Non-inverting op-amp to turn 0-5V into 0-50mV

Go To Last Post
30 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I can't find any circuits online that show less than unity gain for non-inverting. Anyone have any idea how to do it?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

2 resistors?

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Holy crap. That's right isn't it...

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

There might be a situation where an op amp would be appropriate - if you needed to reduce a voltage and have a low output impedance.

Leon Heller G1HSM

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

In which case you would still use the two resistors, and a unity gain voltage follower... (Not a Gain<<1 config).

JC

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hello,
with to resistors you reduce the impedance by 101.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I have programmed a MS Excel workbook which calculates voltage dividers very comfortable.

Just fill in your requirements in the blue cells and choose one of the recommended voltage dividers (green cells) for your application.

NOTES:
- You have to enable macros.
- I have set up the workbook with MS Excel 2007. If you use older versions you >>might<< get problems opening the file.
- When selecting E48 or E96 Resistors, calculation takes a while

Regards
Sebastian

Attachment(s): 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You probably will want to look at the Vishay divider networks with 5 ppm or less temp drift. MPMT-1K/100KCT-ND
is a 100:1 divider. That is the Digikey PN

EDIT: Then use a zero drift amp to buffer it.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I'd first buffer, then divide. Unless you happen to know what the impedance of the source is.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Better yet, buffer, divide and buffer again. Also if you want it really precise, then raise the VCC rail to say 5.5 VDC so the opamp can swing to 5 volts without error. A negative supply on the negative opamp rail will handle the swing to ground. It is the first bufer that needs to swing to the 5 volt rail and the second buffer that needs to follow the 0-50 millivolt all the way to ground. If you do not provide those rails then you could have a big error in your circuit. It all depends on how accurate you want it.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Maybe if we knew a little more about the source & load impedance
and the accuracy required. The OP assumed that an op-amp was required, but that may not necessarily so if the source impedance is low and the load impedance high. (KISS)
In our analog lives we would call this a 40dB. attenuator. :)

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Considering that low end op amps have about 4.5mv max offset, it may intorduce large errors over the 0-50mv range.

If you do not need a low impedance output, just do without the unity gain output buffer.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

LTC2051 Dual Zero Drift OpAmp will get you 3 micro volts offset and no drift. If you run VCC @ 5.1 it will swing to 5 volts. You really need a negative supply as it will swing with an error of up to 15 millivolts of ground. You do not need much negative rail.. just more than that error. Since you are interested in something that had a 5 volt swing and you attenuate it to 50 millivolt swing you can not afford a lot of error here.

Bottom line: If you want it to be accurate you need to buffer it with enough headroom on the supplies, into the attenuator. You might as well use a dual amp and buffer again coming off the attenuator.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Not sure what the load impedance of the UDC2500 Honeywell PLC thermocouple is. I'm hoping it's quite high as I'm about to experiment with a simple voltage divider off of the filtered PWM pin and see what happens. It might be that buffering before and after an op-amp is required.

I also might be able to compensate for all this in software as I plan to have my pc give the AVR the appropriate OCR1A byte.

Right now, I'm using 10-bit fast PWM on the mega644 so my equation for the voltage out is as such:

OCR1A = (uint16_t)((dc_voltage)/(5.09) * 0x3ff);

I can divide that by 100 with a voltage divider downstream (that is high-impedance itself) and send that to the Honeywell PLC to see what happens.

I can map different values of OCR1A to the actual temperature that the Honeywell spits out and make my own non-linear lookup table for OCR1A values to Honeywell temps.

Gonna try that now.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Looks like I'll need to buffer the output of my voltage divider as the Honeywell is taking the output signal and bouncing it all over the place. The Honeywell is not even giving me a temperature value.

@LTC2051: The LTC2051 looks interesting as a possible buffer solution. Good news is that I don't need it to get up to 5V or down to 0V as I want to simulate a temperature range of about 1000 degF to 1400 degF which is a swing from 42mV to 56mV.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
Looks like I'll need to buffer the output of my voltage divider
What resistor values are you using? Are you feeding the unfiltered PWM signal via the divider into the Honeywell thingo?

