Post Go back to editing

Connecting Vref on AD5338R

If I want to use an external reference with the AD5338R, how should I connect this to Vref?  If I tie Vref to the desired reference voltage (5V for example), will I kill the chip?

 

My fear is that by connecting, a 5V rail straight to the Vref pin, this could cause problems since at startup the internal voltage reference generator is on, and driving Vref at 2.5V.

 

I could put some resistance between the rail (5V) and Vref (2.5V) but from experience, when the internal reference is off, the voltage at Vref is slightly less than 5V (presumably due to some leakage current? or even just current through the resistor string?).

 

What is the best way here?

 

(by the way, i understand that for 5V, I could just leave Vref disconnected and set the GAIN to x2, but I want to know another option)

 

Thanks!

Parents
  • Hi Ollie,

    Just to clarify what I meant on my last reply. The AD5338 (without the reference) is the same as the AD5338R in the following aspects:

    • They both use I2C protocol
    • Both are dual channel
    • Both are 10-bit DAC

    They are different in some aspects namely:

    • They use different packages
    • They are not the same die
    • They have different pin-outs
    • They use different software
    • They have different specification parameters

    Apologies if this has caused any confusion.

    To answer your question, there will be a temporary short circuit, the internal die will be heat a little bit, as they try to drive the max current but nothing else, as it will be temporary. This is under the assumption that the reference will be switched from internal to external.

    Regards,

    Mark

Reply
  • Hi Ollie,

    Just to clarify what I meant on my last reply. The AD5338 (without the reference) is the same as the AD5338R in the following aspects:

    • They both use I2C protocol
    • Both are dual channel
    • Both are 10-bit DAC

    They are different in some aspects namely:

    • They use different packages
    • They are not the same die
    • They have different pin-outs
    • They use different software
    • They have different specification parameters

    Apologies if this has caused any confusion.

    To answer your question, there will be a temporary short circuit, the internal die will be heat a little bit, as they try to drive the max current but nothing else, as it will be temporary. This is under the assumption that the reference will be switched from internal to external.

    Regards,

    Mark

Children
No Data