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About The LMP7721 in Transimpedance Amplifier Mode using 0v/5V power supply.

Other Parts Discussed in Thread: LMP7721, LMP7715

We have the following problems with the LMP7721, and we do not know if it is caused by our mistaken design, if some people already have the same experience, please help us.

The following is the condition we are using:

  • We are using a Transimpedance Amplifier made by LMP7721(using a LMP7715 to guard the input) with a T-net(to increase the gain), to convert the current signal produced by the Photon Diode. The schematic is as following:

  • We layout the PCB just as the TI Evaluation Board:

Just as (http://www.ti.com.cn/cn/lit/ug/snou004/snou004.pdf)

  • We using the PLC Sampling Module (differential input, the 0V is not directly connected the PLC power ) to get the signal after amplified, in normally working condition it is about 4.5V,which means we need a very large gain.
  • The power of PLC system is support by a separate switch power.
  • Due to some reason the 0V of our Transimpedance Amplifier can not connected to the GND of the power net.
  • The Power of the Transimpedance Amplifier is support by a battery OR a separate isolated linear power module.
  • The test is in a normal office without other larger electricity devices.
  • During the test, only the dark current of the Photon Diode is used, there is no light on the Photon Diode.

 

The following is the problem we meet:

Condition 1: When we:

  • Using the battery as the power of the Transimpedance Amplifier(5V,0V)
  • With or Without the metal Shield ①.

The output of the Transimpedance Amplifier(Vo+,Vo-, read from the PLC) is about 0.1V and with a very low noise (about 5mV, and even lower with the metal Shield ①)

 

Condition 2: When we:

  • Using the the separate isolated linear power module as the power of the Transimpedance Amplifier(5V and 0V), then linear power module 0V is not connected to the GND of the power net
  • Without the metal Shield ①.

The output of the Transimpedance Amplifier(Vo+,Vo-, read from the PLC) is about 0.1V and with a very high noise about 1V, much larger than the signal of the dark current.

 

Condition 3: When we:

  • Using the the separate isolated linear power module as the power of the Transimpedance Amplifier(5V and 0V), then linear power module 0V is not connected to the GND of the power net
  • With the metal Shield ① connected to the 0V of the linear power module.

The output of the Transimpedance Amplifier(Vo+,Vo-, read from the PLC) is about 0.1V and with a very low noise (about 5mV) of the same lever in condition 1.

 

We are wondering what causes the noise in condition 2 and is there any different with using the battery  or using an linear power module to power the circuit?

  • Hello user4083248,

    Does it respond to light correctly? What happens if you remove the diode?

    Is that 1V "noise" an oscillation? Have you looked at the output with a scope?

    For condition #2, the grounds are not connected, so you are relying on the PLC's input common mode rejection (CMR) to reject any large difference between the grounds.

    The battery would create a true "floating" source. My guess is that the Linear Power Module and the bench supply are line-powered - which means coupled line noise into the supply side of the grounds through the power transformer (or line filters). With no "ground" the 0V of the amplifier circuit are riding along on the coupled noise of the isolated supply. The difference between grounds can be 10's to 100's of volts p-p (at low currents). Measure the difference between the grounds with a AC DMM to see what the difference is...I suspect you will see many volts of difference.

    When the grounds are tied together, the difference is much less (mV's), and easier on the CMRR of the PLC input. But you then have to control where the currents flow in the grounds (ground noise/loops).

    The circuit needs to be shielded...no question there. Was the shield removed, or just disconnected? A floating shield can actually induce noise into the circuit (acting like a big antenna).

    So what you are seeing are effects of the grounding, not the amplifier. This is why you need separate "Chassis" and "signal" grounds.

    Some general comments on the circuit:

    As drawn, your "zero" current is 0V, which will not be 0V because the output cannot swing exactly to 0V.

    Beware of the "Tee" network. It is a compromise between signal gain and noise/drift (S/N ratio). The "Tee" network gives you signal gain, but not transimpedance gain. The best S/N ratio will be achieved with the proper transimpedance resistor value and no (or small) "Tee" network.

    Essentially, the Tee is like having a second amplifier stage after the main 22Meg transimpedance amp. Any errors in the transimpedance amplifier (offset voltage, noise, temp drift, resistor noise) will be multiplied by the Tee network "gain".

    So that 100mV could be the offset voltage gained-up (does it move as you adjust the gain?). Remember that offset voltage can be positive or negative, so do not rely on that 100mV as your "zero current" level as it will change part-to-part as the offset changes. Parts with a negative offset may have a "dead-zone" where the input current has to overcome the saturated "negative" output due to the offset driving the theoretical output below zero.

    Ideally, the zero current is set to a level above 0V to allow for offset and drift, and to properly "zero" the zero current.

    You have a large "Tee" gain range. Do you really need all that range? You would be better off with a larger transimpedance resistor and a lower "Tee" gain (you really do not want to go over 10x). Your S/N ratio will be better.

    Regards,
  • Thank you very much, Mr. Paul Grohe, it is really a great thing that you reply my post!
    I am just a beginner of this field, so some of my questions seem primary and maybe unnecessary, please forgive me…

    ==>Does it respond to light correctly?
    <== Yes, no matter shield or not, battery supplied or linear power supplied, it did works and the signal output changes right with the light (of course with large noise when no shield with linear power).
    ==>What happens if you remove the diode?
    <== I will test latter, do you mean if I leave the diode un-installed OR short the two inputs of the amplifier?

