Op Amp

[metrologist]

Would something like this work if Rs is set up to produce 0 to 50mV and R1=1k, R2=100k?

[Wimberleytech]

A couple of things...
The input resistance (looking into the inverting configuration) is 1000 ohms. How are you generating the 0-50mV?  If is from a current source, then the resistance of Rs needs to be much less than 1000 ohms if you want it to be accurate (0-50mV).

Is the opamp powered with 0 and +5volts?  If so, the 0-50mV input wants to drive the ouput of the opamp from 0 to -5Volts.  Your opamp cannot go below ground as shown.

I have attached a simulation file that will help you think about the problem and experiment.

[metrologist]

I was too brief.



Rs will be very small - probably a shunt from a defective DMM or just a piece if nichrome wire. The voltage across Rs does not seem that important because R1 and R2 will probably be formed from a 100k potentiometer.

I see the function of the op amp as to make the inputs equal voltage. When current I flows, the voltage will be lower on the inverting input, so the amp will supply a positive voltage to balance the inputs.

The op amp circuit will be powered from 0 and 5V, independent of the current source through Rs. I was looking for 0 to 5V out of the amp.

I'm not able to view an asc file.

[Wimberleytech]

OK.  Yes, your circuit should work.

I am attaching the schematic I simulated.  Need an opamp that is happy with single supply and a small negative common-mode input. Op07 that I simulated is not the right choice.  Opamp also needs a low offset voltage...maybe LT1077.

RS = 0.1 ohm
The current varies from 0 to 0.5 amps.

I plotted the voltage on Rs and also the output of the opamp.

[metrologist]

Thanks! I had some OP27. I guess I could create a virtual ground, but it's not becoming obvious to me how the op amp has a reference to know the difference.
I'm sure I have a few LT1077 too.

[Wimberleytech]

... but it's not becoming obvious to me how the op amp has a reference to know the difference.

I have attached a very simple example of a differential input to illustrate the issue.  In order for the bipolars to be biased "on" the common mode input must be large enough to support Vbe and the voltage required across the current source.  A simple solution is to use a pnp input, thus allowing the CM input to go below ground.  Designing an opamp CM to go below and above the supplies is a complicated matter, but can be done.

[metrologist]

I tried a TL1072 and could not get any reasonable output. I tried connecting my RD power supply and step the voltage in either inverting or non-inverting inputs and it would not seem to work, just went to one rail or the other. Connecting a bare wire and touching it would seem to make the output vary in random ways.

I found a TL1078 and tried building the reference circuit on TDS page 1.

http://cds.linear.com/docs/en/datasheet/10789fe.pdf

This has been very frustrating for me. The first problem is I don't know what the chip pinout really is. My chip says 1078, and that should be the pinout shown on the right of TDS page 2. I tried building it twice as shown (I think I made a mistake on the first attempt, so got another chip).

The second problem is, I might still have it wrong because the little +/- in the chip does not match the text labeling.

The third problem is, I don't understand how to connect the circuit. If it is 100x gain, then it looks like only one op amp is needed, and you connect ground and whichever input you want. Otherwise, you connect both inverting and non-inverting inputs to your - and + inputs, respectively.

two days and many hours, this is all I got, a non-functioning blob.

If I connect RD, the output is 4.2V, does not matter what input voltage, does not matter if I connect ground and either input, does not matter if I connect to both inputs. If I connect a 12 inch wire between both inputs and move my hand around that loop, I can make the output vary from around 25mv to 4V. It's haywaire though, sometimes it goes one way or the other.

The only thing I can think of to do now is try the other chip pinout. I tried labeling the chip pins in the image.

[Wimberleytech]

I tried a TL1072 and could not get any reasonable output. I tried connecting my RD power supply and step the voltage in either inverting or non-inverting inputs and it would not seem to work, just went to one rail or the other. Connecting a bare wire and touching it would seem to make the output vary in random ways.

I found a TL1078 and tried building the reference circuit on TDS page 1.

http://cds.linear.com/docs/en/datasheet/10789fe.pdf

This has been very frustrating for me. The first problem is I don't know what the chip pinout really is. My chip says 1078, and that should be the pinout shown on the right of TDS page 2. I tried building it twice as shown (I think I made a mistake on the first attempt, so got another chip).

I am attaching an image showing how I think the connections should be made, given your assumption as to the package type.  Looks like it is different than what you did.

The second problem is, I might still have it wrong because the little +/- in the chip does not match the text labeling.

Not sure what you are seeing re "+/-"  Maybe you could show a picture of the bare chip without soldering.

The third problem is, I don't understand how to connect the circuit. If it is 100x gain, then it looks like only one op amp is needed, and you connect ground and whichever input you want. Otherwise, you connect both inverting and non-inverting inputs to your - and + inputs, respectively.

The schematic you are implementing is an "instrumentation amplifier" which means that it is designed to amplify a differential signal riding on a common-mode signal.  The configuration you describe is OK for amplifying a signal that his no common mode and has the appropriate average level.

two days and many hours, this is all I got, a non-functioning blob.

I have been there...many times...many days.  Don't give up.

