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# ADE5569 Isource (logic output parameter) description

Hi,

ADE5569 datasheet (page 9) contains "Isourse" parameter in Digital Interface - Logic Output section. Digital Interface - Logic Input section contains "Input Currents Port0,1,2" parameter. Isourse = 80 uA max. Input Currents Port0,1,2 = -8.5 uA.

Figure 118 (page 149) shows structure of IO pins.

PINMAPx.x = 0 (Internal Pull-up resistor is used). When I connect Px.x PIN to the ground I get 5 uA current flow (correct). So my question is how could I get 80 uA source current from output pin? Could you please describe IO stage in details? Thank you in advance.

• Hi Ivan

This is how the testing is done:

Input pin:

To measure leakage current for an input pin, set the pin potential to be greater than minimum Vinh (logic HIGH) using Vtest and measure the current. Similarly, set the pin potential to be greater than maximum Vinl (logic LOW) using Vtest and measure current again. This will provide you with bidirectional input leakage current levels. For maximum current levels (worst case leakage current), set 3.3V for logic HIGH and 0V for logic LOW. Your method of grounding the pin and measuring current is correct. Also, connect the pin to logic HIGH as per above description and measure current to see if it lies under the datasheet limits.

Output pin:

To measure output pin sink current, connect a current source as shown above which can provide bidirectional currents (to and from the pin). Keep the pin at 0V and keep increasing the current applied to the pin and measure the potential at the pin using a voltmeter. The maximum current that can be applied to the pin while the pin voltage still stays logic LOW (<Vol maximum) is considered the maximum sink current.

Similarly, to measure source current of an output pin, keep the pin at 3.3V and apply the current in the reverse direction (pull current from the pin) and keep increasing it till the pin voltage stays logic HIGH ( > Voh minimum). This is the maximum source current.

Let me know if you have further questions.

Regards

hmani

• Hi hmani,

Thank you a lot for the detailed description.

Regards,

Ivan

• Hi hmani,

Sorry for disturb you again.

Datasheet, page 149:

Could 80 uA output source current be achieved when internal pull-up resistor is enabled? Or only 5 uA?

Could I source the load when output become open drain (internal pull-up disabled)?

Thank you in advance for explanation.

Regards,

Ivan

• Hi Ivan

For an output pin, the source current spec is independent of the pull-up resistor being enabled/disabled. Mostly an output pin doesn't need a pull-up resistor. But standard GPIOs can be input or output and so pull-ups are available on all pins. Your application will determine if you need a pull-up on the output pin as well. But that shouldn't affect the sourcing capability of the pin.

Regards

hmani

• Thank you hmani a lot.