OPA340 summation circuit

RH;

Rather than equations, can you post the circuit in TINA-TI showing the input voltages and output voltages?

Hi R H,

Welcome to the TI E2E forums!

Here is a picture of what I think your circuit should look like:

You didn't mention putting a voltage supply on V+. It sounds like you grounded it. Your circuit will not work without applying a voltage to V+.

Your equation should be -(R2/R1)(Vin1+Vin2) = Vout, don't forget the negative sign! You are summing inputs into the inverting input of the op amp.

With only a single power supply of 5V, your op amp cannot output a negative voltage, to correct this you must apply a bias to the non-inverting input (the 2.5V seen in the image above), or use bipolar supplies.

Regards,
Chris

Where is the 330 ohm resistor?

Hi Neil,

The app note places it in series with the non-inverting input (R3).

It may act as a protection to limit input current into the op amp if the non-inverting imput voltage goes above +Vcc. I left it out because I don't see that being an issue and removing it removes a small noise source.

Regards,
Chris

Hello again. Thanks for the welcome and the quick reply.

Christopher, in your first reply your diagram has voltage sources at the op amp inputs. Are you telling me that is what I am missing? Or maybe I have not been clear. When I wrote V- and V+, I meant the voltage at the inverting and non-inverting inputs respectively, I did not mean to say I had applied a voltage there. The only voltages I am applying is Vin1 and Vin2. So it is basically a standard summation circuit as laid out in Figure 7.

Also, when you say I need to apply a bias to the non-inverting input for a virtual ground, I have done this with a voltage reference (built from another OPA340) and this outputs 2.5V at the Vcc/2 point in Figure 7.

My circuit is just like that in Figure 7 (except for only two inputs; Vin1 and Vin2). The reason I mentioned voltage values at the inverting (V-) and non-inverting (V+) nodes is because the output was not returning the sum, so I measured the voltages at all the nodes and found that at the two inputs of the op amps the voltages were not as the theory I am familiar with would suggest.

I asked why the two inputs were not at zero because I thought that was my problem, I assumed Figure 7 was correct. Are you saying I need to apply voltages at the inputs as well? If so why, and what magnitude of voltage?

Many thanks,

Rob.

Hi Rob,

Your thinking is correct. When the op amp is working correctly the V+ and V- voltages (voltages at the inverting and non-inverting op amp inputs) should be equal. This is because your using negative feedback and the op amp behavior will be to output whatever voltage (within reason) to make the inputs equal.

The limitation to this is that the op amp output cannot go above Vcc or below ground when using a single supply of Vcc, and usually there is an output voltage swing limitation that limits the op output even more (but the op amp will usually be able to output voltage close to the positive and negative supply rails). In this condition the output is "saturated".

Now, with that limitation in mind here is what I think is happening. In the equation "-(R2/R1)(Vin1+Vin2) = Vout", you are assuming that V+ is grounded. If V+ is not grounded then the equation becomes:

Which, for your resistors values of R1A = R1B = R2 = R1 = 1k reduces to:

Finally, with V+ = 2.5V then the equation is "7.5 - (Vin1+Vin2) = Vout". The summation of Vin1 and Vin2 is subtracted from 7.5V. When Vout is not in the range of 0 to 5V then the op amp output saturates at either rail and the condition V+ = V- is no longer valid. Therefore, Vin1+Vin2 is limited to values of 2.5V to 7.5V.

Regards,
Chris

Rob,

I should of asked you before, are you using the DC coupling capacitors from figure 7 in the Application Report SLOA058?

I left those out of my analysis, but those will also affect the transfer function. If you do use them then I believe your transfer function simplifies to Vin+ - (Vin1+Vin2) = Vout BUT, they will block DC voltages from VIN1 and VIN2 so that you can only add AC signals.

Regards,
Chris

Hi Chris,

Sorry for not getting back sooner.

Yes that is exactly what is happening, thank you very much.

I have only been using DC to test the circuit, although I plan to test with AC as this is what the circuit will be used for. I will therefore add the AC coupling capacitors as per Figure 7. You say this will change the output equation to 2.5 - (Vin1+Vin2). Could you say how you got to this equation?

Thanks again for solving my problem,

Rob.