If I connect RD, the output is 4.2V, does not matter what input voltage, does not matter if I connect ground and either input, does not matter if I connect to both inputs. If I connect a 12 inch wire between both inputs and move my hand around that loop, I can make the output vary from around 25mv to 4V. It's haywaire though, sometimes it goes one way or the other.

The only thing I can think of to do now is try the other chip pinout. I tried labeling the chip pins in the image.

In all problem-solving scenarios, I break the problem down to manageable chunks.  Here is what I suggest you do since you are not certain about the pinout, etc.

Put both amplifiers in unity gain configuration: connect the output to the inverting input on both.  then take a ground-referenced signal and vary it on both non-inverting terminals.  You should see the same signal at the outputs of both amplifiers as the sourcing signal.

If you confirm that, then you confirm the pinout....that is a big first step.

Next, configure ONE amplifier with the 100x gain connections.  Then try it.  You can see where I am going.

Lets know of your progress!!

[danadak]

Your original circuit post is -

1) G = 100
2) 5A * .1 ohms = 500mV * G = 50V
3) You sound like you want to do this on single supply, so OpAmp has
to common mode to ground on its inputs


In other words G is too high. Also makes more sense to do a non inverting design, so
a diff amp makes more sense.

Here is a handbook on the basics -


http://www.analog.com/en/education/education-library/dh-designers-guide-to-instrumentation-amps.html


Regards, Dana.

[metrologist]

Thanks for the support.

To explain using the schema image you posted, pin 3 is labeled "inverting input" in text words. Inside the triangle metaphore, there is a +.  What does that mean? I've never seen op amp input pins labeled +/- and not have the + mean non-inverting input, so why is there text that says inverting input? I also do not understand the significance of the words "instrumentation amplifier" nor differential signal and common mode (here I would wonder what the difference between an operational vs instrumentation amp, and common vs uncommon mode). I'm sure the function is obvious to me, but I do not understand the abstraction of the words.

It gets more confusing as looking at your 10k between outA to -inA would seem it is going from outA to the non-inverting inA (since the other pin is clearly labeled inverting input). I think this is how I had it configured the first time, though.

Is this not an unusual datasheet? Why did they even label the inputs in that sample schema that way? It has to be just plain wrong. Also, this is the first time I've seen a dip (pdip and etc. confuse much, why does plastic matter?) presented with different pinouts, and no clear way to know what you have. I wish I had the obsolete round package here.

I'll have to find another chip and try a simpler follower or something.

BTW, on page 11, fig1a and 1b, aren't those the same? I cannot follow the text preceding that because of it. One works but the other does not, it says...?

For example, a 1mV input signal
will cause the amplifier to set up in its linear region in the
gain 100 configuration shown in Figure 1a, but is not enough to make the amplifier function properly in the
voltage follower mode, Figure 1b.

[metrologist]

Your original circuit post is -

1) G = 100
2) 5A * .1 ohms = 500mV * G = 50V
3) You sound like you want to do this on single supply, so OpAmp has
to common mode to ground on its inputs


In other words G is too high. Also makes more sense to do a non inverting design, so
a diff amp makes more sense.

Here is a handbook on the basics -


http://www.analog.com/en/education/education-library/dh-designers-guide-to-instrumentation-amps.html


Regards, Dana.

I think the input voltage across the shunt is 0 to 50mv. G100 should produce 0 to 5V output?

[Wimberleytech]

Thanks for the support.

To explain using the schema image you posted, pin 3 is labeled "inverting input" in text words. Inside the triangle metaphore, there is a +.  What does that mean? I've never seen op amp input pins labeled +/- and not have the + mean non-inverting input, so why is there text that says inverting input? I also do not understand the significance of the words "instrumentation amplifier" nor differential signal and common mode (here I would wonder what the difference between an operational vs instrumentation amp, and common vs uncommon mode). I'm sure the function is obvious to me, but I do not understand the abstraction of the words.

Oh, I see your confusion.  Well, it is a differential amplifier...there is a noninverting and an inverting input.

I have attached a diagram that should clear it up.

To prove it to yourself,  do a thought experiment.
Ground the inverting input and drive a signal into the non-inverting input.  Is it truly non-inverting?
Next, ground the non-inverting input and drive the inverting input.  Is it inverting?  It goes through two stages...the first is non-inverting and the second is inverting.

[Wimberleytech]

It gets more confusing as looking at your 10k between outA to -inA would seem it is going from outA to the non-inverting inA (since the other pin is clearly labeled inverting input). I think this is how I had it configured the first time, though.

See figure.

[Wimberleytech]

Thanks for the support.

Is this not an unusual datasheet? Why did they even label the inputs in that sample schema that way? It has to be just plain wrong. Also, this is the first time I've seen a dip (pdip and etc. confuse much, why does plastic matter?) presented with different pinouts, and no clear way to know what you have. I wish I had the obsolete round package here.

The datasheet is correct.  Yes, it is a little confusing because they changed pinouts for different packages.

LOL, the old TO-5 cans.  Yes, they were cute but the sockets were more challenging...

I'll have to find another chip and try a simpler follower or something.

