INA317: Output signal behavior

Hi Jagbir,

why do you open a new thread? We are already discussing your circuit here:

Kai

Kai,

Since it's a different issue than the last one about offset, hence I have started a new thread.

Jagbir

Hi Jagbir,

yes, but it's the same circuit.

The input leakage currents of INA317 must see a path to signal ground. See section 8.2.2.5 of datasheet.

Kai

Kai,

I have read the datasheet section and I am curious about what mitigation approach should I take.

I am using the function generator for testing purposes only and my actual application is a Wheatstone bridge. My input resistance is 50 Ohm with function generator and 1000 Ohm with a Wheatstone bridge. I have two questions.

1) For my application, which is a better approach for providing Input Common-Mode Current Path: microphone or thermo-couple.

2)  Why I am observing a constant dc signal instead of a full/rectified sine wave.

Thanks,

Jagbir

Kai,

Can you reply to the 2 questions.

Jagbir

Hi Jagbir,

1) For my application, which is a better approach for providing Input Common-Mode Current Path: microphone or thermo-couple.

For Microphone or thermocouple, you will need to provide current path to ground as Kai indicated, since the Ib for INA317 requires only +/-70pF typical input current. 

2)  Why I am observing a constant dc signal instead of a full/rectified sine wave.

If you feed sine wave into a Wheatstone bridge, you will have identical signals  of Vin+ and Vin-  at INA317 's input (assumed the bridge is balanced),  the output of INA317 will be a constant DC signal, since he difference input signals of INA317 is canceled.    

Best,

Raymond

Thanks Kai and Raymond,

My input strain gauge has a 1K resistance.Does Rb=10K will work for my application or I need to increase/decrease it further.

Thanks,

Jagbir

Hi Jagbir,

first you must know the signal span of your strain gauge. What signal amplitudes do you expect.

Then, the amount of offset voltage is an issue. Sometimes, especially with self-fabricated strain gauge bridges, the offset voltage can heavily exceed the maimum signal span. So, what exactly is your offset voltage originating from the bridge?

Kai

Hi,

If I place a 10 K Rb resistor then my output is always zero and it stops responding to my input.

Here is the information on my strain gauge (SG) :

• sensitivity 2 mV/Volts.
• maximum applicable strain is 30,000 micro-strains.
• Bridge excitation voltage is 3.3 V.
• Full bridge, SG resistance is 1000 Ohms.

Hence,the maximum theoretical value of the signal generated by my SG is 6.6 mV . However, I don't want to break my SG, so I will not go nowhere near 30K micro-strains in my application. For my application, the strain value is 2,000 micro-strains which should give me a signal of 33 micro-volts.The typical offset I see from my bridge is 8-10 micro-Volts

Thanks,

Jagbir

Raymond,

I was applying a sine wave directly to the inverting input of the INAMP that's why I asked why I am not observing a sine wave on the output.

Jagbir 

Hi Jagbir,

I would do it this way:

jagbir_ina317_tlv9001.TSC

But I would suggest to take a strain gauge bridge which generates a much higher output signal!

Kai

Kai,

I am testing my circuit using a external DC power and load resistor but the output is not what I expected, attached is the simulation file.I am not seeing a gain of 5. 

Jagbir INA317.TSC

If I place a 10 K Rb resistor then my output is always zero and it stops responding to my input. Why?

Hi Jagbir ,

Did you configure in the following way? The input has to be a differential signal. A waveform generator is a low impedance device.

Best,

Raymond

Raymond,

As indicated in the circuit my input is DC and  not AC, my question was why the circuit is not providing the expected gain. Can you please answer the question.

My power supply is uni-polar and hence I can not provide a differential supply to the INAMP. How can I achieve the desired gain with the existing uni-polar supply and without using extra circuit for reference .

Thanks,

Jagbir

Hi Jagbir,

you could add the LM7705 to provide a small negative supply voltage.

Kai

Hi Jagbir, 

For input voltage of 100mV with Gain=5 or Vout = 500mV you are violating the requirements of Common-mode Range vs. Output Voltage. Kai suggested a good solution, which it negative bias the voltage rail to appx. -0.23V. 

I know you are using 3.3V, the plot is not in the datasheet, but it will be similar to Figure 23. For your circuit, if you change your input from 100mV to 10mV, then you will see 50mV at the ouptut. 

Best,

Raymond

Raymond ,

If I do not negatively bias my supply voltage then what is the range of acceptable input voltages at a gain of say 500.

Thanks,

Jagbir

Hi Jagbir,

You will not have much input voltage range. Say the differential voltage is 1mV, with G=500, Vout = 500mV, which is outside of the triangle shape shown in Figure 21 with Vref = 0. Your input voltage range may be up to  100uV or so. You should be able to simulate from Tina model. 

Best,

Raymond

Raymond,

Just wanted to ensure that both of us are discussing same terminology. The voltage inside the triangle is a common mode voltage not a differential voltage.

By input common voltage you mean  Vcm = [Vin(+) + Vin (–)]/2 and by differential voltage you mean Vd= [Vin (+) - Vin (–)]

So, for a 5mV differential voltage the corresponding Vcm is 2.5mV which should give Vo as 1.25V, for a gain of 500, which is well within the triangle. But if I simulate this scenario, I don' t get 1.25 volts. Why?

Hi Jagbir,

you violate the common input voltage range:

Keep in mind that the values above are guaranteed while the curves in the figures of datasheet are only typicals.

Kai

Kai,

From your comments, it seems that for any IN-AMP circuit to work reliably either we need to provide a negative supply or we need to provide a +ve dc offset to the reference signal.

I would prefer to provide a -ve supply using LM770 as it will provide more span of the input voltage.

Thanks,

Jagbir

Hi Jagbir,

yes, that's right.

Good luck :-)

Kai

Thanks Kai!!!

I have another question on input bias resistors R7,R10 and the RC filter resistor R3,R4. As you can see that I loose almost 100 mV of my signal at the output, but if I increase feedback resistors or decrease RC filter resistance my error in the output decreases significantly. What are your recommendations for this circuit.

Hi Jagbir,

please post your simulation file.

Kai

+Simulation file.INA317.TSC

Hi Jagbir,

my recommendation is to remove R5, R7 and R10. There's no need for them.

Removing R7 and R10 and especially R5 will heavily increase the sensitivity:

jagbir_ina317_1.TSC

Kai

Thanks Kai,

I am new to TINA.How can I run a resistance sweep like you did and plot the input/output relationship in a single graph ?

Thanks,

Jagbir

Kai

Hi Jagbir,

Enclosed is also some Tina-TI Training videos in your website. 

https://training.ti.com/tina-ti-video-training-series

Best,

Raymond