# Slew Rate Limitations

Discussion in 'Electronic Design' started by Sanjayan Vinayagamoorthy, May 15, 2004.

1. ### Sanjayan VinayagamoorthyGuest

Hello,

I have a couple of questions regarding slew rate limits on opamps.

Suppose I have an opamp with a 1MHz unity gain bandwidth and a slew
rate of 1V/us (It's a very bad opamp, just my example). I place it in
unity gain configuration i.e.:

--------------.
| |
| |
| |\ |
'----|-\ |
| >-----'
----------|+/
|/

Suppose I give the input a step response from 0mV to 25mV. How do I
know if the amplifier will slew rate limit?

What if the amplifier had a gain of -2 (using just resistors), would
that change whether or not it would slew? What about +2? I am just not
sure how I would go about figuring this out.

Any guidance is appreciated.

Thanks,

2. ### Phil HobbsGuest

A voltage-feedback op amp goes into slew limiting when the full bias current
of the first stage is going into charging the compensation capacitor. This
happens when one of the input devices turns off--typically at about
DeltaVin=60 mV for simple bipolar amplifiers, more for FET types and
emitter-degenerated bipolars. You can sort this out approximately from the
unity gain frequency.

A perfect integrator with a unity-gain frequency of f_T responds to a step
input from 0 to V by slewing at pi*V*f_T volts per second. Thus if its
maximum slew rate is SR, the approximate step height required to cause slew
limiting is

V = SR/(pi*f_T).

A 741, with a plain bipolar input, has f_T=1.2 MHz and SR=0.5 V/us, typical,
so that slew limiting sets in at about 66 mV.

5e5/(pi*1.2e6) = 0.066 V

This is a bit of an overestimate, since the degradation is gradual, so the
amp ceases to be linear well before this.

Cheers,

Phil Hobbs

3. ### John LarkinGuest

Ignoring slew rate for a second, you can model your closed-loop opamp
as a pure gain of G (+1, -2, +2, whatever you set it up for) followed
by an R-C lowpass having corner frequency gbw/G, where gbw is the
opamp open-loop gain-bandwidth, 1 MHz in your example.

So R*C = 1/w, where w = 2*pi*gbw/G

So poke your signal into this model, say a step of V volts. The output
will be a step of G*V volts, which passes through the RC to produce an
exponentially-smoothed step, a classic R-C curve. The steepest part is
at the very beginning, where the slope is just G*V/(R*C) volts/second.

If that slope turns out to be greater than the spec'd slew rate, it
can't move fast enough to follow the curve, so it slew limits. Later
on, when the curve slows down, it will again follow the exponential.

Did I get all that right? It feels like there may be some algebraic
simplifications in there somewhere.

I recently had to design a precision dac-programmed slew-rate limiter
for an NMR gradient amp, to keep the customer's signal from trying to
push too much dI/dT into the gradient coil inductance and railing my
amp. Turns out that the loop dynamics is non-trivial to get this to
have clean, crisp corners and be stable over a wide slew range.

John

4. ### Nico CoeselGuest

The slew rate defines the maximum steepness of the signal. It doesn't
depend on the bandwidth of the opamp configuration because whatever
the bandwidth is, the slewrate is always the same.

If you differentiate a sine wave, you'll find the maximum signal
rise/fall occurs at zero crossings.
You can define a sine wave by s=a * sin(2pi * f) (a=amplitude,
f=frequency). If you differentiate this you get: s'=a*f*cos(2pi).
Since cos(2pi)=1 the maximum steepness (=slew rate) is equal to (a*
(1/f)) (in Volts per second). If you want an opamp to produce a signal
of 1V (peak-peak) at 1MHz, the formula will indicate you'll need at
least an opamp with a slew rate of 1V/us. If you want a signal of 10V
you'll need at least 10V/us.