Slew rate

[Bluga]

can someone shed light on Question 5, page 6?

since its a square wave input, how can it have slew rate?

 

[Superdoopercooper]

Bluga,

He's not saying that the SQ. WAVE has slew rate, he's saying the opamp has slew rate.... which means if you put a square wave in... you won't get an infinitesimally small rise-time square wave on the output.... you'll get a sloping rise and fall time of 0.5V/usec. So, you're answer to the question is:

When the input goes from 0.0V to 0.1V, the output of the amp has to swing from 0.0V to -0.5 Volt (due to the gain). Therefore, it will take the amp 1usec to reach -0.5V. Therefore, on the output you have to draw asloping line for 1usec from 0.0V to -0.5V. Then, it is flat at -0.5 Volt until the next input transistion, where it will go from 0.1 to -0.5. Therefore the output will swing from -0.5V to +2.5V --> 3V swing.

A 3V swing will take 3/0.5 usec... = 6usec. Therefore, at the 6usec mark, the output will have only slewed for 3usec... or half the time it needed to. Therefore, only have the swing (1.5V) would have been accomplished. Therefore, the answer to the second part is -0.5 + 1.5 = 1.0V.

The one thing I didn't take into account here is the UNITY-GAIN bandwidth deal at 5*10e6 rad/sec. That is actually 796.2kHz. The square wave is 166.7kHz. However... depending on how the Gain-bandwidth curve of the op-amp is, you could actually run out of gas trying to get a gain of 5 out of the thing at 166.7. We don't know the slope of the GBW curve though... but if I got with what "normally" is... I'm thinking that you'd need a unity GBW product of 5*166.7kHz to support a gain of 5 @ 166kHz. However, since this is "not" sinusoidal, I think slew rate is all you need to worry about --> and which is somehow (that I don't remember) related to the GBW product. I believe the answers above are right. Let me know what the solutions are... it has been 2 years since i've even touched an op-amp equation.

 

[Bluga]

Hi thank you for the reply.

I'm not sure about the timing: why does it take the amp 1usec to reach -0.5V and 3V swing will take 3/0.5 usec?


Also how did slew rate come into play? Thanks!

 

[Superdoopercooper]

I'm not sure I understand your new question? Are you asking "what is slew rate?", or "why do op-amps have slew rates?" I think this is what you're asking, so I'll just address that.

The circuitry has in an opamp has some maximum transistional speed (think rise and fall times... similar to digital circuits)... due to parasitic of the devices. There are parasitic capacitances and there are device effects which dictate how fast holes and carriers can migrate and form channels [of current/charge] in the silicon... all that adds up to a slew rate or rise/fall time limitation.

In an ideal world, op-amps have infinite slew-rate... and instantaneous change on the input will result in an instantaneous change on the output. However, we don't live in this happy magical land. So, in the problem, the professor states that the op amp has 0.5 uV/sec of slew rate. This is the maximum rate of change that the output of the op-amp can sustain.

Since the rest of the op-amp problem is ideal... you know that it has a gain of -(Rf/Ri) = -5k/1k = 5. So, when you put the 0.1 on the input, you get -0.5 out. With the -.5 on the input, the op-amp wants to put 2.5V out. With the change between those two voltages, it is a 3V swing. The op amp takes 1usec every 0.5V that the output tries to move... so, 3V / 0.5V = 6 * 1usec = 6 usec.

I really don't see what I left out here. I thought my first post pretty well covered it. And I think this one may fill any questions you could have if you:

a) understand slew-rate
b) understand op-amps
c) draw it out...

If you are still cloudy, please figure out a more specific/pointed question.