AD633 AD534 square root connection -- linearizer RF detector response

Doing some testing to use an AD534 or AD633 multiplier as a square rooter.  The datasheet for the AD633 nicely show a square root configuration  with the multiplier in the feedback path of and opamp, with the output (W) being connected to the inverting input with 10K feedback resistor. The input for this configuration, E, is the connected to the inverting input of the Opamp, Then output of the Opamp driving the AD633 is split and connected both X1 and Y1, both non-inverting inputs. Both inverting inputs X2 & Y2 are grounded.

A note states that E < 0 is applicable.

Is it critical to have X2 and Y2 be grounded and input on X1 and Y1? Would it matter if X1 was grounded then input connected to X2?

The overall goal is to make the RF input to Volts output response of the ADL6010 more linear. The ADL6010 has a positive output, so I think routing it through a unity gain inverter then connecting  to the AD633+OPAMP square rooter.

A basic excel exercise shows a decently linear response by dividing down the ADL6010 output, the using the W=sqrt(10V*E) where E is the inverted ADL6010 output make a pretty nice response. Ideally it could make made perfect linear but not sure how to do that.

Any advice is appreciated.

Thanks

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    •  Analog Employees 
    on Aug 2, 2018 3:21 PM
    This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

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    EZ Admin
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    •  Analog Employees 
    on Aug 2, 2018 3:21 PM
    This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

    Thank you,
    EZ Admin
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