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Precision Amplifiers Forum
INA122 Off by ~50 mV?
I am trying to amplify a 0 to ~15mV signal to a 0 to 2.5 V range using the INA122 op-amp. I calculated the gain using the equation supplied in the data sheet and adjusted a potentiometer to produce such a gain. However, the gain is well out the range of the specified error. The amplified signal is about 50 mV less than anticipated. The ref pin is currently grounded and the output of the INA122 leads to an ADC. The signal is produced by a pyranometer, which produces a current that I've converted to a voltage with a resistor. The negative output of the sensor is connected to Vin- on the INA122, and is also grounded.
Thank you in advance for any help and my apologies if the problem is trivial, I'm quite new at this.
Are you seeing a 50mV difference across the entire 0 to 15mV input range? You did not mention how you are supplying power to the INA122. In order to ensure proper operation of the part, you must provide power supply voltages that will limit the output swing during expected operation. For your application I would recommend powering the INA122 with at least +/- 3V, with +/-5V or more being better. You will notice if attempting to operate the part from a single supply, the output will not be able to reach 0V and this will cause inaccuracy in your readings.
Also, a potentiometer introduces several possible sources of error in a precision circuit. The reason being that not only do potentiometer values drift significantly over time and temperature, but also pressure applied to the wiper during adjustment can cause errors in the resistance during adjustment. For your required gain of 166.67, I calculate that you would need an Rg of 1237.11 Ohms, the closest 1% resistor value is 1.24k Ohms, which would be a gain of 166.29 (neglecting errors due to tolerance).
One final word, current to voltage conversion using a resistor may not be the most accurate method. Although an ideal current source would have an infinite output impedance, most real-world sensors do not (I was actually under the impression that pyranometers output a voltage, not a current, is yours a unique type?0. This finite output impedance will be in parallel with the resistance used to convert the current to a voltage and can cause an error in the output voltage. For example, below I show two 15uA current sources, one with a 100kOhm output impedance, and one with a 100GOhm (Giga Ohm) output impedance. Both use a 1kOhm resistor to perform current to voltage conversion, but notice the difference in output voltage:
A solution to this issue is to use a transimpedance amplifier rather than a simple resistance for your current to voltage conversion. Below I show an ideal opamp, configured as a transimpedance amplifier with a gain of 1000V/A, notice that the output voltage is now the ideal/expected value (although inverted in polarity) even with the current source's finite output resistance.
This is because the transimpedance amplifier looks like a VERY small impedance to the current source and so the error created by the source resistance in parallel is now much much smaller. Just something I felt you should consider!
Good luck with your project!
Analog Applications Engineer
PA Linear Apps
First of all, I'd like to thank you for your prompt response! I am powering the INA122 with a +5V from a regulated microcontroller power output, which in turn is being powered by a single 6V lead acid battery. Since the output is never able to reach ground does that mean there is an offset in the amplification? Or does that mean I just won't be able to accurately read very low values?
I was not aware that potentiometers drifted so much and I will replace it soon. In fact, that could be the main source of error because sometimes the error drops down to ~20 mV and sometimes it goes up to the ~50 mV I was seeing earlier.
Finally, the pyranometer I have does output a current, and it is bridged with a 147 Ohm resistor as recommended by the datasheet (http://www.licor.com/env/pdf/light/200.pdf). I think I will try building the transimpedance amplifier circuit you mentioned as I am trying to make this as accurate as possible.
Thank you again for all of your help!
That is an interesting device, it looks like it uses a silicon-based photovoltaic rather than a thermopile array for to measure total solar radiation. I recommend that you download Tina-TI (http://www.ti.com/tool/tina-ti) and simulate your circuit to see the effects of different gains and power supply voltages. To see how a 5V single supply effects the amplifier's performance, I simulated the INA122 (5V supply, 1.24kOhm gain resistor) and compared it to an ideal amplifier with exact resistor values:
I performed a DC transfer characteristic simulation which plots the output of the INA122 for an input of 0 to 15mV. From the plot we can see that the INA122 deviates from the ideal for very low inputs as well as inputs on the upper end where it nears the limits of its output voltage range.
