EVAL-ADHV4702-1 using single power supply

Dear Sir/Madam,

I am trying to modifying the EVAL-ADHV4702-1 evaluation board to single power supply (0-220V) mode, to amplify a 0V-2V triangle voltage waveform to 0-200V voltage:

1. For the power management part (the flyback converter on board), I am going to connect both VEE_TRANSFORMER and VEE pins to AGND at P4, and connect VCC_TRANSFORMER to VCC at P1, while leave P5 open. By doing so, I should be able to get a 220V VCC and 0V VEE.

2. For the opamp part, I am going to connect VMID to AGND by P2 to get the full positive amplification from 2V to 200V.

Is this correct?

I see the user guide suggest to connect VMID to (VCC+VEE)/2, which is 110V for my case. But if I do so for the 2V input and 100V/V Av, the output will saturate at 220VCC-110VMID = 110 V, instead of a expected 200V output, right?

Please confirm or correct me if I made any mistakes.

Thank you so much for your help!

Best regards,

Le

  • 0
    •  Analog Employees 
    on Jan 26, 2021 5:15 AM 1 month ago

    Hi,

    For split to single supply conversion, I can see you are on the right track.
    Just leave the GND_TRANSFORMER open.

    For the VMID, considering balance split supply, the VMID is 0V possible reason to have VMID pin and AGND pin to be in adjacent and can be shorted through the P2 male header.
    However, in your case of which you will utilize single supply of 0 to 220V instead of +/-110V, your VMID should be tied at +110V.
    Considering an input of 2V and a gain of 100V/V, you will have an output of 200Vpp.
    Given that your VMID is at 110V, your output will be 100V above your VMID (210Vp) and 100V below your VMID(10Vp).

    In addition, you need to consider the ADHV4702 Input Common-Mode Voltage Range.
    The Spec Table shows +/- 107V for +/-110V supply. This means that you need to have atleast 3V headroom from both rails.
    in your case that you will use 0 to 220V, the minimum input would be 3V.

    Thanks and regards.

  • Thank you for your reply!

    So you mean the correct connection will be as Fig.1 (The reference for Vin should also be VMID?) instead of Fig. 2?

    May I ask why VMID must be connected as (VCC+VEE)/2 instead of other values?

    Thanks!

    Regards,

    Le

  • And may I ask will the circuit in Fig. 3 work (With Vcc = 200V, Vee = -5V, Vmid = 0V)?

  • 0
    •  Analog Employees 
    on Jan 27, 2021 1:21 AM 1 month ago in reply to lelelalala

    Hi,

    Apology for the confusion but I missed out that your input is unipolar (0 to 2V).

    I think the VMID will matter if you have bipolar input.
    In that case, your output can fully swing to both rails with VMID as reference.
    Where as in your case, you have unipolar input and the output will only swing at the positive rail.

    But again, please have at least 3V headroom between your input and rail.

    Below is an example of asymmetrical voltage supply(VS = +110V, -Vs = -40V) and bipolar input(-2 to 2V).
    From the equation VMID = (+VS + -VS)/2, the VMID is 35V.

    As a result, the output fully swing to both rails with VMID as reference.

    And here is a simulation for your circuit where you have unipolar input.

    As you can see, the input and output swing with respect to reference VMID(still at 35V) and this will make your output saturated at positive rail for higher gain or higher input.

    And finally, here is unipolar input simulation with 0V as reference.

    You can maximize the output to your positive rail (note also for VOH or max output swing).

    Hope this helps.

    Thanks and regards.