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LMH6401: VOCM and VICM relationship

Part Number: LMH6401
Other Parts Discussed in Thread: DAC39J84, TINA-TI

To whom it may concern,

I would like to use LMH6401 to buffer the signal from a PMOS current steering DAC (TI DAC39J84). 

I must use DC coupling in my application. 

Now here is the problem: let's assume DAC's output common mode should naturally be biased at common-mode of 0.5V. I want to directly feed this 0.5V common-mode signal into LMH6401. 

This will hurt linearity a bit for sure, but let's ignore that.

I want to use +-2.5V supply on LMH6401 and bias output common mode of LMH6401 at 0V. 

This means that I just need to leave VOCM pin open. 

However, in TINA-TI simulator, I found this result after DC simulation:

Could you kindly explain why?

Also, I want to ask about the input impedance of the amplifier. 

Is the input impedance 50-ohm to gnd each side, or floating 100ohm differentially?

I look forward to your help!

Thank you.

Weiyu

  • Hi Weiyu,

    For interfacing the DAC39J84 output with the LMH6401 input, I would recommend to use the below circuit.

    Attached is the TINA-TI file for the same. DAC39J84_Buffer_LMH6401.TSC

    The LMH6401 input impedance is 100ohm floating differentially. See below.

    Best Regards,

    Rohit

  • Hi Rohit,

    These are exactly what I am confused about.

    1) If I measure the current out of the VOCM pin after DC analysis, when VOCM is connected to an ideal 0 impedance source, I see 10mA (below). How can that node source 10mA current? Where is that current coming from?

    2) If the input port of modeled as shown in your reply, then when a common mode current of 10mA (10mA on each side), the input common mode voltage of LMH6401 should be 0.5V, rather than 0.25V shown in TINA simulation.

    I want to put a potentiometer of 5k Ohm on the VOCM pin of to find tune the output common-mode voltage of LHM6401. But I was unable to simulate that. I get this result:

    This is different from what I see from the data sheet of LMH6401

    Please help.

    Thank you

    Weiyu 

  • Hi Weiyu,

    To answer your questions:

    1. The LMH6401 VOCM pin is high impedance (~ 25kohm) that should source minimal current at its input (< 10 uA). It looks like the TINA-TI model is showing the extra current which is not correct. We will take a note of it, and will plan to fix the model in the next revision.

    2. Yes. You are correct that with a common mode current of 10mA, the input CM voltage should sit close to 0.5V rather than 0.25V as shown in TINA-TI. We have identified this issue earlier as well, and are fixing it in the existing model. That's the reason why I have shown the LMH6401 equivalent input resistance model above, which is the input circuit you will see in reality for the device.

    3. In-order to fine tune the VOCM with respect to the DC voltage at it's output, I would recommend you to use the below integrator circuit.

    4. The input CM voltage is set purely by the output CM voltage of the previous stage. The output CM voltage for the LMH6401 is set independent of the input CM voltage provided by the VOCM pin. If there is some DC offset between the LMH6401 input and output CM voltage, then there will be some DC current flowing across its feedback. However, for your application since the VICM = 0.5V and the VOCM = 0V, the dc current should be low enough such that the device outputs can source it.

    Best Regards,

    Rohit

  • Hi Rohit,

    Thank you for clarifying the software issue.

    For point 3. I know that Figure 72 (shown above) can fine tune the CM voltage, but I wonder if I must use it. Instead, can I use a simple potentiometer, whose total resistance is 10k ohm, to tune the output CM voltage?


    Weiyu
  • Hi Weiyu,

    I think it would be better to use the Figure 72, because the LMH6401 output CM voltage track happens based on the desired Vocm which can be more accurately set from an externally stable DC voltage source. I don't think a simple potentiometer can provide an accurate output CM voltage, which would otherwise need to be hand calibrated.

    Best Regards,
    Rohit