Slew Rate

When audio amplifier slew rates are specified it is usually the 'large signal slew rate', and to achieve this the input stage needs to be driven close to clipping so that it drives the maximum current into the compensation capacitor. A typical arrangement is shown next, with a differential input stage driving a voltage gain stage with a feedback capacitor used for high frequency compensation to maintain feedback loop stability. There is usually a unity gain output stage also, which is omitted here to keep the circuit simple.

To achieve the maximum slew rate using either jfet or degenerated bipolar differential input stages this will need a differential input voltage of typically hundreds of mV, and maybe in extreme cases over 1V. The currents and resistors in the circuit above have been chosen so that a 1V differential input voltage is needed to drive the input stage to clipping and drive 5mA into the compensation capacitor. The 'differential input voltage' , (Vdiff), is just the difference in voltages of the bases of the two input stage transistors, and in the diagram this is Vi - (Vo/20). The current into the 200pF capacitor as a function of differential input voltage Vdiff is shown next as a straight line approximation.

It is assumed that the current available at the amplifier output is sufficient to drive the capacitor from the other end, so the slew rate is determined only by the available input stage current. With a current I charging capacitance C we get the slew rate dV/dt = I / C. For a high slew rate we therefore need a high charging current and a low capacitance. If slew rate is in V / usec then with I in mA we need C to be in nF. With our initial example we have 5mA available from the input stage to charge 0.2nF (200pF)
This gives a slew rate (5 / 0.2) V / usec = 25 V / usec.

Suppose the maximum peak output before amplitude clipping is A, then with a sinewave signal the peak rate of change is a function of frequency, and is found by differentiating A sin(wt), which gives Aw cos(wt), which has a maximum of Aw, where w is the angular frequency (2 pi x frequency). If, for example, the peak output A is 20V and if we increase the frequency until the maximum rate of change is equal to the calculated slew rate limit of 25V/usec then the frequency is around 200kHz, and this is therefore the maximum frequency at which we can get a reasonably undistorted sinewave at the full peak output level. The closed-loop gain falls by -3dB at this frequency, and so we need Vi to be 3dB more than the 1V input voltage required at low frequencies to produce the 20V peak output. We therefore need 1.4V peak sinewave input at 200kHz to get 25V/usec at the output. At 200kHz there is a 45 degree phase difference between input Vi (1.4V pk) and output Vo (20V pk), and the differential input Vdiff is the difference between the 1.4V peak input sinewave and Vo / 20 fed back from the output. Taking the 45deg phase difference into account the difference is 1V peak, as required to drive the differential stage close to clipping and achieve the maximum slew rate. The relative phase angles and magnitudes at 200kHz are illustrated in the following diagram.

That phase shift from input to output is a vital requirement to achieve any slew rate. If there was no phase difference then both input and output would pass through zero at the same time and at that time there would be zero Vdiff, and so the input stage would be unable to drive any current into the compensation capacitor, and the rate of change of output voltage would be zero. For sinewaves and many other music signals the maximum rate of change occurs as the voltage passes through zero, and so Vdiff needs to be at its maximum at the instant when the input Vi is zero. This can only happen if the output is not zero when the input is zero. The phase shift is essential.

If the audio frequency input signal is under the 1V amplitude clipping level there should never be anything near the 1.4V peak at 200kHz needed to reach slew rate limiting, so what is the point of such a high slew rate? One reason sometimes proposed is that input stage distortion can become significant well below the slew rate limit, so having a high safety margin means that distortion at normal levels is very low. This is true, but if we wanted to make input stage distortion lower there is more than one method. We could increase the current in the input stage, using a 20mA total current, then the slew rate limit will be 50 V / usec, and if we work out the distortion at a low level it will be far below the level we had with 10mA current. There will however be drawbacks, such as higher device dissipation, higher current noise and possibly worse offset problems. Suppose instead we kept the 10mA current but replaced each transistor with a complementary feedback pair. The slew rate limit would not be changed much, but the distortion would again be reduced by a large factor. There are therefore alternative ways to reduce input stage distortion which do not involve increasing the slew rate, and there is therefore no necessary link between slew rate limit and distortion, including TID, PIM and other distortions often ascribed to input stage nonlinearity.

One commonly used improvement is to load the input differential stage with a current mirror. This doubles the available drive current and therefore appears to double the slew rate. There are however other effects, it also doubles the gain, and to maintain the original stability margin we need to either double the compensation capacitor, which returns us to the original slew rate and so has no apparent benefit, or we could double the 100R degeneration resistors, then the gain is unchanged, the slew rate limit is doubled, and input stage distortion at lower levels is reduced. The impressive slew rate specification however is in practice unatainable in the example above, it could only be achieved with Vdiff = 2V, and with a 1V peak input signal that could never happen with any possible signal whether sine, square or whatever.

Aiming for a high slew rate limit on the grounds that transients will be more accurately amplified seems reasonable, but turns out to be almost entirely wrong, as shown in Slew Rate Part 2.