Differential Amplifier Circuit Issue

[gtenrreiro]

Hi,

I have a problem making a differential amplifier work properly.

I am using a TL3472 op amp, using the standard differential amplifier configuration ( see attached ).
I am powering the op amp from 5V, and GND.
I also have the non inverting input biased at 2.5V with a voltage divider.
All resistors are 10K.
To test it I am applying the same 12V rms AC signal to both inputs, expecting to get almost nothing on the output, but instead I get a ready weird sine form of about 5V. Can someone help me figure this out?

Thanks

[ikrel]

You're biasing one of the inputs with 2.5v, even though you're inputting the same signal to both inputs one of them will still be 2.5v above the other.

remove the BIAS resistor and you should get the expected results.

[smashedProton]

uhm, this isnt going to be useful at all.  you are powering it with a single supply and you're using ac. And you're exceeding supply voltage.  You need to make a virtual ground at one half of the supply voltage and work from there.  good luck

[Mechatrommer]

I am powering the op amp from 5V, and GND.
To test it I am applying the same 12V rms AC signal to both inputs

this is your problem, 5V single supply and 12Vrms input. ~17Vamp + 2.5Vbias = 19.5V at inp+ is way off the supply rail and common mode range (0-2.8V). dont make that into a habit.

2. Differential voltages are at the noninverting input with respect to the inverting input. Excessive input current can flow when the input
is less than VCC– – 0.3 V.

[jahonen]

I see two errors, first is that your voltage shift resistors need to satisfy condition (R4a || R4b) = R2 to get good common mode rejection, so if everything else would be ok, then with everything else being 10k, they should have value of 20k.

Second error is, like mentioned by others, is that allowed common mode voltage for the opamp is exceeded by a great amount. To fix this without increasing the supply voltage, you'll need to reduce the gain of your circuit. You can try setting R1 = R3 = 56k, R2 = 1k, R4A and R4B = 2k. That should work for 12 VRMS input, but not with great margin. Gain for differential signal will then be 1/56 or about 0.0179.

Regards,
Janne

[gtenrreiro]

Thank you all!!

[kaindub]

This is in complete disagreement with all the above statements.
But I know this is correct.
Firstly you have created an inverting amplifier. the junction of R4a and R4b just creates a virtual ground. R3 now has no effect on the operation of the circuit.

What you need to do
R1 and R3 need to be of equal values.
Place a resistor between the junction of R4a and R4b to the non inverting terminal of the op amp.
call it R5. R2 and R5 need to be equal values.
R4a and R4b should be of equal values and lower resistance than R2.

You will know this is working correctly  if there is no inputs (floating) the op amp is at half the rail supply.

R1(R3) and R2(R5) set the gain

[IanB]

This is in complete disagreement with all the above statements.
But I know this is correct.
Firstly you have created an inverting amplifier. the junction of R4a and R4b just creates a virtual ground. R3 now has no effect on the operation of the circuit.

To take issue with just this point, before considering anything else, can you explain your statement?

R₃, R_4A and R_4B together with V_IN+ and V_DD form a resistor network producing a voltage at the + input of the op amp. By solving this resistor network it should be clear that the voltage at the + input of the op amp will depend on the value of R₃. To take two trivial cases, if R₃ = 0 then the op amp will see V_IN+, whereas if R₃ = open circuit then the op amp will see V_SHIFT. Clearly V_IN+ and V_SHIFT are different, so how can your statement be correct?

[kaindub]

IanB
I made an assumption that the parallell resistance of R4a and R4b was much lower than R3.
That's a fair assumption since R4a and R4b set up a virtual ground (really it needs a capacitor from their junction to ground).
Whilst you are correct in saying that setting R3 to zero will make the non inverting input follow the signal, the circuit is supposed to be a differential amlifier, so R3 has to have some value in order to make the circut work in that function. R3 will alos be of a significant value (10k or more) in order for the input impedance to be high and not load down the signal

I know from experience that the circuit as presented does not work, since I built something similar years ago (I as trying to reduce the number of components - one resistor- but to no avail)


Robert

[Harvs]

IanB
I made an assumption that the parallell resistance of R4a and R4b was much lower than R3.
That's a fair assumption since R4a and R4b set up a virtual ground (really it needs a capacitor from their junction to ground).
Whilst you are correct in saying that setting R3 to zero will make the non inverting input follow the signal, the circuit is supposed to be a differential amlifier, so R3 has to have some value in order to make the circut work in that function. R3 will alos be of a significant value (10k or more) in order for the input impedance to be high and not load down the signal

I know from experience that the circuit as presented does not work, since I built something similar years ago (I as trying to reduce the number of components - one resistor- but to no avail)


Robert

I don't think that's a fair assumption at all.  As discussed above, as long as the elements are equal it works fine (i.e. the parallel impedance of the bottom leg.)


If you need a solid divided virtual ground reference, use a opamp to buffer a resistor divider.  Otherwise the impedance of the divider needs to be taken into account for all your equations, which becomes almost impossible where the ground is used in multiple places (which is why you'd separate the divider out from the rest of the circuit in the first place.)