# Signal response and settling time graph analysis

I have come across a datasheet of an instrumentation amplifier. I have a difficulty in understanding the signal response vs settling time graphs. Can someone please explain figure 28(page 10) of the provided datasheet in the link??

They are showing the response from a step input, and using this test circuit with the gain set to 1000:

The top signal shows the large signal response, and that there is a time constant of roughly 50us, the amplifier takes roughly 80us to finally settle.

I'm not sure what the bottom signal is, but I'm guessing it is the same signal but with the scope in AC coupled mode, to show the settling time in finer detail. Upon closer examination it takes 120-150us to finally settle with a large signal response. It is a little responsible to not lable the signals or put a description on the graph.

I would use this graph instead to determine the frequency response.

• Thank you for your help. Even I have the same doubt baout the bottom signal and also what are the 5v and 5 mv notations..why are they written uptop in a single row..also what the values 0 10 90 100 on the vertical axis – user183710 Jan 3 at 18:03
• The numbers show you the 0% 10% 90% and 100% values, probably for 5V. So it doesn't quite get to 5V. meta.stackexchange.com/questions/126180/… – laptop2d Jan 3 at 18:14
• @user183710 - 5V and 5 mV are on the same line because that's how analog scopes used to do it (and still do). 5V (with an implied "per division) refers to the upper trace, and 5 mV to the lower. AC coupling has nothing to do with it. Vertical axis markings refer to percentage of full scale, with the signal amplitudes adjusted so that the step will go from 0 to 100%. Obviously, this feature was not used for this test, but it was a common way to show pulse response if you didn't want to show the response at two sensitivity levels. – WhatRoughBeast Jan 3 at 18:17
• @WhatRoughBeast so with a square input, how do you get a signal that looks like its AC coupled? (The lower one) Are they subtracting something? I am not familiar with extra features of old scopes, I've only used them on a few occasions. – laptop2d Jan 3 at 18:24
• @laptop2d - Yes, they are subtracting. See the test circuit. Keep in mind that the amp is being tested in an inverting configuration - it makes subtraction very simple. And the percentage display is not generated electronically, it is a printed graticule on the front of the tube, along with the grid. I'm not sure if it is printed directly on the tube or on a thin layer of glass in front - it's been too long since I used one of those, and my beloved Tektronics 2465 died about a year ago. – WhatRoughBeast Jan 3 at 18:53

Settling time, as opposed to frequency response, is a very complicated issue, and is best addressed experimentally.

The problem is that, in effect, every differential amplifier has a pair of transistors, one for each input. When you first apply a difference, the two transistors heat up to different temperatures. This causes them to have different leakage currents, which produces a voltage offset which is an error term, and the time required to equalize temperatures is a major component of settling times, particularly for very small target changes such as .01%. Since the die is very small and the separation of the two transistors even smaller, equalization takes place very quickly - but in tens of microseconds rather than nanoseconds.

For high-precision, high-stability analog ICs, layout of the circuit on the die which minimizes the effects of temperature gradients is critical, and some early IC designers were lauded for their clever solutions.

Settling times to something like 1% are much, much shorter, as the thermal effects on the die are much less significant.

ETA - Oops. As laptop2d mentioned, and I forgot, settling time is measured with a step input on one input, which obviously makes a worst-case situation for the effect I've described.