How can we measure slewrate in an amplifier?

With SPICE or in real circuit?

It's pretty easy to measure on a real circuit. All you need is a function generator and an oscilloscope. Pretty fundamental... You can also look at it in simulation, but unless you're really meticulous you won't have the layout parasitics included, so the simulation will be pretty optimistic (i.e., show an unrealistically high slew rate).

Simulation can still be useful, though. If your simulation shows the circuit doesn't have sufficient slew rate you'll need to change the design so that it does.

So the question becomes: How much slew rate isneeded? I'll answer that below.

Any signal can be reconstructed using a sum of sine waves. The bandwidth of audio is typically considered to be 20 kHz. So the highest frequency your audio system will need to produce is 20 kHz. We can derive the rate of change (i.e., the slope of the curve aka the derivative) for the sine wave like this:

∂/∂t [V

_{peak} * sin(w*t)] = w * V

_{peak} * cos(w*t), where w = 2*π*f.

A sine wave has the highest rate of change at the zero crossing and the first zero crossing happens at t = 0. So let's find the maximum rate of change for the sine wave:

∂/∂t

_{max} = w * V

_{peak} * cos(w*0) = w * V

_{peak}, where w = 2πf.

So the minimum slew rate required of the circuit in order to reproduce a 20 kHz sine wave is:

SR = 2*π*20000*V

_{peak}
For a line level circuit where V

_{peak} is typically 15 V this works out to 1.88 MV/s ... or in more common units 1.88 V/µs. For a power amp specified to provide 100 W into 8 Ω (which corresponds to 40 V, peak) you need: 2*π*20000*40 = 5.02 V/µs.

Or if you prefer to use the 22.4 kHz bandwidth of a digital reproduction chain: 2.11 V/µs for line level, 5.63 V/µs for the 100 W power amp.

Tom