I greatly appreciate your help and I will continue to work at it. I will post an update if necessary.
I have the same pyranometer (LI200SZ) with a 147 ohm attached for measuring voltage.I'm trying to connect to an Arduino, with a TLC251. The max voltage of pyranometer is 10mV, so i need a gain of 500 to utilize the 0-5V range of Arduino.
In datasheet of LI200 they mentions that: The shield of the coaxialcable is positive and the center conductor is negative. This is done becausethe trans-impedance amplifier used in LI-COR light meters requires anegative signal.
This pyranometer has a silicon photovoltaic detector mounted (not a thermocouple type).
My question is in relation to which the circuit for the amplifier should I use: Inverting, Non-Inverting or a Difference Amplifier.
Is the TLC251 a good choice?
I'm not sure what the best choice amplifier is, though it looks like the TLC251 will work. In my application, I ended up using a TLV2372  rail-to-rail single supply op-amp. The type of circuit you should build is called a Transimpedance Amplifier Circuit. More information can be found at  about halfway down the page. Note that the current should be flowing into ground. While I didn't do this, it might be a good idea to use multiple gain stages to help reduce noise. And finally, I believe that TI has transimpedance amplifier ICs; it might be worth looking into. Let me know if that helps!
I'm a little lost.
Before publishing my first review and after doing some research on opamps, I found an article in Make magazine on sensors, and I was building my circuit through this scheme, in which the gain was given by 1 + (R2/R1). Then I discovered that my pyranometer was not the thermocouple type, but the solar type. So I went back again looking for schemes, and I found this site.
In the Transimpedance Amplifier link figure, an Rf and Cf appears, which Cf do you advise? And the Vout = -Vin * Rf right?
I have another question regarding the pyranometer wires.
The black wire that has a resistance is negative (center), and red is positive (shield).
According to the image, I should connect the positive lead of the photodiode (the red pyranometer) to the negative input of the amplifier, is that correct?
Will the Vout come negative? If I need it to come positive, what should I do?
Regarding the gain and resistance to use, I would appreciate if you could confirm the following calculations:
If: Vout = Vin*Rf, Vout (max) = 5V, Vin(max) = 10mV
5V = 10mV(E-3)*Rf ⇔ Rf = 500 Ohm
i.e., I need a 500ohm resistor in order to move from 0-10mV to 0V-5V, correct? For this circuit, the resistance value is the gain value, isn’t it?
Having a resistance in one of the pyranometer wires allows to measure voltage instead of current, but don’t the transimpedance amplifier converts current to voltage? Isn’t the signal coming out of the pyranometer voltage? I'm confused.
I'm a little confused on your classification of positive and negative wires. Just to clarify, from page 2-2 from the LI200 Manual : "The shield of the coaxial cable is positive and the center conductor is negative." I will be using this classification from here on.
Second, you said you have a Li200SZ sensor which, according to , should terminate in bare leads and not have any resistors attached to it. If you have attached the 147Ohm resistor to it, you will need to take it off for the transimpedance amplifier circuit. (BTW, you are correct. The purpose of the transimpedance amplifier is to convert a current into an amplified voltage. Since the Li200 is a current source, you no longer need the 147Ohm resistor)
Onto the amplifier. If you connect the positive lead of the Li200 (the shield) to the negative terminal of the op-amp, you will get a negative voltage output from the amplifier. From what it sounds like, you want a positive output from the amplifier so you can interface with an Arduino. To do this, you will need to connect the positive lead of the Li200 (the shield) to ground, and the negative lead of the Li200 (the center conductor) to the negative terminal of the op-amp. Note that the positive terminal of the op-amp is also connected to ground.
