-developing their equations, since they perform that development for you, already.

The problem you failed to realize in thinking about the divider is that in the usual resistor divider situation, the current is the same in both resistors that are part of the divider. However, in this case, the current in the collector resistor is very much more than the current in the base resistor. So, the underlying assumption made in the usual resistor divider case is false, here, because of the BJT's action in the circuit.

However, I've just shown you how you can recover the idea of a simple resistor divider by adjusting the base resistor's value, appropriately, due to the differences in current. (You also need to account for the base-emitter voltage drop, too.)


Like all transistors, the 2N3904 has a range of current gain so you do not know its base current. Your book simply guessed the typical current gain and guessed the base current.

If Node A is 10V then the collector current is 1(20V - 10V)/4.7k= 2.1mA and if the current gain is 200 then the base current is 2.1mA/200= 10.5uA. But 10.5uA in the 680k base resistor needs a voltage across the resistor to be (10.5uA x 680k=) 7.14V which it is not, since it would be (10V - 7.14V=) 2.86V then the current gain is actually less, the book guessed the current gain is 120.

  • \$\begingroup\$ -1 I don't think you really addressed the OP's question about why the voltage divider approach didn't work. \$\endgroup\$ Commented Aug 31, 2018 at 16:02
  • \$\begingroup\$ The current gain is mentioned in the question actually...and that's all the question has mentioned.The remaining question is the diagram itself \$\endgroup\$
    – G-aura-V
    Commented Aug 31, 2018 at 16:54

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