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DC input impedance in case of AC-lead-off detection

Martin Semkat
Prodigy 100 points

Hi,

with regard to the dc input impedance, in the electrical characteristics table of the ads129x datasheet a distinction is made between current source lead-off detection and pull-up resistor lead-off detection. In the first case, it's 500MOhm, in the latter only 10MOhm. I want to make use of an ac lead-off detection. Now I wonder if that belongs to the pull-up resistor lead-off detection. Having read the AC lead-off detection section, I expect the inner ac signal generation circuitry to reduce the input impedance, but I'm not sure. So, is it reduced to 10MOhm when using the ac generation circuitry?

I want to make use of the internally generated WCT voltage for the negative inputs of the PGAs. According to the datasheet, the WCT terminal should only be used to drive very high impedances (>500MOhm). So, if the input impedance is reduced to 10MOhm through the ac lead-off detection, I cannot use the WCT for the negative inputs. Or can I?

Thank you in advance.
Martin

 

  • 0
    TI__Guru**** 180961 points

    Hi Martin,

    Have you reviewd Tony's application note on ECG Lead-Off detection for the ADS129x yet?  That might be a good place to start.  The document is located here:

    http://www.ti.com/litv/pdf/sbaa196

     

  • 0 in reply to Tom Hendrick
    Prodigy 100 points

    Thank you Tom. Now I have read the application note and learned a couple of things. However, some questions remain.

    My objective still is to apply an ac lead-off detection for a configuration where the minus sides of the channels carry the internally generated Wilson Potential (coming from the WCT) and the plus sides are connected to the ECG electrodes (the connection of which shall be observed). Before going any further into detail, I would like to know if in such a configuration an ac lead-off detection works properly. (The situation is different to that in the application note and in the datasheet, where it is assumed that both at the plus and the minus side an electrode is connected.)

    Also, I still wonder what role do the current sources and sinks play in an ac lead-off detection, if any. As written in SBAS459I, the ac signal is generated by alternatively providing pull-up and pull-down resistors at the channel input. To my understanding, that should be all that there is to be done to implement an ac lead-off detection in the situation where you have an electrode both a the plus and the minus terminal of the channel. However, in figure 59, the switches close to the current sources have the same labels as those which create the ac signal on the channel, so I expect them to switch simultaneously. What is that about?