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For the circuit below, $$β = 80 \quad and \quad V_{BE} = 0.7\;V$$

Enter image description here

I applied KVL to the closed loop on the left -base-emittor loop- and the bigger closed loop on the right -collector-emittor loop- (just the bigger rectangle on the right). I thought that V0 would be equal to the voltage between GND and the wire that the upper 10k is connected to, which is V subscript CE.

So from the first loop I mentioned, $$-1+120\times10^3I_B+V_{BE}=0$$ $$-1+120\times10^3I_B+0.7=0$$ $$120\times10^3I_B=0.3$$

Thus, $$I_B=2.5\times10^{-6}\,A=2.5\,μA$$

From the equality $$I_C=βI_B$$ $$I_C=80\times2.5\times10^{-6}\,A=0.2\,mA$$

Applying KVL to the second loop I mentioned above, $$-V_0-10\times10^3I_C+20=0$$ $$-V_0-10\times10^3\times0.2\times10^{-3}+20=0$$ $$V_0=-2\,V+20\,V=18\,V$$

Thus, $$I_0=\frac{V_0}{10\,kΩ}$$ $$I_0=\frac{18\,V}{10\,kΩ} =\,1.8\,mA$$

But the book says $$V_0=12\,V \;and\; I_0=600\,μA$$

I found some people saying the book is wrong, but I wonder about the answer, and

  • How would you solve this?
  • Is the sentence in bold (above) true?
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    \$\begingroup\$ With no transistor, Vo is 10 volts. Collector current pulls Vo toward ground. The usual practice is to reference all voltages to ground, as you suggest. \$\endgroup\$ – glen_geek May 11 '18 at 22:32
  • \$\begingroup\$ It is better to give each resistor and voltage source a name and insert the values at the last moment. \$\endgroup\$ – Peter Mortensen May 12 '18 at 5:31
  • \$\begingroup\$ Even if a book fails to do it. \$\endgroup\$ – Peter Mortensen May 12 '18 at 5:40
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Well, at just first glance the answer is wrong - if \$I_0 = 600\,µA\$ then \$V_0 = I_0 * 10\,kΩ= 6\,V\$; \$6\,V\ !=\ 12\,V\$ so the answers are contradicting.

Here is the solution:

Base current is what you calculated: \$ I_b = 2.5 * 10^{-6}\,A\$

Collector current: \$ I_b*\beta = 2 * 10^{-4}\,A\$

Now \$ V_0 \$ is simply: \$ 20\,V - i10^4\,Ω \$ (\$i\$ is the current of the top most resistor)

Currents equation: \$i = i_c + i_0 \$ (\$ i_c\$ is collector current)

Next we put current equation into \$V_0\$ equation: \$V_0 = 20\,V - (i_c + i_0)10^4\,Ω\$

\$I_0\$ is \$ \frac{V_0}{10^4}\$

So we combine above three equations into: \$V_0 = 20\,V - (i_c + \frac{V_0}{10^4\,Ω})10^4\,Ω \$

\$2V_0 = 20\,V - i_c*10^4\,Ω\$

\$V_0 = 9\,V \$

and finally

\$I_0 = \frac{9\,V}{10^4\,Ω} = 900\,µA \$

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  • \$\begingroup\$ Yes, that was what I'd done to check the answer too. ^^ \$\endgroup\$ – André Yuhai May 11 '18 at 23:16
  • \$\begingroup\$ Oh, then my mistake was that I just had considered \$I_C\$ to be the current through the top most 10k resistor, I hadn't thought that it would be \$I_0+I_C\$. Everything is clear now. :) \$\endgroup\$ – André Yuhai May 12 '18 at 0:01
  • \$\begingroup\$ Good to hear :) \$\endgroup\$ – DannyS May 12 '18 at 12:34
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The first thing you do is rewrite the problem:

schematic

simulate this circuit – Schematic created using CircuitLab

That's done by simply performing the usual Thevenin equivalent on \$R_1\$ and \$R_2\$ and the voltage applied across them to compose a new Thevenin voltage source of \$10\:\text{V}\$ and Thevenin resistance of \$5\:\text{k}\Omega\$. At this point, it's quite obvious that the output cannot be more than this Thevenin voltage.

The base current is \$I_\text{B}=\frac{1\:\text{V}-700\:\text{mV}}{120\:\text{k}\Omega}=2.5\:\mu\text{A}\$. With \$\beta=80\$, this says that \$I_\text{C}=80\cdot 2.5\:\mu\text{A}=200\:\mu\text{A}\$. The drop across the Thevenin resistance (\$R_4\$ above) is \$200\:\mu\text{A}\cdot 5\:\text{k}\Omega=1\:\text{V}\$.

Therefore, the output will be \$1\:\text{V}\$ below the Thevenin voltage of \$10\:\text{V}\$, or \$V_\text{O}=9\:\text{V}\$.


You can separately analyze it by using KCL at the collector node, using the value of \$I_\text{C}\$ computed earlier (see above):

$$\frac{V_\text{O}-20\:\text{V}}{R_1}+\frac{V_\text{O}}{R_2}+I_\text{C}=0\:\text{A}$$

That solves out as \$V_\text{O}=\frac{R_2}{R_1+R_2}\cdot\left(20\:\text{V}-R_1\cdot I_\text{C}\right)\$

And it gets you the exact same answer.

From there it is quite easy to work out the requested current.


Yes, the book is wrong if that is what it said.

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  • \$\begingroup\$ +1 for thevenin method. Btw, you have mistake in base current - in the denominator you have 1k while it is 120k Ohm. \$\endgroup\$ – DannyS May 12 '18 at 8:03
  • \$\begingroup\$ Well, I have a question. I understood why the Thevenin resistance became 5k but I couldn't really get why that 20V would become 10V as the Thevenin voltage. Can you please explain? \$\endgroup\$ – André Yuhai May 12 '18 at 8:14
  • \$\begingroup\$ @AndréYuhai The Thevenin voltage is the voltage that would appear at the mid-point if the mid-point is unloaded. I think you can see that this would be 10 V. \$\endgroup\$ – jonk May 12 '18 at 19:30
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The answer is wrong. Suppose there isn´t a transistor in the circuit and it´s only both 10 kΩ resistors. Then Vo = 10 V in this situation.

Adding anything in parallel to it will cause that equivalent resistance to decrease, so it´s impossible to have Vo = 12 V, since Vo <= 10 V

As you´ve already figured out Ic = 0.2 mA, then you can apply KVL to both resistors as a loop as follows: 20 - 10 kΩ(Io + 0.2 mA) - 10 kΩ*Io = 0.

Solve for Io, and Io = 0.9 mA, and as a result, Vo = 0.9 mA * 10 kΩ = 9 V

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  • \$\begingroup\$ Then my answer \$V_0=18V\$ is wrong too. \$\endgroup\$ – André Yuhai May 11 '18 at 23:18
  • \$\begingroup\$ @AndréYuhai yes, unfortunately it cannot be greater than 10v. I´ll edit my answer and post the way I did \$\endgroup\$ – Flávio Alegretti May 11 '18 at 23:21

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