Now, the calculations. If the current is flowing into ground like I described in the previous paragraph, Vout = Is*Rf, where Vout is the output voltage, Is is the current produced by the sensor, and Rf is the resistor shown in the diagram in your post. The Rf that you should choose is based on your individual sensor, and what kind of environment you want to use this in. For example, let us assume you have a Li200 calibration constant of (90.00 microamps / 1000 W/m^2). Let's also assume that you would like to use this anywhere in the world, and will therefore see a maximum (assumed) sunlight of 1200 W/m^2. To calculate the current from the sensor at that radiation...(1200 W/m^2) * (90.00 microamps / 1000 W/m^2) = 108 microamps = 108E-6 AmpsNow, to give yourself a little headroom, we'll assume that you want a 4.8Volt output with radiation of 1200 W/m^2. Using Vout = Is*Rf...Vout = Is*Rf --> Vout / Is = Rf --> (4.8V)/(108E-6A) = 44444.44OhmsTherefore, you'll want a 44.44 KOhm resistor to achieve a gain of above. I have been told that it would be good to break the gain into two stages though so you might want to experiment with that.
Onto the capacitor. The capacitor is there to reduce oscillations and noise. You will choose the capacitor based on the frequency you want cut off. Let's assume you use a 1 microfarad capacitor...f = 1/(2*pi*Rf*Cf) = 1/(2*pi*44444.44Ohm*1E-6F) = 3.581 Hz Therefore, if you want to filter out any signal faster 3.581Hz with the above configuration, you should use a 1uF capacitor.
I think that answers all your questions, let me know if you are still confused.
 ftp://ftp.licor.com/perm/env/Radiation_Sensors/Manual/TerrestrialSensors_Manual.pdf http://www.wolframalpha.com/input/?i=1%2F%282%2Api%2A44444Ohm%2A1e-6Farad%29
I have no words to thank you for your help.
I've been doing the calculations with the constant value indicated in pyranometer:
Multiplier: -10,56 (W/m^2)/microamp
Multiplier = -1 / Cal.Const
Cal.Const=-1 / -10,56 = 0,095 micramp/(W/m2)
Through information obtained on NASA website, the value for solar radiation is 1368 W/m^2.
1368*0,095 = 129,96 microamps = 129,96 E-6 amps
Vout = Is * Rf ⇔ Rf = 5V / 129,96E-6 amps = 38473,38Ohms = 38,47KOhms
Despite having made these new calculations, I ended up connecting the pyranometer with values indicated by you. (Rf=44,44KOhms).
I made the connections as you indicated me and I seemed to be getting good results, despite the constant oscillation between values. I did not use a capacitor, these oscillations may be due to this?
I connected the VDD-/GND to the GND and the oscillations decresead, was it right to make this connection?
Another question that I still have is regarding to the capacitor, how do I know how much frequency should i cut off?
The figure below is a scheme that was on the TLC251 datasheet, where a capacitor of 100 pF appears. Should I use this or is this merely illustrative?
Again thank you so much for your previous answer, it really helped me.
I'm glad to hear that I could help you! First, I do not think you need the 100pF capacitor that is shown in the data sheet. To be honest, I'm not completely sure what it does but my transimpedance circuit seems to work fine without it.
However, you do need the capacitor that is shown in the transimpedance circuit (Cf on the above diagram). This capacitor prevents oscillations in the circuit. The value of the capacitor is a design problem: exactly how much noise do you want filtered out? Planet Analog has an article where they show you how to calculate the minimum capacitance you will need to eliminate oscillations . However, I think it is a good idea to filter some noise out, so I used a larger capacitance. In the end, the capacitor value you should use will need to be greater than a certain value, as calculated by . After that, you should choose a value that filters out frequencies that you are happy with (and you can calculate that by using the equation from my previous post). Also, you should also have VDD-/GND hooked up to ground. I'm surprised that circuit worked without that.
Finally, I haven't read the datasheet for the TLC251 completely, but I'm not sure what Offset N1, Offset N2, and Bias Select do. You should make sure you understand those functions before using the IC in your application.
Let me know if that helps!
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