Electronic – Meaning of speed in analog circuits

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I'm often confused what the meaning of 'speed' is in analog circuits. To me it seems that the higher the cutoff frequency, the 'faster' the circuit is. In other words, a higher frequency input signal can be processed.

The reason I'm confused is that I just read about the 'speed' of an analog circuit being affected by the current. To my knowledge, the current doesn't affect the time constant (R or C) and as a result doesn't affect the bandwidth, so how can for e.g. greater current result in a 'faster' circuit? Is it something to do with the rate of change of capacitance voltage increasing i.e. \$I_c = C\frac{dV_c}{dt}\$ ? Is this related to the slew rate?

Could it be that a higher current increases the transconductance and therefore the unity-gain bandwidth?

In the digital world, the meaning of speed is very straight-forward but there seem to be too many different meanings of speed in the analog world.

Best Answer

Take an example of a simple common-emitter BJT amplifier: -

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For low frequencies, the voltage gain is fairly well approximated by RL/RE but, as frequency increases, the internal miller capacitance across the base-collector region starts to impose negative feedback and reduces gain.

If RL and RE are made ten times smaller (as a result of operating at an emitter current that is ten times greater), the frequency that which the miller capacitor starts to impose unwanted effects gets pushed up to a higher frequency.

It works the same with hanging an external load on the output - this usually means a little bit extra capacitance gets put in parallel with RL and this means the -3 dB operating frequency gets lowered. If RL and RE are much smaller (increased IE) then this can be somewhat alleviated.

So, here's an example where operating at a higher current gives a wider bandwidth.

Using a totally different example of a common collector colpitts oscillator: -

enter image description here

Fundamental to the working of this oscillator is the "hidden" resistance inside the emitter of the BJT (rE) - it is necessary to be there or the circuit won't oscillate (Barkhausen criteria not met). BUT, if the losses in the inductor are too high (such as in UHF oscillators) and rE is too large in value, the circuit won't oscillate. So, in order to reduce rE, collector current is increased by biasing the base a bit harder.

Because rE = 26 mV/Ie, a bigger collector/emitter current means a lower rE.

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