I know this might be a bit too specific of a question, but I have trouble calculating \$V_o/V_i\$ in the given scheme.

I was given a BJT-transistor and had to find an equivalent AC-scheme (i.e. eliminating conductors and DC sources).

Now, I am asked to calculate \$V_o/V_i\$ as seen in the image.

Can anyone give some tricks where to start etc. and/or perhaps give a solution? There are no numerical values, so the solution should be in 'symbolic' form. The part which confuses me is the 'current controlled current source' and the resistor \$R_B\$ (more precise, how to bring it in to account).

enter image description here

  • \$\begingroup\$ Where's i_B in the circuit? \$\endgroup\$ – Andy aka Jan 7 '16 at 21:14
  • \$\begingroup\$ Between the top two left points, left to right (is that clear, else I'll make a new picture) \$\endgroup\$ – Robin Haveneers Jan 7 '16 at 21:28
  • \$\begingroup\$ The equivalent diagram represents one of the classical BJT amplifier schemes: Common emitter with voltage controlled current feedback (Resistor RB). The calculation would not too complicated in case RB>> Rc||Rl. However, without this simplification, it would be rather involved. \$\endgroup\$ – LvW Jan 7 '16 at 21:31
  • \$\begingroup\$ The exercise states "Let's assume the transistor is correctly set up with \$V_{cc}, R_b, R_c \$ . Not really sure what that's supposed to mean.. ( \$V_{cc}\$ obviously left out because this is the AC-equivalent.) \$\endgroup\$ – Robin Haveneers Jan 7 '16 at 21:35
  • \$\begingroup\$ R_B is a feedback resistor. I remember generally how to solve these problems, but haven't done it in ages. You start by opening the feedback loop. Determine the open-loop gain (gain with R_B open). Then you need to determine the feedback type and feedback factor. Then use feedback factor to calculate the closed loop gain. Wikipedia might help. en.wikipedia.org/wiki/Negative_feedback_amplifier \$\endgroup\$ – mkeith Jan 7 '16 at 23:10

"Can anyone give some tricks where to start etc"

OK - here is a rough description of my approach.

  • At first calculate the gain Ao from base to collector using the known formula transconductance x resistance Rtot (total at the collector): Ao=-gm*Rtot. Transconductance gm=beta/Rpi.

  • Rtot=Rc||RL||Rbc with Miller-resistor at the collector node Rbc=Rb/(1-1/Ao), Ao negative!

  • Hence: Ao=-gm/(1/Yc)+gm/(1/YL)+gm*(1-1/Ao)/(1/Yb) . This equation must be solved for Ao (Ao appears twice!).

  • As the last step, you calculate the input resistance Rin=Rpi||Rbb at the base node using the Miller-Theorem (at the base) with Rbb=Rb/(1-Ao).

  • Finally, you can compute the resulting gain A=Ao*Rin/(Rin+Rs)

EDIT_1 (comment): Vo/Vi=Ao because Vi=V at the base node.

EDIT_2: Because you specifically ask for the exact role of Rb:

The classical procedure to calculate the output voltage Vo leads to a negative value (with respect to Vi) because the gain Ao is inverting. However, there is another unwanted part of Vo which is NOT inverted. This part is not yet included in my above calculation. This part of the output voltage arrives DIRECTLY at the collector node (not via the transistors collector current) and is found using the simple voltaged divider rule Vi*Rtot/(Rtot+Rb). As you can see, this (positive) part disappears for Rb>>Rtot.

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Although conventions for currents and voltages are arbitrary indeed, I prefer looking at a circuit of which I 'understand' and feel familiar with the notations and polarities. Therefor I'd like to redraw the circuit, defining the BJT's base current as \$ i_b \$, replacing \$ R_B \$ and \$ i_B \$ with \$ R_{fb} \$, and \$ i_{fb} \$ respectively, as they pertain to feedback. In the end these choices will not change the outcome \$ \displaystyle \frac{V_o}{V_i} \$.


simulate this circuit – Schematic created using CircuitLab

With Eric Best's approach $$ i_{fb} + i_{R_o} + \beta i_b = 0 $$ hence $$ V_o / R_o + (V_o-V_i)/R_{fb} + \beta V_i/R_\pi = 0 $$


$$ V_o \left( \frac{1}{R_o} + \frac{1}{R_{fb}} \right) = V_i \left( \frac{1}{R_{fb}} - \frac{\beta}{R_\pi} \right) $$

and indeed

$$ \frac{V_o}{V_i} = \displaystyle \frac{\frac{1}{R_{fb}} - \frac{\beta}{R_\pi}}{\frac{1}{R_c} + \frac{1}{R_l} + \frac{1}{R_{fb}}} $$ which after replacing \$ R_{fb} \$ back with \$ R_B \$ becomes $$ \frac{V_o}{V_i} = \displaystyle \frac{\frac{1}{R_B} - \frac{\beta}{R_\pi}}{\frac{1}{R_c} + \frac{1}{R_l} + \frac{1}{R_B}} $$ which is exact confirmation of Eric Best's result.

