I want to solve the following example circuit not using node analysis or other methods but calculating input impedance/output impedance and gain of the circuit. The circuit has no AC source and is purely resistive to keep things simple. I shared my findings below and explain my reasoning. My experimentations failed. Can you point out my mistake and help me understand?

input impedance while Vsource is intact = 193 ohm input impedance while Vsource is grounded = -315 ohm output impedance while Vsource and Vinput is intact = 125 ohm output impedance while Vsource and Vinput is grounded = -333 ohm output impedance while only Vinput is grounded = 66 ohm

Gain while Vsource is intact and no load (open circuit gain) = 3.66 (test input voltage is 1V)

When input voltage is 2V, the output voltage is 4V.

The reasoning:

We learned that BJT amplifiers can be shown like this in the picture below and I want to believe that similar thing can be applied to DC circuits. However, it does not. I do not know what to ground and what to keep and how I will construct simplified model like in the picture below.

the circuit

The reasoning

  • \$\begingroup\$ When calculating input impedance the source must be regarded as being 0 volt. \$\endgroup\$
    – Andy aka
    Feb 12 at 17:04
  • \$\begingroup\$ @Andyaka You mean the power source, don't you, rather than the input signal source? Just so we are clear. \$\endgroup\$ Feb 12 at 17:50
  • \$\begingroup\$ It's called Vsource in the question so I called it source. Nice and logical I thought. \$\endgroup\$
    – Andy aka
    Feb 12 at 17:56
  • 1
    \$\begingroup\$ It should be possible in a circuit simulator to insert ammeters in the input port and output port. Then the load resistor can be varied from effectively infinite to effectively zero. This treats the two port network as a "black box" to find its input resistance and output resistance. If Vsource (Vs) Vs = kVin where k is a constant then the gain factor may be constant independent of Vin. But if Vs is a fixed independent voltage then the gain may be a function of Vin. In general the black box treatment of a two port network assumes no independent sources inside the black box network. \$\endgroup\$ Feb 12 at 23:13

1 Answer 1


A small signal model is a differential equation one, applied so that you can simplify analysis and use linear algebraic tools. (Solving stacks of non-linear equations is hard.)

Assume the purplish curve below (I'm no art major, nor am I female and blessed with demonstratively better color discrimination, so I've honestly no clue what to call it) is the actual non-linear behavior curve for some device:

enter image description here

If, for some reason, you happen to know a priori that the current operating point for this device is located where the bluish (same excuse) star is located, then you can place a little tangent line there (shown in black -- I think I can get that color right.) That tangent line isn't correct, of course. But at least it is simple. And, if you don't stray too far left or right of the bluish star then the black tangent line isn't terrible. It works well enough for tiny changes away from the operating point.

This is what small model should always mean to you. A simplification that allows non-linear devices to be treated as linear devices for the purpose of small changes about some operating point.

It's almost never applied to linear devices like ideal resistors because they are already linear. There's no need for a small model view.

In your case, for example, you can simplify both left and right "ends" by taking their Thevenin equivalents:


simulate this circuit – Schematic created using CircuitLab

Here, since your source resistance is zero, \$V_{_\text{OUT}}=\frac{V_{\text{IN}}\,R_2+V_2\,R_3}{R_2+R_3}\$. If you want to know how \$V_{_\text{OUT}}\$ changes with respect to small changes in \$V_{_\text{IN}}\$ then just take the derivative and find that \$\text{d}\,V_{_\text{OUT}}=\frac{R_2}{R_2+R_3}\,\text{d}\,V_{\text{IN}}\,\$. So the gain, so to speak, is \$A_v=\frac{\text{d}\,V_{_\text{OUT}}}{\text{d}\,V_{_\text{IN}}}=\frac{R_2}{R_2+R_3}=\frac13\$. That's the slope (or tangent.)

Of course, you still need to know the initial operating point. Here, since resistors are entirely linear you can pick any starting point. It's easier to just assume that \$V_{_\text{IN}}=0\:\text{V}\$ as a good starting point (for the bluish star) and find that in this case the output voltage is \$V_{_\text{OUT}}=\frac{0\:\text{V}\,R_2+V_2\,R_3}{R_2+R_3}=5\:\text{V}\frac{R_3}{R_2+R_3}=3\frac13\:\text{V}\$.

So know you know the equation is \$V_{_\text{OUT}}=\overset{\text{gain}}{\frac13} V_{_\text{IN}}+\overset{\text{offset}}{3\frac13\:\text{V}}\$. The gain is \$\frac13\$ and the offset is \$3\frac13\:\text{V}\$.

Let's move towards your suggested model, though.

The impedance seen by \$V_{_\text{IN}}\$ is \$R_{_\text{IN}}=R_1\mid\mid\left(R_2+R_3\right)=\frac{6000}{19}\:\Omega\$. The impedance seen by \$V_{_\text{OUT}}\$ is \$R_{_\text{OUT}}=R_2\mid\mid\left(R_1+R_3\right)=\frac{7000}{19}\:\Omega\$. So those two values are established and can be used to comport with your model.

The problem is that there is also an offset. And your model doesn't account for that fact.

We can get the input and output resistances but we cannot use your model to generate the large-scale output. We can use the gain to work out how things will change with small changes in the input voltage. But we cannot use your model to work out the large-scale voltage output. Only the relative change of the output given a change at the input.

That's normal enough. If all you want to know is how the output changes with respect some given change at the input, then you have the gain factor already in hand.

If, instead, you want the output operating point for any input voltage then you are asking a large-scale model question and no longer a small-scale model question.

And there your model fails.


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