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LMP7721: Clarification on CMVR, PSRR, and CMRR. How do these parameters tell me what combinations of supply voltages to use?

Camille Leclerc
Prodigy 120 points
Part Number: LMP7721
Other Parts Discussed in Thread: TINA-TI

Apologies in advance if these are mundane specs or queries… I have only briefly worked with ideal op-amp theory in the past (roughly 4+ years ago). If my following queries should be on a separate forum post(s), please let me know and I will break up these queries.

I have been configuring a photodiode amplifier that uses the LMP7721 op-amp (for now). It is my first time designing/simulating a transimpedance amplifier circuit and so I have been trying to read-up the following forum/documentation to understand the significance/importance of op-amp parameters listed in the typical datasheets:

  1. https://electronics.stackexchange.com/questions/684615/how-to-determine-whats-a-suitable-opamp/684622#684622
  2. LMP7721 datasheet
  3. TI’s “Understanding Op-Amp Specifications” document (www.ti.com/.../sloa011b.pdf
  4. e2e.ti.com/.../5019621

TLDR: I am having trouble interpreting the relation/significance of the input common-mode voltage range (CMVR), common-mode rejection ratio (CMRR), and power supply rejection ratio (PSRR). How do these values help with selecting the appropriate supply voltage values/range to power an op-amp (such as the LMP7721)?

Detailed comments/questions:

For CMVR, my interpretation of this parameter from pp. 11-13 of document 3., leads me to believe that it is a limitation to the voltage applied to the supply pins (V+ or V- ) based on the voltage applied to VIN- and/or VIN+   where (for example) if V+ >> VIN+ then the op-amp won’t amplify the signal or operate at all… however, the LMP7721 datasheet (p. 5) simply quotes the CMVR as a max (1.5 V) and min (-0.3 V) values. Are these max and min CMVR values what V+ must be less than for the op-amp to operate/amplify a signal?

For PSRR from document 3 mentions it is a ratio between the output voltage and supply voltage “Vcc”. Is this just another way of saying what the supply voltage needs to be for proper operation of the op-amp? The datasheet of the LMP7721 (p. 5) mentions that it is tested for 1.8 V ≤ V+ ≤ 5.5 V, V- = 0 V, VCM = 0, which is the supply voltage range (Vs) quoted for the LMP7721. (Also, how is VCM = 0 when V- ≠ V+ ?)

For CMRR, the best I can gather from document 3., is that the offset voltage of the input pins (VIN- and VIN+) changes due to the amount of gain that is occurring. However from p.5 of the LMP7721 datasheet (and from 4.) I interpret this parameter as the range of voltages applied to the supply must obey (V-) < Vcm <(V+)-1.1V (for +/- 2.5V supply) for the op-amp to operate appropriately. Does this parameter imply that for combinations of (V+ or V- ) that do not adhere to the inequalities obey (V-) < Vcm <(V+)-1.1V the op-amp simply does not operate?

I tried using TINA-TI simulations along with information on a past TI forum post (e2e.ti.com/.../5019621 to learn/experiment with various supply voltage configurations that work for my TIA. I attached the same TINA-TI file to this post. Below are snippets of my TIA design and oscilloscope reading of Vout for an inputted 15 pA square wave (IG1). From this experiments/simulation I concluded/ came to the question of:

 Why do I need to apply a negative voltage to the negative supply rail (V-) to get a valid input from the LMP7721? And how does this relate to CMRR, and CMVR?

Supplied +/- 1.5 V to the supply rails (Vcm = 0). It obeys my interpretation of CMRR as -1.5 V < 0 < 0.4 V. as such the oscilloscope reads and actual square wave signal.

Supplied +/- 1.0 V to the supply rails (Vcm = 0). It does not obey my my interpretation of CMRR as -1.0 V < 0 < -0.1 V. the oscilloscope still reads and actual square wave signal.

Supplied + 2.5 V & 0.0 V to the supply rails (Vcm = 1.25). It does not obey my interpretation of CMRR as 0 V < 1.25 V < 1.4 V is valid yet the oscilloscope does not read a square wave signal with ~ 9uV amplitude.

Supplied + 3.0 V & - 0.5 V to the supply rails (Vcm = 1.25). It does obey my interpretation of CMRR as -0.5 V < 1.25 V < 1.9 V is valid and the oscilloscope does read a square wave signal.

Acknowledgment: the TINA-TI file used here is an edited version of a file found on the following forum: e2e.ti.com/.../lmp7721-clarifications-on-input-protection

Photodiode simulations.TSC

  • +1
    TI__Guru 68541 points

    Camille,

    For +/-1V supply (2V total) you should use Vsupply of 2.5V specification to determine Vcm and Vout ranges - see below.

    Since 0<Vcm<1.4V on 2.5V single supply, this means -1V<Vcm<-0.1V for +/-1V supplies.

    Also, 0.15V<Vout<2.2V on 2.5V single supply is equivalent of -0.85V<Vout<+0.7V for +/-1V supplies.

    Thus, by grounding the non-inverting input as shown below, you violating Vcm voltage by 0.1V.  what is the current range you need to measure?

    Also, even though the circuit has 62 degrees of phase margin at its effective bandwidth, the phase dips to 13 degrees around 10kHz.

    This results in marginally stable circuit as shown by transient analysis below.

    Thus, you need to recompensate the circuit by increasing Cf to 10pF - see below.

    Re-running transient stability simulation shows no overshoot - see below.

