Signal conditioning for AC excited bridge circuit.

Hi,

yes, this is possible and is done from time to time. The key here is to generate a sine of very exact and stable amplitude 

Or you also measure the amplitude of excitation signal and do some amplitude controlling or division of amplitudes (excitation signal and bridge signal) in software.

For very noisy signals you can even use the method of "synchronous demodulation". -> Lock-in amplifier.

But in very most standard applications these sophisticated methods would be technical overkill

Kai

kai klaas69 said:

The key here is to generate a sine of very exact and stable amplitude

Could you suggest the best possible ways of doing this? I am aware of some oscillator topologies or filtering of PWM output to achieve the sin wave but not sure what should be a suitable choice. Also, does TI provide any integrated solution that might make the task easier.

Hi A BC,

Q: removing all DC offsets through AC coupling the inputs of instrumentation amplifier or using a DC restorer circuit. Is this a feasible solution for getting rid of errors due to offsets? Are there any design considerations to keep in mind?

Can you tell us the type of application you are working with?

If you just want to get rid of DC components in AC signals, you may use capacitive coupling in such application. What is input differential signal amplitude and frequency? What are the required gains etc.? 

Best,

Raymond

Raymond Zhang1 said:

If you just want to get rid of DC components in AC signals, you may use capacitive coupling in such application

Yes, I wish to remove any DC offsets associated with the signal conditioning circuitry from the AC signal. The input differential signal will be 10mV p-p, frequency is not decided yet, would chose something that would allow the best overall performance. The signal will be read out by an ADC, so gain can be anything between 200-500 depending on reference of ADC used. Capacitive coupling is what I am considering, but I also came across this (TIPD191 Instrumentation Amplifier with DC Rejection Reference Design | TI.com) which seems more advantageous.

Hi,

I think Raymond wanted to know what the bandwidth of your signal coming from the bridge is. Not the frequency of the excitation signal. His idea is, that when the bridge signal is already AC, you can just use AC coupling.

But I think our bridge signal goes in the frequency down to DC, right? In this case you cannot use the method with the DC servo shown in the TIPD191.

So, please tell us what the bandwidth of your bridge signal is.

Kai

It is a purely resistive bridge so the signal from bridge should be the same frequency as the excitation signal.

Hi A BC,

Adding to Kai's comments, I created a crude simulation and tried to understand the design requirements. 

In the example below: an AC excitation reference signal with 10Vp is used. I used resistors to construct a Wheatstone bridge, R9 is likely your AC frequency induced transducer.  The transducer may have its own response BW to external influenced conditions on top of the excitation AC frequency. We need to have the characteristics plot or description. 

If you are using AC excitation reference driving signal, the input should not have DC component at the input of instrumentation amplifier. The instrumentation amplifier is only interested the differential input signal, and common mode signals will be rejected. In this case, CMRR = 145dB below 60Hz. 

As Kai pointed out earlier, you do need to have a precise and stable AC reference excitation voltage. 

Sure, INA819 will have some DC components (due to Vos) after the input signals are amplified, but this can be readily removed as well (outside of the discussion). 

INA819 AC Excitation Wheastone 07202021.TSC

Best,

Raymond

Hello Raymond,

Thank you..this is what we also have in mind. I would like to know if there are any considerations while selecting the value of R and C for coupling (apart from cut off frequency ofcourse) in terms of what should be the order of values of R and C to achieve the desired cutoff. For example a 15Hz cut off frequency can be achieved by using 100k resistor and 100n capacitor or 10k resistor and 1u capacitor etc. In this case which would be the better choice. What are the factors that are usually considered while selecting the combination of values? I haven't ever really been able to find a straight answer regarding this.

Hi A BC,

Since you are only interested AC excitation signal, you should consider to Null the wheatstone circuit at the excitation frequency and amplitude. The technique should be very similar to standard wheatstone bridge circuit null method, except you are working with impedance at a given frequency. 

1. Your transducer should have a nominal working operating range in R, C, amplitude at a given frequency. Verify the impedance figure with LCR meter. Also, you will need to understand the frequency characteristics of resistors used in the bridge circuit.  In addition, you need to understand your transducer's sensor's model, if it is R||C or R+C or some other complex RC or L network. 

2. Since Wheatsone is a ratiometric measurement, there are two common sensor placement, see the image below. 1. For instance in the image below, Ra and Rs (Rx)  needs to be matching pair at the given excitation frequency, where one is the reference and other one is your transducer with and without load. 

3. Regarding the selection of R and C, it depends on if R and C are in series or in parallel or other complex network. Once you have the transducer's pspice model, you can simulate it and see which configuration is more sensitive to external condition changes (that is the one you'd prefer). Of course, you need to match the impedance of the wheatstone bridge. 

4. The instrumentation amplifier only works with differential input signals, and it will reject common mode signal appears across at its input. INA819 has an excellent CMRR to reject input Vcm signal. At 145dB (Gain=100), it is able to reject 1/10^(145/20) = 56 nV/V. If the input has Vcm near 5Vpk, the input signal will be 5V/*56nV/V = 280nVpk, which you need to null out the bridge circuit < 280nVpk as an example. 

If you have other questions, please let us know. 

Best,

Raymond

Hello Raymond,

Thanks for your detailed answer. However, I meant R and C values of the AC coupling as shown in the figure, not the bridge.

Also, could you suggest what is the best way of generating the stable excitation ac signal?

Hi,

A BC said:

It is a purely resistive bridge so the signal from bridge should be the same frequency as the excitation signal.

yes, of course, but at what speed the resistors in the bridge are changing? Down to DC? And what is the maximum speed? I ask because the frequency of excitation signal has to be way higher than the maximum frequency of bridge signal. 

Kai

Ah okay, it is around 0-500 Hz. 

kai klaas69 said:

frequency of excitation signal has to be way higher than the maximum frequency of bridge signal

Approximately how much higher? Would 2 times be enough?

Hi A BC,

I meant R and C values of the AC coupling as shown in the figure, not the bridge.

You mentioned that the application's Wheatstone bridge is purely resistive. I'd like to ask why you want to remove DC components of input signals. I am still not fully understanding your transducer's characteristics and application requirements. What is your application?

Instrumentation amplifier (IA) will reject the common mode voltage appears across the input of the amplifier. I demonstrated the CMRR calculation in INA819. Majority of the DC common mode voltage including DC is rejected from the input of IA. Only differential signals at the input will be amplified by the IA. 

Also, could you suggest what is the best way of generating the stable excitation ac signal?

As Kai mentioned previously, the best way to generate a stable AC signal is via phase locked feedback loop. Of course, the output of phase locked loop has to be able to drive the voltage amplitude and current of Wheatstone bridge a load. 

Best,

Raymond 

A factor of > 10 would be good

Kai