Posted 12.14.2008 8:31 am

26 Comments

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.]

Get a Trackback link

- Pingback: What about ADSR envelope of the tones transmitted by pickups? on October 16, 2012

## 25 Comments

Ray Moth, December 20, 2008:

Ken,

Thanks for that. It’s the first readable explanation I’ve seen on slew rate and I think I’m beginning to “get it”.

I’m a tube audio hobbyist and an active member of Audio Asylum and diyAudio Forums. I’ve seen quite a few warnings about slew rate, but never an explanation of what it means. It was mainly in the context of warning against the use of a low-current tube such as 12AX7 or 6SL7 as a long tail pair phase splitter driving output tubes directly, such as in the Mullard 5-20. A higher-current tube such as 12AT7 or 6SN7 was recommended instead, because of lew rate – bur not knowing what slew rate was supposed to mean, I could only guess.

Ken, December 22, 2008:

ray, i’m glad you enjoyed/got something out of the article.

See AlsoOp Amp Specifications- Slew RateUnderstanding Slew Rate Maximum Frequency And Its Impact On Circuit Performance | SoundScapeHQ1 Move for Every MLB Team Before the 2024 Trade DeadlineUnderstanding Slew Rate in Op-Amp and How to Measure and Calculate it for your Designsthe real danger comes when you wrap some loop NFB around slewing stages… basically the amplifier’s error signal will become HUGE when the slewing occurs, and this can push the operating points even further into non-linearity.

basically you end up with one big mess.

the compounding effect of a slewing NFB amp is a lot less of a concern with modern diy designs, which are typically open loop. the stage still slews, but it’s not made worse as the amp valiantly tries to make output match the input (and absolutely cannot).

slew-limited stages will turn sine waves into triangles if you view the output with an o-scope. the other telltale sign is that the output cleans up considerably when you reduce the test frequency, even a little bit. in fact, putting a pole or HF rolloff previous to the slewing stage is one way to “treat” it.

by contrast, the amp that is just clipping and distorting from being overdriven will not clean up as the frequency is lowered to the midband. the only way an amp in that situation will decrease in THD is when the input signal amplitude is reduced.

J. Fenlon, February 19, 2009:

I’m an electonics major in my second year of school. I’ve been pouring over websites and digging through my text book for quite some time trying to get my slew rate numbers to work out properly. Your explination really cleared things up for me. I appreciate it greatly. Thanks Ken!

mohamad, May 3, 2009:

hello and thanks for youre xplanation

I had a question: what about a square wave?

in that case, SR of rising edge and falling edge are not same. why? and how?Ken, May 4, 2009:

mohamad,

in the case of a square wave with asymmetrical rise/fall rates, i would look at your circuit’s ability to sink and source current.

generally most circuits are active/passive pull up/pulldown. in other words, take a typical common cathode/emitter/source topology, with an RC coupled load. in such a circuit the load can be actively pulled down, but only passively pulled up.

say your stage had a slow(er) pull up… for greater symmetry, you could increase the standing/quiescent current through the stage, and use a smaller load resistance, thereby increasing the current available from the + rail to be passed through to the load when the active element “turns off.”

also, a push-pull circuit could be used. for all but a few directly coupled OPT-less circuits, with tubes this means using center tapped transformers (since no “complimentary” devices exist as they do with SS).

even then, true symmetry is hard to attain because of subtle differences in the “complimentarity” of those SS parts (as you probably know, a hole does not behave exactly like an electron, and vice versa).

finally, if this is for music applications, one must ask one’s self if the added complexity is worth the decrease in even order distortion. does the new circuit trade off low, even order distortion for higher, odd + even order distortion? if so, look (and preferably LISTEN) carefully to determine if the change is justified.

marie, August 7, 2009:

Hi Ken,

Thanks, nice explanation. Are the terms slew rate and rise time interchangeable?Jason, January 21, 2010:

i am a little confused with a part of this,or maybe i am helping finish it. when you say that the valve has the ability to sink and source. i interpreted to mean that i would have to swing at least 5 * i peak as i go to v peak. ok heres the math; (simple as i am)

your quoted

I = C * 2 * pi * F * Vpeak

and using your figures

I = 0.598mA

now using a little transposition and ohms law to determine the value of RL

Vpeak / ( 5 * I ) = 7K578

now i have a problem in the design of my driver.

if i selected a 6sn7 like quoted above by Ray and using all logic and any decent text book RL must be less than or equal to 2 * ra

however a 6sn7 has an Anode resistance of around 7 – 8 K

therefore it will not drive an EL34

i also tried real figures

Cin 15.2pF

mu 10.5

and

I = C * 2 * pi * F * Vpeak

I = 0.3mA

15 / ( 5 * I ) = 9K972now i would like to say that either i have got it wrong or a 6sn7 is not capable of driving an EL34.

if i have this wrong then i need help, i dont think that it is wrong. i also hope that your reply will resolve this matter or it will not do any good for one of the best posts that i have seen, a real eye opener.

