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LMP7721 Evaluation board Transimpedance config.

Other Parts Discussed in Thread: LMP7721, TINA-TI, LMP7715

Hi TI,

This is a follow-up question from (http://e2e.ti.com/support/amplifiers/precision_amplifiers/f/14/p/328519/1145207.aspx#1145207) original post.

I have another question regarding the LMP7721 eval. board.

I have changed the original eval. board in to trans-impedance configuration from non-inverting config. My question is, can we use same input bias current test that shows in this video (http://focus.ti.com/general/docs/video/Portal.tsp?langId=en&entryid=0_65zy5r36) for TI config? 

 

Because, when I tried to measure the input bias current with the same procedure, evaluation board now gives around -1.5V (when input is connected to the ground. When I disconnect it from the ground, it gives near 0V. Is this something wrong with my board?

Your suggestion regarding this will be greatly appreciated.

 Thanks

~Ishara

  • Hello Ishara,

    The current test in the video was a "floating input" test. The 7721 was configured as a follower. This is the "stock" configuration as shipped.

    A long, grounded  wire was inserted into the input and pulled out. The bias current charged the known input capacitance and the slope determined the bias current. I = (dV/dT) * C

    This would not work for the trans-impedance amplifier. Shorting the "input" (inverting input) node would just ground both inputs...an "illegal" condition.

    Note I am saying "input node" to refer to the measurement "input", which is the junction of the amplifier inverting input pin and the feedback resistor, to differentiate from the amplifiers input pins.

    The "leakage" current would be measured with the input node not connected. Any leakage current into the "input"  node would cause a fixed offset from the ideal zero current value (minus offset voltage). -I = Vout / Rf, or more accurately -I = (Vref-Vout) / Rf, Where Vref is the voltage in the positive input.

    Remember that transimpedance amplifier is a current-balancing circuit. Any current into or out of the "input" node causes a voltage drop across the feedback resistor. The output then responds to correct it so that the two amp inputs are back at the same voltage. The amount the output moves is proportional to the current added/subtracted from the "input" node.

    So assume you have a transimpedance amp with a 1G resistor on a split supply (Vref = 0V). With nothing connected to the input node, you should see a theoretical 0V on the output.

    BTW: It is handy to know the 7721's actual offset voltage, which can be found by temporarily shorting-out the 1G resistor to create a follower. The offset at room should be less than ±200uV.

    With the 1G resistor back in place, measure the output voltage. The value read on a DMM may be several mV due to the integrated noise of the resistor. You may need a filter or averaging for best results. The guard buffer input should be connected to the amplifiers non-inverting input pin (Vref).

    With nothing connected to the input node, if you measured +10mV on the output, then 10mV/1G = -10pA flowing at the "input" node. This would be 10pA flowing OUT of the input node.

    Remember that input bias current is just one of the many sources of leakage. Make sure the board is clean (as shown) and handle it by the edges.

    Regards,

  • Hi Paul,

    Thank you very much for your help so far.

    I have to clarify another point regarding LMP7721 evaluation board. For the trans-impedance configuration, how can we calculate the feedback capacitor value?( I am using 10GΩ resistor as the feedback resistor.)

    Thanks Again

    ~Ishara

  • Hello Ishara,

    The compensation capacitor value will depend on the total capacitence on the input (including sensor capacitance, layout capacitance, cabling capacitance and amplifier input capacitance - it all counts!).

    There are several appnotes & articles on the subject:

     http://www.ti.com/lit/an/sboa055a/sboa055a.pdf

     http://www.ti.com/lit/an/sboa122/sboa122.pdf

     http://www.ti.com/lit/pdf/snoa515

    The simplest formula is:

     Cf = SQRT ( Cin / (6.28*GBW*Rf))

    Where:

     Rf = Feedback resistor (in Ohms)

     Cf = Compensation Capacitor across Rf (in Farads)

     Cin - Total input capacitance (includes sensor, fixture and amplifier input capacitance)

     GBW is the bandwidth of the amplifier (in Hz)

    Sub-pF values for CF are not uncommon for Gohm resistors. You may be surprised about how much difference 0.2pF to 0.5pF can make! So don't rule it out... Below 0.1pF is usually not too feasible unless some drastic layout precautions are made. Most resistors have 0.3pF to 0.5pF strays across them to begin with.

    No matter how many lengthy equations or simulations you run, the actual value is almost always found by tweaking the actual value in the actual circuit. So be prepared to "tweak" the value. Tweaking should be done with pulsed signals so that settling can be observed. Be sure to test the design over the entire input range and operating conditions, too, as some sensors (photodiodes) can change their capacitance over the operating range.

    Regards,

  • Hi Paul,

    Thanks for your help so far. I have another point to clarify.

    I am using LMP7721 Eval. board TI config. I tried different compensation capacitor values to get a stable signal. What I noticed, when I increase the capacitor (CF)  value it help me to reduce the noise but, at the same time sensitivity of the Op-amp reduced. ( output voltage for input fA signal). Is this expected for TI configuration or something wrong with my set-up? 

