# current sink/source in ADS114S06B analog/digital converter

Here is the datasheet.

It seems from the graphs on page 17 that the I/O pins will sink or source as much as 7.5 mA:

However on page 9 the table indicates an expected sourcing of 1 mA (I say expected because this are their test conditions) and a maximum sinking of 1 uA:

How do I establish a sourced / sunk current for my power source calculation ?

The outputs are tested for voltage when sourcing or sinking 1mA output current. The graph tells you what is the typical voltage difference between the output and the supply rail (sourcing) or ground connection (sinking) when delivering a current between 0 and 7.5mA.

The inputs are tested for correct operation with 1uA being sourced into or sunk from the pin as an input.

You need to use the characteristics of the thing you are driving with your outputs before you know how much current the load will draw from them (up to the maximum of 7.5mA)

Looking through the data sheet, I can't see any specification of input capacitance, which presumably means it's negligible compared to the board tracking. We often take 10pF as a typical for 'a bit of track and an IC pin' if there's no capacitance spec. The current this takes to charge and discharge varies linearly with the frequency, so it's negligible for controls, and significant for a fast clock. If you're running (for instance) a 5v 10MHz clock into 10pF, then that's a charge of 5v * 10pF = 50pC each edge. So at 10M edges per second, that's a current of 10MHz * 50pC = 500uA.

• Not sure I understand the last paragraph. I am driving the ADC with a MCU. The MCU outputs 10 - 20 mA on a pin. Does this mean I should place a resistor on the output line so the current does not exceed 7.5 mA (the absolute maximum is given at 10 mA for this ADC) ? – kellogs Sep 19 '18 at 12:02
• The MCU can output up to 20mA on a pin, but only if the load demands it. If the load is an input pin on your ADC, then it will only draw up to 1uA static current, but significantly more to charge its stray capacitance when it's making a transition. The 1uA figure is so low, you can ignore it. The current demand of your load is then given by the charging and discharging of its capacitance, which will vary with drive frequency. What I'm doing here is 'using the characteristics of the thing I'm driving' to see what sort of load it is, so what current it will draw. – Neil_UK Sep 19 '18 at 12:37
• I certainly want to determine those: characteristics, load type and ultimately current draw. Is there a guide somewhere that would help me ? – kellogs Sep 19 '18 at 12:45
• looking through the data sheet, I can't see any specification of input capacitance, which presumably means it's negligible compared to the board tracking. We often take 10pF as a typical for 'a bit of track and an IC pin' if there's no capacitance spec. Calculate the dynamic current into that, at your drive frequency. – Neil_UK Sep 19 '18 at 12:55
• Ok, care to put this comment in the answer ? I'd like to vote it in – kellogs Sep 20 '18 at 14:12

The first graph is telling you that when the output is high (nominally 3.3 volts unloaded) if you add a load that takes 1 mA, the output might fall from 3.3 volts to typically about 3.25 volts. These are not guarantees.

The table is telling you that they guarantee that the lowest voltage that the output pin will fall to with a 1 mA load is 0.8 x Vdd or 2.64 volts but they don't rule out that it may be still 3.3 volts.

It's the difference between typical performance and guaranteed performance levels. A similar story for the output pin sinking current when trying to maintain 0 volts at the output.

• Ok, I understand. I still don't know how to calculate how much current will this ADC draw on its digital IO pins though. My biggest doubts right now being the apparent 7.5 mA sinking capabilities of the part (shown in fig. 34) while the last line on the table specifies a maximum input current of 1 uA. Are those two currents different ? How ? – kellogs Sep 19 '18 at 12:13
• The 1 uA is for when IO is set as an input. – Andy aka Sep 19 '18 at 13:24
• and the [0 - 7.5] mA from figure 34 is not ? – kellogs Sep 19 '18 at 13:50
• Figure 33 and 34's title is digital pin output voltage!! – Andy aka Sep 19 '18 at 15:36