BTW, on page 11, fig1a and 1b, aren't those the same? I cannot follow the text preceding that because of it. One works but the other does not, it says...?

For example, a 1mV input signal
will cause the amplifier to set up in its linear region in the
gain 100 configuration shown in Figure 1a, but is not enough to make the amplifier function properly in the
voltage follower mode, Figure 1b.

Well done!!!  You found an error in the datasheet.

See attached correction.

[danadak]

I think the input voltage across the shunt is 0 to 50mv. G100 should produce 0 to 5V output?

The shunt shown is .1 ohms, max current is 5A, so Vmaxshunt = 5A * .1ohms = 500m mV input to amp.

Voutamp = .5Vmax * 100 = 50 volts.

Unless I am missing something.

Your shunt of 50 mV @ 5A = .01 ohms.


Regards, Dana.

[metrologist]

I think the input voltage across the shunt is 0 to 50mv. G100 should produce 0 to 5V output?

The shunt shown is .1 ohms, max current is 5A, so Vmaxshunt = 5A * .1ohms = 500m mV input to amp.

Voutamp = .5Vmax * 100 = 50 volts.

Unless I am missing something.

Your shunt of 50 mV @ 5A = .01 ohms.


Regards, Dana.

Yes, my shunt would need to be 10 mohm. You might have been looking at Wimberleytech's circuit.

[metrologist]

Thanks for the support.

To explain using the schema image you posted, pin 3 is labeled "inverting input" in text words. Inside the triangle metaphore, there is a +.  What does that mean? I've never seen op amp input pins labeled +/- and not have the + mean non-inverting input, so why is there text that says inverting input? I also do not understand the significance of the words "instrumentation amplifier" nor differential signal and common mode (here I would wonder what the difference between an operational vs instrumentation amp, and common vs uncommon mode). I'm sure the function is obvious to me, but I do not understand the abstraction of the words.

Oh, I see your confusion.  Well, it is a differential amplifier...there is a noninverting and an inverting input.

I have attached a diagram that should clear it up.

To prove it to yourself,  do a thought experiment.
Ground the inverting input and drive a signal into the non-inverting input.  Is it truly non-inverting?
Next, ground the non-inverting input and drive the inverting input.  Is it inverting?  It goes through two stages...the first is non-inverting and the second is inverting.

In this diagram, the reference circuit is reduced to a simple op amp. Does that mean I need to add a feedback circuit to the inverting or non inverting input, or can I simply just apply 50mA from my power supply to the inverting and non inverting inputs and see 5V output?

[Wimberleytech]

In this diagram, the reference circuit is reduced to a simple op amp. Does that mean I need to add a feedback circuit to the inverting or non inverting input, or can I simply just apply 50mA from my power supply to the inverting and non inverting inputs and see 5V output?

No it does not.  The gain is already established by the four resistors and two opamps, so no feedback is needed to set gain.

Lets reset to your original goal: Amplify a 0 to 50mV signal to 0 to 5V.  That can be done with one amplifier (LT1078) and two resistors as shown in the attached schematic.

[bson]

You need to make sure the two inputs stay within the common voltage limits.  The lower limit of the TL072 in particular is Vss + 4V!  This means if one end of Rs is grounded it can't possibly work.  For your particular scheme you need either a RR input op amp, or you need to bias the input.  However, a TL072 in particular isn't very useful with a single 5V rail because of its high low end common voltage limit.  You only get about 1V of input range...

[metrologist]

I was off rebuilding the circuit again. When I power it with no input, I measure 4V output. If I short the two inputs, I measure 4V output. If I connect a lead to the non-inverting input, the output goes to mV. If I connect my power meter to the inputs, it outputs 4V no matter what input voltage setting.

this is TL1078 and I'm still trying to build the reference circuit on TDS page 1. There is not point in considering anything else if I cannot even get this simple thing to work.

[Wimberleytech]

You need to make sure the two inputs stay within the common voltage limits.  The lower limit of the TL072 in particular is Vss + 4V!  This means if one end of Rs is grounded it can't possibly work.  For your particular scheme you need either a RR input op amp, or you need to bias the input.  However, a TL072 in particular isn't very useful with a single 5V rail because of its high low end common voltage limit.  You only get about 1V of input range...

He is using a LT1078, and measuring wrt ground, 0 to 50mV.  The LT1078 accepts input as low as ground.

[metrologist]

I updated my photo and labeled the pins. This one is working the same as the first one, and pretty much the same as the second one too.

I've been having far too many problems to continue with this sort of thing.

[Wimberleytech]

I updated my photo and labeled the pins. This one is working the same as the first one, and pretty much the same as the second one too.

I've been having far too many problems to continue with this sort of thing.

The 1M resistor connected to pin 2...where is the other end connected?--it is hard for me to tell.

[Wimberleytech]

Next, ground pins 1 and 3, and measure the voltage on every pin of the package and report back.

[metrologist]

I bent pin 7 straight up and just over the top of the IC. Both 10k and the 1M are soldered to it (that big blob).

Grounding both inputs results in 3mV output.

OK, I see the 1M needs to connect to pin 8.