Opamp Output Problem

[muashr]

Hi,

Background: The Opamp OPA547 is used in 2 identical but separate units of power supplies. One unit is functioning properly whereas the other one is defective. After comparison it was figured out that every pin of these two opamps has the same value/voltage except the input & output pins. There are 7 pins in the opamp.

Problem: The theory says that opamp just amplifies the difference in voltages applied at its input terminals. That is, if both inputs were of equal (i.e. 2mV & 2mV or 5mV & 5mV) values then the difference would be 0 and output would also be 0. However, what is noticed here is that one opamp with inputs (.6V & .6V) and the otherone with the inputs (0.9V & 0.9v) produce different outputs (instead of producing same voltage at the output). Please see the figure. Please also Keep in mind that all of the pins except for those mentioned in the figure are used but are not mentioned in the figure since they have the same values.

In case anyone interested further in knowing about the pin values then they are also mentioned in the following.
Opamp 1
1: 0.6v
2: 0.6v
3: -5V
4: -5V
5: 10.2V
6: 5v
7: -1.7V
Opamp 2 (Defective)
1: 0.9v
2: 0.9v
3: -5V
4: -5V
5: 10.2V
6: 7.3v
7: -1.7V

 

[audioguru]

Moderator, this is a duplicate thread.

 

[alec_t]

The two opamps will not be identical in practice. Have you taken opamp input offset difference into account?

 

[muashr]

The two opamps will not be identical in practice. Have you taken opamp input offset difference into account?

There is a mistake in the schematic. The feedback is -ve. I made a mistake while drawing the schematic. What you say about the fact that same Opamp with two different sets of same input voltages (0.6V+0.6v & 0.9V+0.9V) provide different outputs. Shouldn't the output be same? That is, in order to get 5V for both opamps is it neccessary to apply 0.6V at the inputs of both opamps?

 

[audioguru]

You did not show the resistor values so we do not know the voltage gain.
If the resistors produce a voltage gain of 10.0 then with +0.60V input the output should be +6.0V plus or minus amplified offset voltage.
If the resistors produce a voltage gain of 10.0 then with+0.90V input the output will be saturated as high as it can go which might be +8V since the positive supply voltage is not high enough for it to be +9.0V.

For an output of +5.0V with an input of +0.60V then the resistors must set the voltage gain at 5V/0.6V= 8.3333 times.

 

[audioguru]

Why are there TWO threads about this?
MODERATOR, please combine the threads.

 

[muashr]

You did not show the resistor values so we do not know the voltage gain.
If the resistors produce a voltage gain of 10.0 then with +0.60V input the output should be +6.0V plus or minus amplified offset voltage.
If the resistors produce a voltage gain of 10.0 then with+0.90V input the output will be saturated as high as it can go which might be +8V since the positive supply voltage is not high enough for it to be +9.0V.

For an output of +5.0V with an input of +0.60V then the resistors must set the voltage gain at 5V/0.6V= 8.3333 times.

Since the both power supply units are identical therefore the corresponding resistors and opamps are also same (Therefore, they must have the same gain!). That is, the reason I did not mention them. All I need to know is whether or not different sets of equal input voltages results in different output voltages. If this were the case then the problem is almost solved because then the IC (DAC) feeding one of the inputs must be changed.
What is your opinion on this?
Note: I have compared the outputs (which are inputs of opamps) of DAC of both of the power supply units. In one of the supply the DAC is feeding the opamp with 0.6V whereas in the malfunctioning power supply the DAC is feeding the opamp with 0.9V. I also want to mention the voltages/values on all of the pins of both DACs are same except the outputs.
What you think now? Should the DAC be replaced which would means that Opamp in the malfunctioning supply will also produce 5V if at its input 0.6V instead of 0.9V is applied.

 

[audioguru]

Why are there TWO threads about this?
MODERATOR, please combine the threads.

Oh, it is on the other website (EDA board).

 

[audioguru]

Since the two DACs have different output voltages then obviously the DAC in the malfunctioning one is defective and both opamps are probably fine.