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  • \$\begingroup\$ But what if you want to find Vo/Vs instead? \$\endgroup\$ – G36 Dec 25 '18 at 7:31
  • \$\begingroup\$ @joe electro Yes, very good. In my solution I was just keeping the attitude/conventions of the original question. In one of my comments I wrote "it is more usual, in my opinion, to choose the opposite polarity than that chosen in the diagram.". Your solution fully corresponds with common conventions. \$\endgroup\$ – Eric Best Dec 25 '18 at 13:21
  • \$\begingroup\$ @G36 See my EDIT (continuation of my original answer), where you can find my solution concerning Vo/Vs. \$\endgroup\$ – Eric Best Dec 25 '18 at 21:26

The current controlled current source (CCCS) represents the amplification property of the BJT and in this case, it means that the current through this source is given as \$ \beta \cdot I_b \$ (as it is stated in the diagram, \$ I_b \$ being the BJT's base current). However, realize that \$ I_b \$ is the current flowing only through \$ R_\pi \$ (downwards - this is determined by given relation to the direction of the current through CCCS drawn in the diagram) because \$ R_B \$ is an external element (not part of the BJT)! See the Andy aka's question "Where's i _ B in the circuit?" and the Robin Haveneers's answer "Between the top two left points, left to right" - which must be wrong from the reason mentioned above.

So, this base current \$ \displaystyle I_b \$ is given as \$ \displaystyle -V_i / R_\pi \$ (I have applied the \$ \displaystyle V_i \$ polarity used in the diagram, the same later below with the \$ V_o \$ polarity).

Let's define \$ R_o = R_c || R_l \$ (just to simplify it a bit)

Then we can write:

  1. For the current flowing through \$ R_B \$ (to the left; its direction is arbitrarily chosen by me):

    • \$ \displaystyle I_{R_B}=\frac{V_i-V_o}{R_B} \$.
  2. For the current flowing through \$ R_o = R_c||R_l \$ (upwards and to the left, its direction is arbitrarily chosen by me again):

    • \$ \displaystyle I_{R_o} = \frac{V_o}{R_o} \$.
  3. For the current flowing through the CCCS (downwards, already stated above, its direction is determined in the diagram):

    • \$ \displaystyle I_{CCCS}=\beta \cdot I_b = -\beta \cdot \frac{V_i}{R_{\pi}} = -g_m \cdot V_i \$,

the transconductance \$ g_m \$ being defined as \$ \displaystyle \frac{\beta}{R_\pi} \$.

Since according to KCL: \$ \displaystyle I_{R_o}=I_{R_B}+I_{CCCS} \$, we obtain:

\$ \displaystyle \frac{V_o}{R_o} = \frac{V_i-V_o}{R_B} - \beta \cdot \frac{V_i}{R_{\pi}} \$,

and, after some rearrangement:

\$ \displaystyle \frac{V_o}{R_o} + \frac{V_o}{R_B} = \frac{V_i}{R_B} - \beta \cdot \frac{V_i}{R_{\pi}} \$

From here it is already very easy to express the required gain:

\$ \displaystyle A_V = \frac{V_o}{V_i} = \frac{\frac{1}{R_B} - \frac{\beta}{R_{\pi}}} {\frac{1}{R_B} + \frac{1}{R_o}} = \frac{\frac{1}{R_B} - \frac{\beta}{R_{\pi}}} {\frac{1}{R_B} + \frac{1}{R_c} + \frac{1}{R_l}} = \frac{\frac{1}{R_B} - g_m} {\frac{1}{R_B} + \frac{1}{R_c} + \frac{1}{R_l}} \$

EDIT (Dec 25 2018):

If we want to know the relation between \$ V_o \$ and \$ V_s \$ (as G36 mentioned in his comment), we can continue the following way:

Using KCL, we can write: \$ \displaystyle I_{R_B} = I_b + I_{R_S} \$ (with \$ I_{R_S} \$ flowing downwards; chosen by me)