Jason, January 21, 2010:

oops my bad, bit of a confused description

interpreted to mean that i would have to swing at least 5 * i peak as i go to v peak.

should read;

interpreted to mean that the drivers current would have to swing at least 5 * i peak as the driver swings to v peak.

Ken, January 26, 2010:

jason, i am a bit confused by this line:

“now using a little transposition and ohms law to determine the value of RL

Vpeak / ( 5 * I ) = 7K578”if the current required to drive the grid capacitance is half a mA, then just make sure your driver plate current is a few times higher than that (for headroom), and call it done. in other words, pull about 3mA through your driver. for the best linearity from the driver, size your Rp to be as large as possible by balancing off Rp and B+.

as i specifically mentioned in the article, introducing plate and load impedance is NOT germane to a discussion of

*slew rate*and will quickly lead to confusion.Jason, January 30, 2010:

see i said that i needed help.

so the ability to sink and source current is just this simple?

1, calculate the required current from above

2, multiply by 5 or more

3, choose driver valve that can idle at current.

4, use a large enough anode resistor to keep things linear.

so simple that i got it wrong and interpreted it to mean amount of current change needed at the same time as Vpeak.should i try to keep the current above the required 5 * i along the loadline from V p-p ?

Ken, January 30, 2010:

well obviously as the valve turns off you’re going to see plate current drop to a minimum and plate voltage rise to a maximum. as long as the IDLE current is a couple of times bigger than Islew current you’ll be ok… the Ip doesn’t have to exceed 5*Islew ALL the time.

Jason, February 2, 2010:

thanks for that.

basically then, the larger the load resistor and the higher the voltage it is supplied from the more linear and the slew rate is covered.kany, February 8, 2010:

hi Ken. thanks for a wonderful explanation.

but what is a slew current ? consider this line, “the slew current (of a driver circuit) must be large to drive the input capacitance of large pass device and prevent large signal oscillations”Jason, February 17, 2010:

i think thats what i was trying to say in my post.

cross trainer, July 29, 2010:

Hello,

how long does it take the output voltage of an op-amp to go from -10V to +10 if the slew rate is 0.5V/microS?

Really great post, enjoyed reading it. Thanks,

CrossKen, August 2, 2010:

well, that’s 20vpp, at half a volt per uS, it’ll take 40uS.

shiva, March 25, 2011:

hello

are slew rate and offshot the same ?

thanksvijya, May 1, 2011:

hii

thanx a lot for the explanation…after a long and hectic search i got such a meaningful answer…Keddy Kekana, March 5, 2013:

i don’t understand the slew rate thing but now i do and its all thanx to you…continue with the good work

Keddy Kekana, March 5, 2013:

meant i didn’t understand the slew rate thing but now i do and its all thanx to you…continue with the good work

Justin, December 5, 2013:

Hi, you are really know-how. And could you please further explain triangle wave case and why there is a factor of 2/pi when transfering sin to tri? And which one is more useful for ckt design? Thanks.

Justin

Justin, December 5, 2013:

Actually I see many textbooks show that a sinuwave input with a marginal SR will give a triwave output. For example, Prof. Sansen argues SR=Vout,max*4fmax. It just makes me more confused. Thank you in advance.

Justin

ram, April 10, 2014:

KEN what is difference between pull up/down resistor and termination resistor. And please help me to explain i got more confusion about it.., thanks in advance

Christian Riegels, April 12, 2014:

Hopefully you are still around to answer this question.

If so, I want to point out that the rise time (10%-90%) of a 20khz sine wave is going to be 14.76microseconds. (1sec/20khz)((2*arcsin(.8)sec)/(2pi))

If you need your 15Vpeak on the grid of your triode strapped el34, this corresponds to a slew rate of 0.94V/uS. (arcsin(.8)*15Vpeak/14.76uS)

das, December 22, 2014:

hi….i observed that simulated slew rate is different from the theoretical slew rate of the amp…? why …can u help me?

### Recent Comments

- Jagraj on Impreza Manuals
- Roger Bjørneseth on Impreza Manuals
- Dave Hager on Impreza Manuals
- Paul Bowes on 04 impreza service manuals
- Ben on Impreza Manuals
- Donald A Shore on 04 impreza service manuals
- Jason Seibert on Impreza Manuals
- John Halliday on Impreza Manuals
- David on Choke-input power supplies, Part 1 (Henry Pasternack)
- OBD Tracking Device on Impreza Manuals
- Mike on Building Your Own Tube Amp
- Scott on Impreza Manuals
- Rajkumar on Building Your Own Tube Amp
- Ravi on Building Your Own Tube Amp
- Harold on Impreza Manuals

### Search