# slew rate explained

Ok, I’m going to do my best here. I remember it taking me a while to wrap my brain around this the first time *I* encountered it, so here goes…

First–*Slew Rate* is nothing more than a term used to describe how quickly the potential on a circuit node must change with respect to time. As a result, it always has units of V/S, or volts per second. You may see Volts per uS (microsecond) more commonly, since typical values of slew rate result in many hundreds of thousands or millions of volts per second. An important concept at this point is that slew rate represents the rate of change of voltage.

How do we determine the slew rate? Well, first, we’ve got to be talking about either triangle waves or sine waves. It makes no sense to talk about the slew rate of a square wave, since theoretically the slew rate of a perfect square wave is infinite. Most of the time we test our amplifying circuits out with sine waves, so let’s use that as a first demonstration.

Hold on; here comes the trigonometry…

The instantaneous amplitude of a sine wave is given by the equation:

V = Vpeak sin(wt)

where:

- Vpeak is the peak potential of the sine wave
- w is omega, also known as the angular velocity, or radian frequency, and is equal to 2*pi*F
- t is the instantaneous time of interest.

If we are interested in the slew rate, or the rate of change of V, we need to differentiate that equation with respect to time. Consulting the old tables that gives us:

dV/dt = w Vpeak cos(wt) dt

Now that rate will be a maximum when the [cos(wt)] is at a maximum, and that is when t=0, since cos(0) = 1. (We know this intuitively by picturing the sine wave, and considering where the amplitude is changing the fastest… it’s at the zero crossings.) We can then drop the cos(wt) which leaves us with:

dV/dt = w Vpeak dt

Let’s assume that we want to find out the slew rate of a 20kHz sine wave. At 20kHz, the angular velocity (w) is 2 * 3.14 * 20,000 or 125600 radians per second. Therefore, the slew rate for any desired peak output voltage will be 125600 * Vpk.

That’s all well and good… what does it mean?

Say you’ve got a single ended output circuit, using an EL34 in triode mode. The grid is biased at -15Vdc at idle. To develop full power output you want to create a peak input amplitude of 15Vpk, so that the grid just touches the 0V point. To develop that max output at 20kHz you will need to move the grid at a rate of 2 * pi * 20,000 * 15V, or 1884000 volts per second. Big goddamned number, eh? That’s why it’s easier to say 1.88V/uS.

It should be obvious that changing either the maximum frequency OR the maximum amplitude will alter the requisite slew rate, which is why it is a LARGE SIGNAL phenomenon.

How does this relate to building amplifiers?

Well, you don’t want to limit the slew rate in your circuit. The circuit must be designed to slew quickly enough to preserve the original waveform shape during transmission. What slows it down, so to speak, is capacitance, so we have to include the capacitances involved.

Let’s go back to our fictitional triode mode EL34 example above, biased at -15Vdc on g1 at idle. Let’s assume that the Cgp is around 10pF (which is higher than it would be when run as a pentode, but we’re talking about triode mode here). We know that the maximum voltage gain (mu) of the stage will be about 20 (this is given by the “triode mode mu” or “amplification of g2 wrt g1″). In a real circuit it won’t be this high, but this is not a page about Miller effect–it’s a page on slew rate and slew rate limiting. So the real-world, worst-case scenario of input capacitance will be 10pF * 21, or 210pF.

210pF represents the total capacitance of the g1 circuit node in a triode connected EL34 single ended amplifier. This capacitance is what must have CURRENT pushed into it and pulled out of it in order to change the potential. We know to develop full power from the circuit at 20kHz we need to slew at 1.88V/uS. How do we relate that to current?

The equation relating current and capacitance is:

I = C * dV/dt

Hey! Look at that! The old slew rate term pops right up! In fact, to determine the amount of current (in Amps) we need, all we have to do is multiply the slew rate (in V/S) times the capacitance (in Farads). This is equivalent to multiplying the slew rate in V/uS to capacitance in uF. In our example we have 1.88V/uS and 210pF, or 0.00021 uF to work with. The value for current is then 0.00021 * 1.88 = 0.0003948A, or about 0.4mA of current.

This means that 0.4mA of current will be used in slewing the 210pF g1 node 15Vpk @ 20kHz.

As Steve Bench points out in the other article on slew rate, a nice headroom would be about five times this value, to provide a cushion for the large peak to average ratio of music. That leaves our final answer as 2mA. If the driver stage of the amp is able to sink and source more than 2mA of current, we will be in fine shape.

Combining all the equations together we get:

I = C * 2 * pi * F * Vpeak

Notice how driving impedance NEVER ENTERS THE CONVERSATION. This is because it has nothing to do with slew rate or slew rate limiting, and only enters the discussion when we branch it out into frequency response and/or filters. The only things that matter here are the variables in the above equation–current, capacitance, frequency, and voltage.

[This is by no means finished, but it's all I have time for at the moment. Please send me some feedback if parts are unclear or need better explanations--I am sure I will be adding to this page soon.]