 

[JimB]

audioguru said:
Why are there TWO threads about this?
MODERATOR, please combine the threads.
Oh, it is on the other website (EDA board).

I sometimes think that there should be a site called something like "All About EDA Tech Online"

During a quick look at the other sites, I see the same questions from the same people, being answered 0over and over again by the same people.

JimB

 

[audioguru]

Some answers to questions on other websites here and there are wrong and the corrections are in different places which makes it confusing.

 

[alec_t]

Another cause for unexpected outputs from your opamps is the fact that one input (which you intend to be the inverting input) is fed via a resistor whereas the other input is not. The input bias current will therefore affect one input more than the other and so add to or subtract from any inherent input offset (which varies between samples of the same opamp type .... check the datasheet, which specifies an offset anywhere from -5mV to +5mV).

Edit: If an opamp has identical voltages on its inverting and non-inverting inputs then its output should be 0V. You are not seeing that. Adding an input offset-nulling arrangement would overcome the output error.

 

[audioguru]

The voltage gain of the opamps is only 8.33 times.
The tiny current in the resistors amplified 8.33 times will not cause the input voltages to the opamps to be different.
The tiny input offset voltage amplified 8.33 times also will not cause the input voltages to the opamps to be different.

 

[MrAl]

Hello there,

My understanding of this question up to this point is that the original drawings were wrong in that the non inverting and inverting terminals were drawn as swapped. To correct this we swap the two and then we have simply two regular inverting amplifiers with two resistors. So we are really starting with just two plain inverting amplifiers.

My guess is that you are MEASURING the two inputs and then concluding that since they are the same for both circuits the outputs should be the same. But that's not how op amps work. The inputs could be very NEARLY the same, yet have different outputs, and this is very normal and even necessary. So the two LOOK like they are the same, but they are in fact a tiny bit different, and that difference gets amplified and appears at the output, and the feedback resistor and other resistor form a voltage divider that provides a feedback voltage that is ALMOST the same as the non inverting input terminal voltage but NOT exactly the same. This slightly different voltage is probably the cause of the confusion.

The resistor ratio A1 for the first amplifier can be calculated from:
5/(A1+1)=0.6

and the resistor ratio A2 for the second amplifier can be calculated from:
7.35/(A2+1)=0.9

Solving these for A1 and A2, we see that:
A1=22/3, and
A2=43/6

Note that if A2 was 44/6 (close to 43/6) these two would be exactly the same, and indicative of the resistor values being say 22k and 3k for example, or any other two that are the same ratio and are not too low in value.

So lets figure out the mean and see what happens...

((22/3)+(43/6))/2=29/4

so between the two we have an average ratio of 29/4 in both amplifiers, and this is not much different from the more accurate gains for the two A1 and A2. So we'll call this A3: A3=29/4

Now we know that the input of the first amp is 0.6v, and the second amp the input is 0.9v, so we'll calculate the output for both amps:
Vout1=0.6*(A3+1)=0.6*(29/4+1)=4.95 volts
and:
Vout2=0.9*(A3+1)=0.9*(29/4+1)=7.425 volts

Already we can see that we have something that is quite close to the real life measurements, so the circuits appear to be working close to the theoretical.

Add resistor tolerances to this calculation and a little input offset, and we might see exactly what was actually measured in the first post.

Taking more accurate measurements would reveal more about the circuits. It could be that one circuit has worse resistor values. To find out, swap all the resistors from one amp to the other and vice versa and see how that affects the real world measurements.

 

[alec_t]

Post deleted.

 

[MrAl]

How did you get the 8.33 value, AG? The differential input is supposedly zero, and the values of the resistors are not stated.

Hi,

See post #14. The average gain between the two is 8.25 so that's almost the same. So the circuits appear to be working more or less OK.

 

[audioguru]

I calculated the gain of the first opamp as 5.0V output/0.6V input= 8.333333. The output of the second opamp might be saturated as high as it can go near the positive supply voltage.