Maximum input voltage

Question

Hey, 
 I have a development AD7608 development board, the datasheet specifies that the board requires a input voltage between 7 and 9 V. When looking on the schematic it seems like the inputs are connected to two different capacitors (C23, C24), a voltage divider and the VIN of the ADP7104ARDZ-5.0 chip. 
 Based on the BOM for the AD7608 eval kit, the capacitors have a rating of 16 and 50 V. 
 Based on the datasheet of the ADP7104ARDZ-5.0, VIN can range between 3.3 V and 20 V. and the Enable pin can be anything from -0.3 V to VIN (relative to ground). 
 What is it that decided the voltage range of 7-9 V on the development board, what is it that stops me from being able to connect 12 V on the input (connector J3).

ADIApproved · Accepted Answer

Hi sindrekvam

I am sorry for the late reply. 
 The 7V to 9V VIN range is based on the typical performance of the device. Please refer to Figure 17 and Figure 20 on page 9 of the ADP7104ARDZ-5.0 datasheet. 
 
 As shown, the VOUT is stable when VIN is approximately between 6V and 10V with loads ranging from 100&micro;A to 500mA (Figure 17). 
 The dropout voltage starts to increase with a current load of 10mA or greater (Figure 20). 
 Considering these typical performance characteristics, VIN is set to 7V to 9V to provide additional headroom while ensuring the device's performance is guaranteed. 
 Please review the aforementioned figures for your VIN application. Note that for optimal LDO efficiency, VIN should be close to the output voltage. 
 Thank you. 
 Regards, 
 Red)

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Maximum input voltage

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