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TIDA-01471: IEPE Signal Conditioner with single source ,single ended to differential ended low pass filter

Electronx Electronx
Intellectual 305 points
Part Number: TIDA-01471
Other Parts Discussed in Thread: THP210, OPA320, LM7705

Hello, I want to design an iepe sensor signal conditioner. This signal conditioner that I will design will not include gain and attentautor parts. The signal from the iepe sensor will pass through the coupling capacitor and then pass through the low pass filter(gain :1 . After filtering the signal, this low pass filter will convert the single ended signal to  a differential ouput  signal and give it to the ADC input.

When I make the configuration in the picture and apply a 10khz +-1 V signal, I can get a differential signal with a slight phase shift.

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However, when I put a coupling capacitor at the beginning and add 12 volt dc offset to the ac signal, it does not filter the offset(hp filter with coupling cap) and I get a result like this? I'm having a hard time understanding where my mistake is.

A second question: Is there a phase shift after the tida-01471 reference design passes through the low pass filter? 

Can you help me with my design, thank you.

  • 0
    TI__Expert 3520 points

    Hello Electonx,

    - The phase shift is a characteristic of any LPF or HPF, it represents the group delay, and it is frequency dependent. you can deduce the phase shift from the Bode plot of the filter response. In general this phase shifts should not be critical for vibration analysis as far as I know.

    - The second circuit has the issue that you didn't set the DC value of the AC coupled input node  (left of R6). you can think of it simply like that: when there is no AC input. the left node of R6 is floating. it will take the voltage level of the right node. as R2 is connected to the highZ input of the THP210, it will not also pass any current. the node between R6 and R2 is hence a high impedance node, the only low impedance path is through R4, and R8, and they don't carry any currents. so the output at OUT- will simply determine the DC level of this input. The circuit then will take the state that will fulfill the following conditions (output = gain x input, and the node between R2 and R6 = OUT-), that's why you see 8V at the input, as well as differential 8V at the output.

    - The fix for this should be to add a resistor to ground after C6. there is one trick here, The resistor added has to satisfy:

    1. the resistor along with C6, will constitute the high pass filter. the R is determined by the required cutoff

    2. on top, the high pass filter will act as a potential divider of the input resulting in attenuation. 

    2. the resistor added to one input need to added to the other input to have the same DC value. note that (VCOM)*(R8+R4)/R6 is the DC value of the input. you need to make sure that this value is within the input CM range of the THP210.

    Unfortunately, in this circuit as there is no buffering after the filter, you can't independently control the attenuation, the HPF cutoff, and the input DC value. adding a resistor, and a buffer after the cap will allow you to decouple those paramters.

    hope this helps you get along with the circuit.

    Kind Regards,

    Ahmed

  • 0 in reply to Ahmed Noeman
    Intellectual 305 points

    Hi Ahmed !, thank you for your feedback, I tried what you said, I added 159 ohms to both sides for the 1000 Hz cutoff signal in db. It is better than the old one, but when I apply a 100mV signal, a dc offset still occurs as in the graph. Do you have any other advice?

  • 0 in reply to Electronx Electronx
    TI__Expert 3520 points

    The R8 should be the only path to ground. you have shorted it.

    on the other hand, 159 Ohm is a bit low, but if you have the right cap it might work if you have the overall chain gain as required.

  • 0 in reply to Ahmed Noeman
    Intellectual 305 points

    Thanks for your quick response, but in this case, it causes the signal to weaken. For example, when I apply 100 millivolts, the output gets a value like 78.8mV. Is there another solution?

  • +1 in reply to Electronx Electronx
    TI__Expert 3520 points

    you can simply compensate for the attenuation seen here, by increasing the gain of the filter stage. However, note that the attenuation is function of input frequency. you can also create a biasing network on both sides, but this will not change the fact that you will have some attenuation.

    I think there are more than DC level shift to be considered in this design:

    - you need to design your HPF at the input to fulfill several parameters (input impedance, cutoff frequency, and attenuation factor):

    1) currently your cutoff is relatively high (1kHz), typical designs has sub 1Hz cutoff

    2) also your resistor value is relatively small 159 w.r.t the cap (1uF). this means you have attenuation relative to R/(r+Xc)

    3) moreover, effective input impedance drops with higher frequencies, leading to attenuation, if source impedance is taken into account, your will have magnitude errors due to the low input impedance, and you might even overload the sensor. typical input impedance is over 100kOhm

    >> what I would recommend, start by determining the input impedance you want to achieve at high frequency (like 10kHz), this will set the resistance you would need for the HPF, then with the target cutoff frequency you determine the capacitance.

    >>once done, you would find the resistor probably of a high value, and can't drive the relatively low impedance filter you have here, this means a buffer is probably needed between the HPF and the filter. (here I can recommend OPA320, check the new design https://www.ti.com/tool/TIDA-010249, it is low voltage though, and would need attenuation before)

    >> if you used a buffer, and decided that you input common mode is zero, as you selected here, then you need to make sure the buffer can output true zero. this can be ensured by using a RRO buffer, using bipolar supply, or using a simple negative charge pump (like LM7705) as a negative power supply for the buffer.

    >> if you don't want to use bipolar supply, then you can set input common mode to non-zero value. this would require shifting both inputs. you can use the buffer in difference amplifier mode to shift inputs, and make sure to apply the shift on both input. the shift value can be either Vcm generated by the ADC, or internal VREF (you might need to buffer it though) this would look like the following figure

    >> adding of the buffer would allow you to set independently : your input impedance level, your HPF cutoff frequency, your attenuation (this can be compensated at later stages), as well as setting the input common mode.

    I do recommend to go through the following documents

    https://www.ti.com/lit/an/slyt749/slyt749.pdf

    https://www.ti.com/lit/pdf/tidud62

    https://www.ti.com/lit/pdf/tiduf13

    to get more insight into the considerations for such signal chain.

    Kind Regards, 

  • 0 in reply to Ahmed Noeman
    Intellectual 305 points

    Hi Ahmed, thank you for your help, information and resources. I understood that the tida-01471 design is a single source oriented design. After the dc offset of our input signal is deleted, we have a +- signal, and in the gain section, dc bias is added again to prevent the signal from being clipping . Additionally, the input is an active high-pass filter with adjustable 1st-order gain.

    Our input impedance for our signal conditioner is the capacitor impedance + the resistor we put between the input and gnd(or reference ). (1/(2πfC) + R)

    Although increasing the input impedance increases the accuracy of the signal measurement, it increases parasitic noise and deteriorates the signal quality. For this reason, we must keep the input impedance at an optimum value.

    In this case, first the input impedance was determined as 256kohm and then the cut-off frequency at -3dB was determined as 0.6217Hz. Thus, our capacitance value is 1uF.

    So here I understand better why I should use Active filter.

    Thank you!