    ==>Is that 1V "noise" an oscillation? Have you looked at the output with a scope?
    <==Yes, it is oscillation periodically. The 1V noise is not shown on the scope, but I guess the 0V is connecting to the GND of the power net through the scope(the resistor value is 0 between the probe clip and the power plug GND of the scope). Connect the '0V' to the power net GND will eliminate much of the '1V noise' but for some reason it is not so practical in the application field(Just as you said it is not easy to control the currents flow in the grounds).

    ==>For condition #2, the grounds are not connected, so you are relying on the PLC's input common mode rejection (CMR) to reject any large difference between the grounds.
    <== There is no information about the CMR of the PLC module, it is said to be "have differential inputs and possess a common, internal ground potential." We are not sure whether it causes this problem. We use a AC DMM measure the output, it shows a similar result(an AC noise).

    ==>The battery would create a true "floating" source. My guess is that the Linear Power Module and the bench supply are line-powered - which means coupled line noise into the supply side of the grounds through the power transformer (or line filters). With no "ground" the 0V of the amplifier circuit are riding along on the coupled noise of the isolated supply. The difference between grounds can be 10's to 100's of volts p-p (at low currents). Measure the difference between the grounds with a AC DMM to see what the difference is...I suspect you will see many volts of difference.
    When the grounds are tied together, the difference is much less (mV's), and easier on the CMRR of the PLC input. But you then have to control where the currents flow in the grounds (ground noise/loops).
    <== We have the following connectors in the Power Module: L N, GND, V-, V+. The L,N,GND is connected to the power net and the V-is connected to the '0V', and V+ is connected to 5V of the amplifier. The following is the result by AC DMM:
    GND to V-:5.66V;
    GND to V+:5.66V;
    V-to V+: nearly 0V.
    You are quite right, Linear Power Module is not so "stable floating" like the battery. Is there any other type of Power Module can work just like a battery?
    AND
    |==>You wrote "With no "ground" the 0V of the amplifier circuit are riding along on the coupled noise of the isolated supply."
    Does it mean the coupled noise of the isolated supply causes the very large output noise of the amplifier? In my inappropriate understanding, if the V+ and V- is floating together synchronously refer to the GND of the Power Module, how the it effect on the PCB of the amplifier(I mean the route of the noise, for example by the induction between the PCB and other devices/wires connected to GND)? It may also solve my puzzlement about how in Condition 3 with a shield connected to V-(0V) can remove this noise. And I think is not get in the PCB through the PLC because we use the AC DMM measure the output, it also shows a large AC noise when no shield with linear power.

    ==>The circuit needs to be shielded...no question there.
    <== You mean if I continue use the Linear Power Module, a shield is necessary (connect to V-,V+ or GND?)? What about if we use a battery?

    ==>Was the shield removed, or just disconnected? A floating shield can actually induce noise into the circuit (acting like a big antenna).
    <==Yes, we did the test, a disconnected shield will sometimes introduce more noise than that without shield.

    ==>So what you are seeing are effects of the grounding, not the amplifier. This is why you need separate "Chassis" and "signal" grounds.
    <==I agree.

    Some general comments on the circuit:
    As drawn, your "zero" current is 0V, which will not be 0V because the output cannot swing exactly to 0V.
    Beware of the "Tee" network. It is a compromise between signal gain and noise/drift (S/N ratio). The "Tee" network gives you signal gain, but not transimpedance gain. The best S/N ratio will be achieved with the proper transimpedance resistor value and no (or small) "Tee" network.
    Essentially, the Tee is like having a second amplifier stage after the main 22Meg transimpedance amp. Any errors in the transimpedance amplifier (offset voltage, noise, temp drift, resistor noise) will be multiplied by the Tee network "gain".
    So that 100mV could be the offset voltage gained-up (does it move as you adjust the gain?). Remember that offset voltage can be positive or negative, so do not rely on that 100mV as your "zero current" level as it will change part-to-part as the offset changes. Parts with a negative offset may have a "dead-zone" where the input current has to overcome the saturated "negative" output due to the offset driving the theoretical output below zero.
    Ideally, the zero current is set to a level above 0V to allow for offset and drift, and to properly "zero" the zero current.
    You have a large "Tee" gain range. Do you really need all that range? You would be better off with a larger transimpedance resistor and a lower "Tee" gain (you really do not want to go over 10x). Your S/N ratio will be better.

    <==Thank you for the comments, I now can understand the disadvantage of the Tee network, I think it will just amplify the noise like a second level amplifier. I will use a 1G transimpedance resistor (also with 1pF feed-back capacitor) instead of 22M, but also keep the Tee net about 5x, because:
    1. We have to adjust in a large range the gain of this amplifier according to the field condition.
    2. A test using a stable light source have been performed, we record the noise level and found it smaller than the ‘natural ’noise(this noise can’t avoid by circuit design) of the light source we want to monitor.
    About the 100mV “zero current” problem, thank you for reminding, I will evaluate and try to improve it accordingly.