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Let’s use an example where the device is configured to use 1 channel with an LED current of 100mA. The peak AFE current is = 3.0mA +(1.5mA * #CH) + (5.7x10^-3*i

_{LED}). For the described example, the peak AFE current would be ~5.1mA. To properly select the size of the decoupling cap to provide the peak currents during sampling you would use the following equation:C = (t

_{AFE}* I_{AFE})/dVWhere:

- C is the decoupling capacitance
- t
_{AFE}is the amount of time that the AFE is on. This will be equal to (t_{LED_Offset}+ (t_{LED_Period}* #Pulses)) - dV is the maximum allowable drop in the supply voltage and must be less than 0.1V
- I
_{AFE}is 5.1mA as calculated above

See the datasheet for a detailed description of how to calculate the current draw of the ADPD103 and the variables listed in the equations.

Let’s use an example where the device is configured to use 1 channel with an LED current of 100mA. The peak AFE current is = 3.0mA +(1.5mA * #CH) + (5.7x10^-3*i

_{LED}). For the described example, the peak AFE current would be ~5.1mA. To properly select the size of the decoupling cap to provide the peak currents during sampling you would use the following equation:C = (t

_{AFE}* I_{AFE})/dVWhere:

_{AFE}is the amount of time that the AFE is on. This will be equal to (t_{LED_Offset}+ (t_{LED_Period}* #Pulses))_{AFE}is 5.1mA as calculated aboveSee the datasheet for a detailed description of how to calculate the current draw of the ADPD103 and the variables listed in the equations.