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INA125: INA125P giving a different reference voltage than what is being given in the data sheet

Part Number: INA125

For the last six months, I have been using INA 125 P amplifier. The problem I am facing is that in figure 6, page number 13 in the datasheet, pins 14 and 4 have been connected and due to this connection, we would be getting a reference voltage of 2.5V (as is shown in the figure). Experimentally, I also made the same connection but I am getting a reference voltage of 1.44V. Why this anomaly is happening when everything is correct in my circuit? Please help.

  • In reply to CAPTAIN SWING:

    If you're trying to get 10V across ~100 Ohms then you need to source 100mA and to get that you'll have to start looking at power op-amps. However, a power op-amp is not going to have very good dc performance. I don't have anything to offer that is going to have an on-board reference capable of driving that bridge. If you don't want to boost the current with a transistor, then you could consider using a device like the OPA191 to buffer the reference. This is an op-amp with a 65mA max output current and good dc precision. My recommendation would be to run the bridge with a lower voltage (closer to 2.5V) and split the gain into multiple stages if needed.

    Regards,

    Zak Kaye
    Precision Amplifiers Applications 

  • In reply to Zak Kaye:

    Hai Zak,
    In figure 4 of the datasheet of INA 125, the transistor that you people have shown is TIP29C. But two different types of TIP29C's are available in the market. In one of the TIP29C's, the hFE is 75 and in the other, hFE is 15. Rest all the parameters are equivalent or same.

    Can you suggest, out of these two, which one would be better for solving the problem that I am facing.

    Secondly, could you elaborate more, that how putting a transistor, can increase the bridge impedance to 2.5 kilo ohms without making a significant potential drop, as is in the case of a high value resistance. i.e it would increase the impedance but there would be significant potential drop across it, thus further reducing the voltage required to excite the bridge.

    Regards,
    Captain
  • In reply to CAPTAIN SWING:

    Generally speaking the hFE of the transistor is not critical and will tend to fluctuate pretty heavily. A larger hFE just means you have greater dc current gain, so you don't have to drive as much current into the base to get the desired output current. You should be fine with either transistor, but since we are unsure what the drive capability of the reference is, if you want more assurance then choose the version with higher hFE.

    Adding the transistor does not increase the impedance of the bridge, it increases the impedance seen by the op-amp and supplements the output stage of the op-amp with a transistor capable of higher current output. You don't necessarily need to use the TIP29, you just need a transistor that can tolerate the maximum current you intend to source to your bridge and that has a breakdown voltage greater than your supply. As long as you make the connection as shown in the datasheet then the transistor is in the feedback loop of your op-amp, and thus does not contribute any significant error to the reference voltage across the bridge.

    Regards,

    Zak Kaye
    Precision Amplifiers Applications 

  • In reply to Zak Kaye:

    Hi Zak,

    After yesterday's reply, I tried with a different transistor in place of TIP29C. The transistor that I used was 2N2222.  My first question is that I couldn't understand the circuit diagram of figure 4, involving the transistor. (which I have attached herewith). Emitter and base portion is clear but in the collector region, only V+ is written. I want an excitation voltage of 2.5V. So, what should the appropriate value of V+. and where to connect V-?  Is V- connected to the pin 4 as shown by blue line in the diagram?

    Please suggest

    Regards,

    Captain

  • In reply to CAPTAIN SWING:

    V+ is the positive supply that you intend to pull the additional current from. This will likely just be the same supply you are using for the INA125, but you should still make sure that the supply can provide the current you need to source to the bridge. The supply must be greater than your bridge reference voltage because some voltage must drop across the collector to emitter for the transistor to function. This voltage will be the difference between your V+ and the bridge voltage, so you must make sure you remain within the power dissipation limitations of the transistor.

    Regards,

    Zak Kaye
    Precision Amplifiers Applications 

  • In reply to Zak Kaye:

    Hi Zak,

    First of all, thank you very much. You had been replying all my queries one after the other.

