This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

# ADS1296R: AC Lead-off: possible current source types?

I have a question regarding the possible current source options that exist for doing AC lead-off measurements on the ADS1296R.

pg. 40 of the ADS1296R datasheet says "AC Lead-Off: The AC signal is generated by providing pullup and pulldown resistors at the input with a fixed frequency." This makes it seem like pull-up/pull-down resistors are the only possible way to generate AC current. When we look at the register 4 bit options, however, it seems like there is also a current source option for doing AC lead-off.
My questions are:
1. Can register 4 be set to a value of 0x0d? i.e. a 24nA (48nAp-p) AC lead-off measurement using a current source.
2. Can you clarify the options?
For AC lead-off: register 4 bit 4 must always be 1 (pull-up resistor mode)? Or can it also be 0 (current source mode)?
For DC lead-off: register 4 bit 4 can be either 0 (current source) or 1 (pull-up resistor). Is that correct?
Thanks so much for your help
• Hi Lina,

Thank you for your post and welcome to our forum!

That sentence on page 40 is not totally accurate. Both the AC and DC lead-off excitation may use the resistor pull-up/pull-down method or the current source method. AC lead-off excitation is formed using the same method as the DC lead-off excitation. To produce the AC square wave, the current sources or the pull-up/pull-down resistors are swapped between the positive and negative channel inputs.

When LOFF[4] = 0, bits [3:2] determine the magnitude of the current sources. When LOFF[4] = 1, bits [3:2] are only used by the lead-off current source on the RLD amplifier output. All other channel input pins will connect to a 10-M pull-up/pull-down resistance and the current flowing through the electrodes will depend on the supply voltage and other impedances in the input signal path.

Best regards,
• Hi Ryan,

Thanks so much for your help!

One other thing... could you just confirm whether or not what I wrote was correct:
when register 4 is set to 0x0d is the resulting AC square wave 48nA peak-to-peak? Or is it actually 24nA peak-to-peak?

Thanks,
Lina
• Hi Lina,

The current waveform will be 48 nA peak-to-peak. The magnitude of the current source does not change from DC to AC method. So in one half of the period, the current will be 24 nA from INxP to INxN. In the second half of the period, the current direction will flip (24 nA from INxN to INxP).

Best regards,