Hello,
I am working on a low power application, where I want my sensor to be turned ON only for a certain period of time, so that it should not consume power when not in use (rather when the ble in in sleep mode). So, I thought if I could provide supply to the sensor through GPIO Pin itself. The sensor would consume around 500uA of current. I thought of using a transistor first, then ended up concluding that it would need more power.
Please let me know, is it possible to draw that much current through the GPIO pins?
or
Is there any other solution to this problem?
Thank you!
Device:
Hi DarshanDodalm,
请见表11:数字输入/输出utput - DC Characteristics in the DA14585 datasheet in order to find the GPIO characteristics. Why don’t you try to put the sensor in power down mode once the DA14585 sleeps? In that case you should sent the “power down “command in your sensor, so that you can reduce the power consumption.
Thanks, PM_Dialog
Actually, the sensor I am speaking of is a bare sensor(not a proper embedded system is interfaced with it) connected directly in a voltage divider and then to the DA14585 ADC pin. Now, about the Datasheet, it says, max 250uA can be consumed. However, the basic kit uses an LED (KP-1608SGC) that has current rating of 20ma (according to datasheet of KP-1608SGC) and the LED is interfaced to the pin of DA14585, this put me into thinking ,if I can use a gpio pin as a current source to my voltage divider circuit. I am not much into electronics, please help me with this.
Hi DarshanDodalm,
If the table 11 “fits” with your sensor, then you might use the GPIO to power it down. However, my recommendation would be to send the power down command in order to put is in the lower power mode.
Thanks, PM_Dialog