Hi, I am wondering what is the current limit on the AIN inputs for the AD9650,

I am designing a high speed sampling system that has a +-10V peak pulse so I need dc coupled inputs.

Best Regards

Lasse Eriksson

  • 0
    •  Analog Employees 
    on Dec 2, 2014 12:16 AM

    Hi Lasse,

    The AD9650 datasheet specifies an allowed voltage range for the input.  For normal operation, the AD9650 accepts a 0.9 V common mode voltage typically, but the common mode voltage can be increased as shown in Figure 67 of the datasheet to accommodate a 2.7 Vpp differential input.  To have a 2.7 Vpp differential input, each of the differential pins is allowed to swing 1.35V which is +/- 0.675 V. 

    So the max input is when the VIN+ pin = Vcm + 0.675; and the VIN- pin = Vcm - 0.675V.

    While the min signal is the VIN+ pin = Vcm - 0.675; and the VIN- pin = Vcm + 0.675V.

    Table 6 in the datasheet shows Maximum (not operating) range of the input pins to be -0.3 to AVDD + 0.2V.  Damage to the device will occur outside of this range.

    I'm not sure how you planned to accommodate the +/- 10V peak pulse, perhaps with a resistor divider network?



  • Hi Anthony.

    Yes I have read the datasheet about the maximum voltages.

    For many adcs if the voltage can be higher then the absolute value then there is a diode inside that limit the voltage and you can limit the current to the input to a maximum value(10mA).

    So there are no such value in the datasheet, so there for I ask if there are a diode in the adc?



  • 0
    •  Analog Employees 
    on Dec 3, 2014 8:57 AM


    The input voltage can exceed the operating range but must always stays within the Absolute Maximum Range. Beyond the Maximum Range the ESD diodes will become active and could damage the device.  Do not rely on ESD diodes to shunt current as this can cause damage to the device.  Figure 68 in the datasheet shows an approximate input circuit that show the ESD diodes on the Input circuit.

    Specifically, regarding the question of what would be the max input current, I suggest using the Operating range of the device.

    Using the following calculation gives an approximate value for a maximum current of ~ 740 uA.

    If clock is 100 MHz, Cin = 11 pF (Vin+ to GND), input range is 2.7Vpp differential or +/- 0.675 V single ended

    I = c dV/dt ; so max current allowed at the input is 11 pF * (0.675 V * 100 MHz) ~ 740 uA single ended.