    Looking forward to hear from you.

    Thanks

    ~Ishara

  • Hi Ishara,

    What value and type of capacitor are you using? Standard ceramic capacitors may have high parallel resistance/leakage. You want to use polypropylene types or, ideally, Teflon capacitors.

    You can make a nice Teflon capacitor out of a few inches of Teflon dielectric coax (RG-188). Connect the outer braid to the output and the inner conductor to the inverting node. Start with a few inches and cut it back to the proper value.

    Remember, adding the capacitor to "filter" the signal is really just increasing the time constant. Are you waiting long enough??

    Regards,

  • Thanks for the prompt reply Paul. 

    You are right. I used ceramic capacitors(C0G/NP0) with capacitance 0.1, 1, 10, 15, 20, 30, and 100 pF. 100pF gives the lowest noise but sensitivity reduces significantly. ( I have an aerosol sensor, which charged particles condensed on to an electrode. This generates a low current signal (fA level). LMP7721 uses to measure this signal). Your explanation makes more sense now. 

    I recorded the data for a long period. So, time constant should not be a problem. But, what are the values for typical time constant . (I think it is in milliseconds, right?)

    Another question. I ordered two evaluation boards but, when I was doing soldering solder pads( provided for Cf) was damaged of the one board. is there anything TI can do to repair that? 

    Also, it is really difficult to find a supplier for 1T(or above) 1206 size resistors. Do you have any suggestions? Newark and Allied do not carry stock in this range.

    Thanks again and really appreciate your help.

    ~Ishara

  • Hi Ishara,

    The feedback resistor pads were a compromise. You can get up to 550G in 1206. To put the pads for a 2512 resistor would have added too much surface area. The majority of the resistors used would be 1206 or smaller - and you can stretch 0402's onto 1206 pads...but the 2512 just would have been too big. You can't please everybody...:^(

    You can mount the large 1T resistors on their side...it also reduces the stray capacitance to the board...tricky, but possible. Not great for production, but good for prototyping (and easier to remove). Solder one end down (an intentional "tombstone") and solder a thin wire from the other terminal to the other pad. Ideally the wire should be on the low impedance (output) side.

    Or you could get two of the 549G resistors and tombstone them together...it also reduces the series capacitance by half....IF you can get them...Ultra-high resistance values are not a high volume item and distributors generally don't stock them. You may have to go directly to the manufacturer and ask for "samples".

    Sorry... I do not have any 'spare' boards here...I sent them all to the warehouse. Just tack the cap across the top of the resistor (if possible, use a 1206 cap - larger body will reduce leakage).

    Regards,

  • Hi Paul,

    I have a question regarding LMP7721 Eval. Board. trans-impedance configuration

    In the manual, page 9, it says "For dual supply setups, Vbias is usually at ground. In this case, R2 is jumpered and R1 is left open." which you have drawn in the page 15 schematic also. However, when we do a simulation of this cct with TINA-TI (I have attached a schematic file for your reference), if Vbias at ground potential, guard connection not driven to the same potential at input node. I am wondering, what is happening there? 

    If Vbias is floating, we can get the guard potential to input signal potential.

    Thanks

    7288.LMP7721_Eval_simulation.TSC

    ~Ishara

  • Hi Ishara,

    The simulation is correct.

    There is a -9uV difference between ground and the guard. The LMP7715 has a (optimistic) offset of -9.96uV,  so you are seeing the addition of the buffers offset voltage.

    Also, the input node (LMP7721 negative Input) is not going to be at exactly 0V - since the LMP7721 has it's own offset voltage (55uV in the simulation).

    If you were to use "prefect" op-amps with no offset or bias currents - then the input and guard would match exactly. There is no such thing as the perfect op-amp.

    Vbias cannot float. The amplifier needs to know that "zero" is, and Vbias provides that. With no input current, the LMP7721's output should be equal to Vbias (±offset voltage, and bias current through RF).

    If you want to go to single supply, then you would ground V- and set Vbias to some voltage above ground. Remember that there is an inversion with the transimpedance circuit, so a positive current into the input will cause the output to go negative (trying to go below ground). So you would have to set Vbias near V+ (but less than the 1V below V+ CMVR limit).

    If you want the lowest bias currents, keep the input common mode at least a volt away from the rails.

    Regards,

  • Hi,
    My question is related to the material posted above. I want to use the lmp 7721 eval as a ph preamp, but I only want to amplify one side of ph range, the side above 7 which is 0mv to ~250mv. The probes are very high impedance typically. The reason for this is I want to use this as a preamp for an industrial controller that typically runs in a narrow ph window so my output will not change very much. Running this as a non differential amp will greatly simplify design.
    Can you suggest a configuration for the eval board? The documentation does not match, and I'm not sure what steps to take, although I am reviewing your answers in this thread carefully.