    In order to improve your understanding of datasheet specification, please review TI Precision Labs material under following link:

    https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html

    Photodiode AC Stability simulations.TSC

  • 0 in reply to Marek Lis
    Prodigy 120 points

    Hi Marek,

    Thank you very much for your continued support on my queries. I really appreciate the link you provided for the precision-labs-op-amps tutorial! I will be going through that in the next few days BlushWould there be a similar video playlist for using the TINA-TI software? I see that you are able to measure bode plots to check the gain and phase margins of 1/beta and fGBW­­­ and I would like to learn how to do that.

    Below are my response/clarifications for the supply voltages and Cf discussions:

    For the supply voltages:

    So I was correct that for the +/- 1 V supply voltages used the CMRR becomes -1V<Vcm<-0.1V?

    But what actually tells me what Vout is going to be is the relation (V-) + 0.15 V < Vout < (V+) – 0.3 V, where the 0.15 V and 0.3V is from the AVOL where 0.15 V is from 0.15 V – (0 V), and -0.3 V  is from 2.2 V – 2.5 V V­+? If the answer/clarification is found in the precision-lab link you shared then don’t bother answering this, I will go through the document shortly after sending this reply.

    For the TIA stability:

    For the TIA, I am trying to amplify a modulated square wave current of 100 fA – 15 pA (pessimistic estimate). The first stage of amplification (the TIA) is intended to convert the 15 pA modulating at 10-50 kHz current to ~10 uV square wave (hence why I am using 600 kOhm Rf).  

    I followed the following document to choose my Cf value:(https://www.ti.com/lit/ug/tidu535/tidu535.pdf?ts=1708000631412&ref_url=https%253A%252F%252Fwww.ti.com%252Fsitesearch%252Fen-us%252Fdocs%252Funiversalsearch.tsp%253FlangPref%253Den-US%2526searchTerm%253Dphotodiode%2Btransimpedance%2Bamplifier%2526nr%253D1538)

    I originally picked Cf (or C­1) to be 2.65 pF (I used 2 pF in the TINA-TI file to see how the outputted square wave changed on the oscilloscope) as I want my fp (desired bandwidth) = 100 kHz and using the formula (9) for f1 provided on p. 6 of the above link (assuming CIN = 11 pF + 135pF +11 pF +11 pF, where the 11 pF are the input and differential impedance of LMP7721 and 135 pF is the Cj of the SFH 2440 photodiode I am using) f1 > fp. Using a Cf of 10 pf would make f­p ~ 26 kHz which is okay… but unideal for the TIA first stage amplification. I could lower gain i.e, Rf of the TIA to improve stability as well as I do plan on using a cascade of op-amps to achieve my overall amplification/goal.

    My overall goal is to make a cascade of op-amps to amplify my peak 15 pA square wave modulated photodiode signal to 10 – 100 mV (also square wave modulated) while maintain a bandwidth (f­p) of minimum 100 kHz.

  • 0 in reply to Marek Lis
    Prodigy 120 points

    Quick follow-up,

    The Precision Lab tutorials has resolved my queries above. This is an amazing set of lectures and topics to get me to understand the realities of analog circuit designs with op-amps. Just going through the first 3 modules resolved my queries for this post.

    Thanks, Marek for sharing this!

    Thank you Ian Williams, Ying Zhou, and Art Kay for creating these lectures. You are all awesome.

    Below is the link for the tutorials:

    https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html

  • 0 in reply to Camille Leclerc
    TI__Guru 68541 points

    Camille,

    Great to know you enjoyed TI Precision Labs tutorials.  Having said that you have big challenges facing you in what you try to do.

    1. The cut-off frequency of your feedback filer is 27kHz, thus you may not pass 50kHz signal.

    2. If you try to do so with RC time constant of 6us, you will get a distorted waveform as shown below.

    3. Also, even if you add a low pas filter, R1||C1, with fc of 500kHz, the total output noise will be much higher than your10uV output signal - see below.

  • 0 in reply to Marek Lis
    Prodigy 120 points

    Thank you for the follow-up, Marek!

    I think for a future iteration of the circuit design I will lower Rf slightly and add a second stage of amplification so that I can get more bandwidth from the TIA. I will also look at some other Op-amp available for the TIA stage. 

    Once I have finished all the modules from the precision Lab I will attempt this circuit design again Slight smile

    Appreciate your insight, feedback, and assistance so far! You have saved me a lot of time and headaches with troubleshooting.

  • 0 in reply to Camille Leclerc
    TI__Guru 68541 points

    Camille,

    A higher bandwidth will result in higher noise and thus lower resolution - see below:

    There is a way to increase the overall TIA gain without adding another gain op amp - see below.

    However, increase in gain will also increase the output rms noise - see below.

    All in all, the total noise is dominated by the thermal noise of 600k resistor (98nV/rt-Hz) and not LMP7721 broadband noise (7nV/rt-Hz).

  • 0 in reply to Marek Lis
    Prodigy 120 points

    Marek,

    Thank you for the additional insight!

    I was under the impression that pink noise was the dominant noise in this system so it would go down as frequency increased. Not much I can do about thermal noise however... unless it is a form of leakage current a guard ring/trace could remove. 

    I would like to minimize the noise between 10-100 kHz i.e., the sampling frequency of my photodiode. If you have any additional insight/suggestions on how to do that I am all ears. Since the dominant noise is coming from the 600 kOhm resistor gain, my gut is telling me using an additional op-amp stage will not reduce the noise efficiently.

    I will see if the Noise module in the TI precision-lab lectures will help with optimizing my noise performance.  

    Best,

    Cam

  • +1 in reply to Camille Leclerc
    TI__Guru 68541 points

    The only way to lower the total noise dominated by thermal noise of a resistor is by lowering the resistor value or lowering the effective bandwidth.