• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 49 subscribers
  • Views 890 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

simulating the LMH5401 and not getting what I expect

Timothy Starr
Expert 2445 points
Other Parts Discussed in Thread: LMH5401, LMH6552

I’m trying to simulate an LMH5401 and I’m using the circuit from app note tiduba4 as well as the datasheet.  I don’t get your results at all.  

 If you plug in numbers into the gain equations for the LMH5401 the resulting gain does not equal the simulation.   I’m using TINA and TI’s SPICE models.  If I SPICE their oscilloscope circuit, I don’t get any gain at all.  I only get attenuation.  If the input is 0.25V and the input divider action cuts it in half, which you would expect for a matched source and load, the output is then something like 40mVpp.  The positive output might be a volt but the negative output is also almost a volt.  This is for the LMH5401.  

 I can also get results that work with the LMH5401 but I have to use feedback resistors of 2K-Ohms in order to get a gain of 12V/V.  I have a small input resistor, like 36 Ohms and a feedback of 2K-Ohms.  I am negatively biasing the positive input and feeding the input into the negative input.  My signal only goes negative from 0 to negative 0.25 or -1 or -4.

 I need to know how biasing the input shows up in the output..  You claim that setting the bias voltage to half the input signal range will cause a bipolar output.  I agree with this and I can simulate this but the voltage is never half the input swing.  It’s always somewhere else.  I can’t figure out why there’s a difference between what youand I would expect and the simulation.

 I’ve attached a simulation file.  Could you please help me to figure this out?

 The other issue is input impedance.  I can calculate and simulate and get agreement on input impedance because I see the voltage get divided in half when I match the amplifier input to the source impedance.  However, again, this happens not by calculation agreement but more by trial and error with huge gains.  

Thanks and Best Regards,

-Tim Starr on behalf of GG@ESLMH5401-VS2.TSC

  • 0
    TI__Genius 13935 points

    Hello Tim,

    I've looked over the attached schematic.  I did not design this reference design, but it was designed to be a portion of an oscilloscope input.

    What is the purpose of the negative bias on the positive input?  As you can see it has created a differential offset on OUT+ and OUT-.  We usually try to avoid this.  When I set VS1 to 0V I get results that look more reasonable.  Did you have a reason to set it to -2V? 

    I think a key part of the answer to your question is that the gain and input impedance in SE/DIFF mode is partially an active impedance.  This impacts the gain, the input impedance and the DC operating points. 


    For single ended inputs the gain and input impedance are hard to calculate by hand.  One of our engineers has created a spreadhseet to help with these calculations.  I have found this to be very helpful.

    Regards,

    Loren3542.FDA single to diff resistor calculations with BW and Noise calculations to share.xlsx

  • 0 in reply to Loren Siebert 1
    Expert 2445 points

    Hi Loren-

    Could you please provide the gain equations for a unipolar, negative-transition, inverting, SE-to-DIFF LMH6552 and LMH5401, including the common mode output voltage.  Then, I can compare these to my models, make any corrections as needed and then be better able to predict how a circuit will work.  I want the equations to predict the individual output pin voltages.  I can get the differential voltage output by subtracting the voltages.  I don’t care if there isn’t 1 single equation as long as there’s a process, start-to-finish, for going through multiple equations.  Please be clear about if the Rf includes any internal resistors.  Same for Vout-P and Vout-N results, include or make clear if any output resistors are involved.

     The next thing I’d like is to understand why I need to use double the gain for solving my Rt for the LMH5401.  If I plug in my desired gain, say 12, I don’t get the correct answer.  If I plug in 24, I do.  This seems to be the same in their spreadsheet.  What does GAIN mean in the context of my application?  Here’s an example:  My signal is 0.25V.  The source impedance is 50 Ohms.  The amplifier input is 50 ohms.  So, I’d expect the signal to be 0.125V at the input to the amplifier.  I need 1.5V out.  So, 1.5/0.125 = 12.  If my target gain is entered as 12 in either spreadsheet, then the result is wrong.  If I enter 24, it’s correct.  I use your circuit for this and not my circuit.  So, it seems that your app note doesn’t agree with your spreadsheet.

     I think your process for finding the circuit values should be reversed.  I already know my Rt.  It’s the parallel combination of the other 2 amplifiers (LMH6552) that are hanging on my signal.  I have 3 amp’s in parallel and I pick the output (via a relay) to send to the ADC.  So, I need a 50-ohm total load impedance for my signal.  So, each amp is designed to be 150 ohms input impedance.  So, in simulation, I put 75 ohms (150 /2) where Rt goes and then simulate from there.  Also, for spreadsheet calculations, I iterate until I get Rt = 75 ohms and the correct impedance and gain.

    I might use the LM5401 as the third amp if the Gain-BW doesn’t work out for the high-gain LMH6552 (amplifying 0.25V/2 up to 1.5V) case.  In simulation, the 0.25V amp rolls off too soon.  So, I pre-set my Rt to 75 ohms and then calculate the rest.  So, I have to iterate my Rf and Av until I get the correct Rt. 

    -Tim Starr on behalf of GG@ES

  • 0 in reply to Timothy Starr
    TI__Genius 13935 points

    Hi Tim,

    I understand your challenge.  I run into similar issues as well.  Unfortunately input impedance is usually the last parameter calculated because it depends on all of the other components. 

    This application note by Jams Karki is pretty much the gold standard for differential amplifiers:

    www.ti.com/.../sloa054d.pdf

    This app note is 28 pages. The pages of interest to your questions are 8 to 14 and 17 to 20. In particular pages 9 and 10 have the equations for Vout(+) and Vout(-).  Then the single ended solutions are derived starting on page 17 using the schematic of Figure 21. 

    As for the discrepency of 2x I think that is a mainly a matter of application definitions.  In our lab we use 50 Ohm test equipment and the source resistance occurs before the indicated power or voltage of the measurment so we do not include Rs or the 6dB loss in our calculations.  In this case we use Rf/Rg as the definition for gain where Rg does not include termination resistance or source resistance.  When you include the loss of the source into the equation the gain is cut by half.  You are correct that in the broadest sense this reduced gain is the actual gain of the circuit (and indeed is the noise gain that is used for loop stability and noise analysis as well).  We typically use the amplifier gain (and not the net gain with source losses) because this is how our evaluation boards are tested and for the purposes of the specifications of the datasheet the voltage gain of the evaluation board does not include the loss in the source resistance. 

    Best Regards,

    Loren