Suppose we have an amplifier which uses a non-linear device to amplify voltage or current. Now suppose we connect the output of amplifier A to the input of amplifier B. We replace the transistor with its equivalent circuit made of only linear elements (small-signal analysis) preassuming the biasing of the transistor of amplifier A is fixed by the DC source of the circuit. Why do we assume that the biasing of the transistor of amplifier B is fixed by the same DC source when amplifier A amplifies its input signal resulting in a much larger input signal for the next stage?

  • 1
    \$\begingroup\$ Your question needs clarification. Why is only amplifier A using a non-linear device? Why do you assume that amplifier B is biased in the same way as amplifier A? If I understand your question at all, you are correct in that the design of amplifier B has to take into account that its input will be a higher level than that of amplifier A. \$\endgroup\$
    – Barry
    Mar 12, 2023 at 22:52
  • \$\begingroup\$ I do not understand whether you are worried about DC biasing or about input signal? When we talk of DC biasing, it is computed in the absence of an input signal. A small diagram will help explain your question better \$\endgroup\$
    – sai
    Mar 13, 2023 at 4:21

4 Answers 4


In a system of cascaded amplifiers A and B, let's say that the gain of A is 1000.

If your input signal (to A) has an amplitude of, say 1μV, you might consider that to be small, but the input to amplifier B would be 1000 times larger, 1mV in amplitude, which is definitely not small. Your concerns would be valid, since that would certainly be enough to upset the biasing of various non-linear devices in B.

The answer to your question is: if you manage to disturb the DC operating point (biasing), then the input was not small.

In the context of small signal analysis, we don't consider 1μV to be small, or even 1nV, since any element with sufficient gain can produce a significant deviation of potentials and currents from their quiescent value.

In this context "small" means so small, that no potential, and no current anywhere in the circuit is ever further than an infinitesimally small amount from it's DC operating point.

What happens when that is not the case? Well, if amplifier A is producing a signal significant enough that potentials or currents deviate by more than an infinitesimally small amount from their quiescent values, then your input signal is already too large to be called a "small signal".

That is why we use the term "infinitesimal". It means as close as it's possible to get to zero, without actually being zero. A value so small that even if you multiplied it by a number approaching infinity, it would still be infinitely close to zero.

Then, you would need elements in the circuit to have infinite gain, to be able to produce any deviation from the DC operating point, which is not feasible.

Small means really, really small, a perturbation so small that it's impossible for anything in the circuit to respond with anything that isn't itself considered small.

Technically, biasing sets conditions at a particular place on a curve off \$V\$ versus anything else, say \$I\$, and the relationship between \$V\$ and \$I\$ can be very non-linear, like this (a diode's IV curve):

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However, any small section of that curve (inside the green markers, for instance) \$V(I)\$ looks like a straight line, linear, if you constrain your position along it to a very small region:

enter image description here

This is the principle of calculus. As long as the function \$V(I)\$ is continuous, and \$dI\$ and \$dV\$ are infinitesimally small changes, the gradient doesn't change at all either side of point \$(I, V)\$, say \$(I+dI, V+dV)\$ or \$(I-dI, V-dV)\$, and you may consider that section of curve to be linear, with constant and equal gradient at all points along it.

By biasing your diode, or whatever element, at some point along its curve, the relationship between \$V\$ and \$I\$ there is perfectly linear, providing you don't deviate further than an infinitesimal distance either way along the curve.

Since our "small" signals are intended to deviate from this "operating point", to take advantage of this region of linearity, we have to ensure that those signals are small enough that no point anywhere in the circuit, even following an amplifier with gain one million, or trillion, is perturbed enough to exit its own region of linearity.

In the true sense of calculus, \$dI\$ and \$dV\$ are the amounts of deviation permitted. Those values are infinitesimally small, as close to zero as it's possible to get, without actually being zero. We are talking about a \$dV\$ that makes 1μV, 1nV and even 1pV look gargantuan.

A large answer, about small.


If the signals are no longer small signal, then small signal analysis won't be very accurate, and may not be useful at all. That applies to a one-stage amplifier as well as each stage of a multi-stage amplifier.

However if have a 100uV signal and amplify it by 20 to 2mV it's still a pretty small signal compared to bias levels in most cases.


What you do is you first analyze the amplifier performance assuming the small signal (i.e., linearized) model. Then you take the resulting signal amplitudes at (and between) various points in the circuit and you check to see if the small-signal model is still valid.

Note that this requires some judgement: if you have a transistor stage where the entire input signal shows up across the base-emitter junction (i.e., there's effectively no feedback), then depending on what you consider to be objectionable, you may start seeing objectionable amounts of distortion at \$V_{be} \pm 2\mathrm{mV}\$, or even less. But in a stage where there's heavy feedback, because the feedback will reduced distortion the variation around \$V_{be}\$ that's still acceptable could go up considerably, even to the tens of millivolts.


Excellent question, I feel like this is not emphasized enough in textbooks.

The short answer is: small-signal analysis becomes a good approximation of the amplifier characteristics when the loop gain of the negative-feedback amplifier in question is large.

Why? because a negative-feedback amplifier actually amplifies the error signal, not the input signal itself. The first stage output of the amplifier is actually the amplified version of the input error signal.

You might remember that one of the golden op-amp rules is V(+) = V(-). This is equal to saying V(+)-V(-)=0. In other words, that the amplifier's error signal is 0. This is equivalent to saying that the amplifier has an infinite open-loop gain, which would translate to an infinite loop gain.

This is obviously impossible with a practical op-amp, but it's still a damn good approximation for large open-loop gains.

Nonetheless, we know that the output signal is large, and probably a lot larger than the output of amplifier's first stage. How does the small-signal approximation hold for the subsequent stages then?

Well, the larger the swings in the subsequent stages, the worse the small-signal approximation becomes. So, you might argue that the small-signal approximation is best used for the 1st and 2nd stage, while for the 3rd stage will probably not hold to large output swings (most amplifiers don't have more than 3 stages).

You can also say something is an small signal judging by the bias voltage and variation around it. If you have some typical common-emitter amplifier with a collector held at a few volts, and your signal swing is few millivolts, you might consider approximating this with small-signal analysis. People still do so with higher swings, mostly to get a feel for what's expected, you can't expect accurate results doing so.


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