    I have connected the V- of the power source to the ground and the V+ to the collector as shown in figure 4 of INA 125 datasheet. So, now, I get the excitation voltage as is given in the datasheet. i.e.(2.4786  volts)

    But now, I face another problem, my Rg (between pins 8 and 9) is 1kilo ohm. So, I get an amplification factor of 65. But when I make Rg equal to 120 ohm, I do not get the amplification factor of 500. It gets saturated at 1.18 volts. My input voltage to  the amplifier is 11mV. So, after amplification, the voltage should be 5500mV but what I get is 1180 mV. My supply to the INA 125 and transistor is 5 volts. Why is this happening? Please help.

    Regards,

    Captain 

  • In reply to CAPTAIN SWING:

    Hi,

    No problem, that's what we're here for!

    Is this an ac or dc input signal? With a gain of 500 your bandwidth is limited to 900Hz, so if your signal is changing any faster than this there will be attenuation. Additionally, the output swing of the INA125 is only within about 0.8V of the positive rail (1.2V min), meaning for a typical device you can't get closer than 4.2V to the rail (or 3.8V min). If you are making the connections as shown in previous posts, then your negative output terminal is raised to 2.5V to begin with, meaning you could only accomodate an increase of about 1.3V to 1.7V before you run into the output limitations. I believe this is the root of your issue.

    Regards,

    Zak Kaye
    Precision Amplifiers Applications 

  • In reply to Zak Kaye:

    Hi Zak,

    The signal that I want to amplify is a dc input signal. So, I think there won't be any question of bandwidth for my problem as frequency is zero Hertz. I didn't understand the second and third lines of your reply.

    So, once again, I am just telling you that my Rg is 121 ohm. So, AF is 500. My input voltage to the amplifier is 10-12 mV DC. So, after amplification, the voltage should be 5000-6000 mV DC. But the voltage that I get after amplification is less than the theoretical value. The supply to the transistor and the amplifier is 5 volts.

    Just I would give an instance.

    Input voltage to the amplifier=10 mV DC

    Output amplified voltage=3.1 V DC

    For this instance, I would just give the voltages of each pin with respect to ground

    So, can you suggest me how I can get the desired output amplified voltage that matches with the theoretical value.

    The connections are all same as discussed in the previous posts.

    Regards,

    Captain

  • In reply to CAPTAIN SWING:

    There are a couple issues here and I will try to explain as best I can:

    1) The voltage you have measured at pin 5 suggests you have not tapped the transistor at the emitter, but rather the output of the reference voltage op-amp. The 2.5V transistor output is the accurate reference that you want to use. However, if you are only applying a dc signal that will never change polarity, you don't even need to bias the output to a pseudoground as done in figure 6, and you can have your load reference to ground. See below for a comparison:  

    2) You can't expect your output voltage to exceed, or even equal your supply voltage. No op-amp can truly swing all the way to the supply rail. Biasing pin 5 means that pin 10 must go to IAref + 5V, which means you need an even greater supply voltage. So if you have 2.5V at IAref, then the output needs to swing to 7.5V to produce a 5V differential output. On top of that, worst case the INA125 can only swing to within 1.2V of your positive rail, meaning in that same scenario you would actually need a supply voltage of 8.7V to get a 5V output.

    The short answer is that if you increase your supply voltage to about 10V, then you shouldn't have any problems.

    Regards,

    Zak Kaye
    Precision Amplifiers Applications 

  • In reply to Zak Kaye:

    Hi Zak,

    I made the necessary corrections like connecting pin 5 to the emitter of the transistor, so that the pin 5 is always at a voltage of 2.5 above the actual ground. I also could understand the important facts in your last reply. But now the problem is that I am able to get an amplification factor of 65 and 500 but not 1000. I would give three instances-

     For all the three instances, supply voltage to the amplifier is 15.6 volts and to the transistor is 5.2 volts.

    Similarly, for a gain of 500

    For a gain of 1000,

    In the second and third instance, we can see that the measured amplification factor is in great variance from the theoretical amplification factor. If saturation of voltage was to take place, then it should have started at 10 volts or so but why in third instance, the voltage gets saturated at 6.008 volts

    My main purpose is that I want to amplify the signal into 100 times or 1000 times. Please help why saturation takes place so early. In the data sheet, it is written that we can amplify to an extent of 10000 but here getting a gain a consistent gain of 500 and 1000 is becoming difficult.

    Please help

    Regards

    Captain

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