What is the need for amplifier stage in a data acquisition hardware?

Question

Below is a typical data-acquisition block diagram:

As you see above after the multiplexer there is an instrumentation amplifier IA.

This IA stage can be shown in more detail as below:

Since this is a differential amplifier IC, its input stage must be a differential pair topology as follows:

Now my questions are:

1-) All the data-acquisition boards I used measured the actual voltage applied, and did not amplify them. Normally if one sets the daq input range to -/+10V, and if one applies 2V with to daq's inputs, the daq will meaure 2V. Here my first question is, what is the need for the IA stage if it doesn't amplify? Is that common mode rejection? And what if the daq used for single-ended inputs? Would the IA still have a use? Why not directly coupled to ADC converter?

2-) If I swing the input voltage to the daq from -10V to 10V this is a large change. Wouldn't it cause non-linearity in the IA's differential input stage(large-signal analysis)? How is this be linearized?

Answer 1

You would include a unity-gain buffer on the input side of the ADC to increase input impedance, as the ADC is likely to have lower impedance than an Op-Amp.

This avoids readout to be dependent on the source impedance, and also allows impedance matching on the sink if needed (i.e. you can add a termination resistor and the resultant input impedance will be close to the resistor value).

This is especially important for higher-frequency signals, where the ADC input impedance would be frequency-dependent, while a termination resistor will have only little capacitance and inductance, giving a flat frequency response.

You can optionally configure for a different gain factor if it makes sense in the application.

Answer 2

1) To isolate the ADC and signal source. The non-inverting input of the opamp has huge input impedance, so it doesn't load the signal source.

2) There is a negative feedback to the each input stage opamp. This acts like a proportional controller, the output from opamp is compared to the input signal and the output is corrected if they differ with regard of its amplification or attenuation .

Answer 3

1) Even if the IA stage does not amplify, it does provide other functions:

• buffer the input signal so that any disturbances present at the ADC's input cannot reach the multiplexer.

• provide a well defined input impedance for the multiplexer and/or the circuits connected to the multiplexer.

• provide a buffered signal for the ADC. The ADC might require a (small) input current, it might have an infinite input impedance. With the IA in place as long as the IA can drive the ADC's input impedance, there is no issue.

2) The IA consists of opamps with feedback. Such circuits suppress the non-linearity of the circuits (in the opamps) by having excess loopgain.

Let's say that the opamp has a voltage gain of 10000 (10k) at a certain frequency, but you configure the feedback so that the opamp with feedback has a gain of 10 then the excess loopgain is 10000/10 = 1000 which means a factor 1000 supression of non-linearities. So 10 % distortion would be reduced to 0.01 % distortion.

Also that differential pair will not have more than a few mV across its inputs. So it does not need to work with 10 V or 2.5 V, it cannot handle that anyway. Opamps work linearly only when they have a large gain which causes the input voltage (of the differential pair) to be quite small. For example: if an opamp needs to output 2 V and has a gain of 10000 then at the input the voltage is only 2 / 10000 = 0.2 mV.

Answer 4

The DAQ input stage may well attenuate the signal by something like 2:1 or 4:. Not many modern ADCs will accept a +/-10V signal directly. You would have to look further into the circuit diagram to determine that. Typically you might have different range settings which would change the amplifier gain, and regardless of the range you select the ADC would see a fixed input such as +/-2.5V or +/-5V. So if you select +/-10V the programmable gain amplifier (PGA) would have a gain of less than 1.

Q1: An instrumentation amplifier has very high input impedance (and usually low bias current) on both inputs. That allows differential or single-ended input with the former having both inputs high-Z. It's particularly important to have high impedance inputs if they are to be multiplexed, otherwise the mux switch resistance could affect the readings, but it's always a good thing since it won't load the input signals. Differential input allows rejection of common mode noise.

Q2: The nonlinearity of the differential pairs used in such an application is minimal. The open loop gain of a precision op-amp is so high that the change in the differential pair input voltage (at DC anyway) is only microvolts for a full-scale output swing. They are very linear over that range. Even if they were not, the effect is a percentage of microvolts and your signals are much higher because the closed-loop gain is a small fraction of the open-loop gain (at DC).

I mention "at DC" a couple of times, because things may be considerably worse at high frequencies where the op-amp open-loop gain is less and nonlinear effects such as slew rate limiting can come into play.

Answer 5

There isn't one universal answer. Consider the requirements for each input individually.

1. If the ADC accepts -10 to 10V and you are using a range of 0-2V, then you are using only 10% of the ADC's input range - and effectively throwing 3 bits of resolution and accuracy away. If that meets your needs, then you don't need amplification. If it doesn't, then you can amplify the signal to make better use of the input range and reduce the errors introduced by the DAQ process.

2. The non-linearity will be a function of the input slew rate and the bandwidth of the amplifier. Amplifiers are easier to make than ADCs so it is likely that the ADC itself introduces greater slew rate or settling time problems than the amplifier - and that the sample rate is high enough to represent the signal. Part of your task is to choose both an ADC and an amplifier that are fast enough to meet the input requirements : OR condition the input signal (reducing its bandwidth and slew rate) to avoid problems with the amplifier and more importantly, ensure your sampling system meets the Nyquist criterion.

This isn't generally a problem in the schematic above, because:

• If you are handling high speed signals, you won't generally be multiplexing inputs into a single ADC.
• You control the multiplexing rate : simply ensure it isn't too fast to allow the amplifier (if used) and ADC inputs are stable for the specified sampling time.

Answer 6

Supposes your DAQ board uses a delta-signal ADC with Cin of 10pF, sampling at 10Miliion samples per second; some 24 bit systems are like this.

The (average) input current is Fsample * Csample * Vsample = 10Million * 10pF * 10 volts = 10*10*10 microAmperes = 1 milliAmpere.