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LMH6644: What does Vcm=Vo=V+/2 mean? andRL=2kohm to V+/2?

Part Number: LMH6644

What does Vcm=Vo=V+/2 mean?  andRL=2kohm to V+/2?

From http://www.ti.com/lit/ds/symlink/lmh6644.pdf  P6

  • Hi William, 

    This is saying that the common-mode and the load resistor are terminated to a voltage that is set mid-way between the supply voltages. For example if the supplies were V+ = 2.5 and V- = -2.5 then Vcm = 0, or if they were V+ = 5 and V- = 0, then Vcm = 2.5. 

    Regards, 

  • Hi Jacob Freet:

    all limits ensured for V+ = 3 V, V– = 0 V, VCM = VO = V+/2, VID (input differential voltage) as noted(where applicable) and RL = 2 kΩ to V+/2.

    but what's the parameter below if V–=0.85V or V– get other Non-zero voltage?

    Can you give me the schematic of the following test conditional parameter model? I want to apply the parameter details when V– is not equal to 0.

  • Hello William,

    The parameters should be equivalent if the negative supply is taken below ground. The main factor will be the difference between your supplies. For example if you used V- = 0.85V and V+ = 2.15, the difference in your supplies will still be 3V and so you can use the parameters in table 7.5 for your design. Note that the common mode will now have shifted to 0.65V. 

    Best,

    Hasan Babiker

  • Hi Hasan Babiker:

    Could you give me the schematic of the following test conditional parameter model?

    i don't clear about parameters in table 7.5.

  • Hey William,

    I built up a Tina model with the test conditions on table 7.5. You will notice that the gain may be different based on which parameter you are looking at, I just went with Av = +1. Note that MID in this case will be 1.5V. Hope this clears things up!

    5226.WilliamYan.TSC

    Best,

    Hasan Babiker