Further, using the voltage polarities from the original diagram (notice the \$ V_s \$ polarity in relation to \$ V_i \$), we obtain:

\$ \displaystyle \frac{V_i - V_o}{R_B} + \frac{V_i}{R_\pi} + \frac{V_i + V_s}{R_S} = 0 \$

We already know the relation between \$ V_o \$ and \$ V_i \$ (\$ A_V \$), so it's easy to express either the "overall" gain \$\displaystyle \frac{V_o}{V_s} \$ or just the partial one \$\displaystyle \frac{V_i}{V_s} \$, substituting for either \$ V_i \$ or for \$ V_o \$ in the above equation and rearranging it to get the required voltage ratio:

  1. \$ \displaystyle V_i = \frac{V_o}{A_V} \$

    \$ \displaystyle \frac{V_o}{V_s} = \frac{1}{\frac{R_S}{R_B} - \frac{1}{A_V} \cdot [1 + R_S (\frac{1}{R_B} + \frac{1}{R_\pi})]} \$

  2. \$ \displaystyle V_o = {A_V}{V_i} \$

    \$ \displaystyle \frac{V_i}{V_s} = -\frac{1}{1 + R_S (\frac{1}{R_\pi} + \frac{1 - A_V}{R_B})} \$

(Notice: \$ V_s \$ is a voltage of an ideal voltage source (thus it's the internal voltage of it) and \$ R_S \$ evidently represents here the internal resistance of that source. It determines an attitude to the above derived formulas - I understand the only reachable/accessible voltage is \$ V_i \$ in fact.)

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  • \$\begingroup\$ I suggest to write \$ R_l \$ instead of \$ R_I \$, as my guess is that \$ R_l \$ is the load's resistance. In the same line I'd suggest to use \$ R_o \$ ('o' for 'out') instead of \$ R_{tot} \$; it's shorter and me seems more to the point. Just some minor suggestions, still going through your calculations. I also assume the current \$ I_{Rb} \$ is flowing to the right, instead of to the left. \$\endgroup\$ – joe electro Dec 24 '18 at 5:17
  • \$\begingroup\$ If you define \$ I_{R_{tot}} \$ upward and to the left, it should read \$ -\frac{V_o}{R_{tot}} \$. Regarding point 3.: \$ I{CCCS} \$ I'm getting confused now. Please illustrate your answer with a drawing indicating the polarities and current directions. \$\endgroup\$ – joe electro Dec 24 '18 at 5:21
  • \$\begingroup\$ May I inquire your voltage polarity convention? Is the positive polarity located at the tail of the arrow? Because that would lead to \$ I_b = \frac{V_i}{R_b} \$ if \$ I_b \$ is defined downward through \$ R_\pi \$. Otherwise I think it should be that \$ I_b = - \frac{V_i}{R_b} \$. Also assuming that with 'b' (small capital) you mean the BJT's base, and with 'B' you mean the feedback resistor. \$\endgroup\$ – joe electro Dec 24 '18 at 5:29
  • \$\begingroup\$ @joe electro I have made some changes according to your advice (I -> l, tot -> o), but: 1. You wrote: "I also assume the current \$ I_{R_b} \$ is flowing to the right, instead of to the left." - No, the current direction can be chosen arbitrarily, what I did. The result direction is given by its sign (if positive, then it is identical with the chosen one, if negative, then it is opposite). (to be continued) \$\endgroup\$ – Eric Best Dec 24 '18 at 21:44
  • \$\begingroup\$ @joe electro (cont.) 2. You wrote: "If you define \$ I_{R_{tot}} \$ upward and to the left, it should read \$ −\frac{V_o}{R_{tot}} \$.“ – No again, have a look at the polarity of \$ V_o \$ in the diagram (it means \$ V_o \$ is located at the lower connection of the resistors in fact, being referenced to the higher connection of them). It is a convention that the arrow points from a point with a „higher voltage“ (or potential, if you like) to a point with a lower one. (to be continued) \$\endgroup\$ – Eric Best Dec 24 '18 at 21:45

Use KCL/KVL. There are only two unknown currents, so just need two meshes for KVL. The current source should not be in either of those two meshes. BTW, are you sure that \$I_B\$ does not flow through \$R_\pi\$?

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  • \$\begingroup\$ I agree with Chu - the current \$ I_B \$ flows for sure only through \$ R_\pi \$ - this is the actual base, \$ R_b \$ is an external element! \$\endgroup\$ – Eric Best Dec 24 '18 at 0:31

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