I envisage that you should have the PWM pin going a filter\opamp wich would feed the divider.

I'm guessing that the output of the above would be clean DC and could feed say a 1K load being made up of a 10R and 990R resistors giving you less that 10R source impendance.

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

From what we can allude you have a K type thermocouple that
you are trying to interface to something (an AVR)?
The Honeywell information is redundant!

Quote:

I want to simulate a temperature range of about 1000 degF to 1400 degF which is a swing from 42mV to 56mV.

In other words, reading between the lines you are trying to obtain a DC voltage from 42mV to 56 mV from a +5V power supply?

Quote:
the Honeywell is taking the output signal and bouncing it all over the place

I don't understand what you are saying?

JS's solution should work fine as your power supply should have a very low output impedancd and any digital voltmeter will have a very high input impedance. NO OP AMPS REQUIRED You can use a potentiometer in the bottom part of the attenuator & get a fully variable voltage which you can trim
to the exact value.
I gather this whole thread has gone of the rail by the fact that everyone assumed that you had an interfacing & measurement problem.

Now interfacing the final Thermocouple will be a little more tricky, but you will need gain rather then attenuation.

Are you actually building something or is this just a academic exercise?

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

@LDEVRIES: Actually, I'm doing the opposite. I'm faking a thermocouple using an AVR. I'm trying to produce the signal that a real type K thermocouple would produce.

I'm taking the PWM output of the AVR and running it through a resistor and then into a capacitor with the other end tied to ground. This makes a DC signal from the PWM signal.

Next thing I do is take that DC signal and run it into a voltage divider circuit to bring it down by around 100X or so. I'm using a 100kohm in tandem with a 8M ohm resistor to do the division.

That new signal with the current pwm duty cycle I'm using results with 45 mV which ought to be interpreted as about 1100 degF. That signal is going into the Honeywell PLC.

The problem is that the signal is stable on my scope when it isn't tied to anything. When I then tie it to the Honeywell, it goes all over the place so it seems the Honeywell is low impedance hence my need for a buffer.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Thanks for that. It was not clear to me (and probably not to many others ), what

Quote:
Circuit required - Non-inverting op-amp to turn 0-5V into 0-50mV
actually means and what it eventually will lead to! :)

More then likely the configuration will not operate as you expect, because the Honeywell PLC has a conditioned input for a thermocouple, whereas you are assuming that it probably behaves like an ADC input.
When you connect your "simulator" output to the thermocouple input, you are probably upsetting its bias conditions (most likely because your AVR is referenced to an entirely differnt ground supply). Thermocouple interfacing is not as straighforward as you might think!

It is a complicated way of getting a small variable voltage that you have chosen to implement! :o What is the need to use a micro to produce a small variable voltage like that?

A circuit completely isolated from ground made from a 1.5V primary cell & some resistors as previously suggested may work OK?

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
A circuit completely isolated from ground made from a 1.5V primary cell & some resistors as previously suggested may work OK?

This known as a Run-Up Box, like the one described at the bottom of page 2 at this link:
http://www.dcnz.com/resources/tutorials/improvising3.pdf

Stan

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
Run-Up Box
Thanks Stan..That term is a new one on me! Good little .pdf that, for Wannabee Instrumentation Engineers.
Alternative name for pdf "If you can't do the way it should be done, do it the way it can be done! :)

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

How about using an actual thermocouple coupled to an AVR controlled heater resistor?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Quote:
How about using an actual thermocouple coupled to an AVR controlled heater resistor?

With a PID control loop...now we are talking!

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I think you may need to significantly lower your divider resistance. Try changing from 100k ans 8M to 50 and 4K ohms.

It all starts with a mental vision.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Some crazy ideas there! I need to simulate 1300 degF so I'd prefer not heating a block of metal to that just to get the thermocouple output.

Anyhow, the break-out box pdf shows a 9-volt battery and a voltage divider. I'll see if that works. If it does, that ought to prove that my circuit will work if I use an output buffer.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

The battery with a voltage divider works fine so next step is to order that buffer with my circuit.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Why? If it's not broken, don't fix it!

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Because I want to be able to change the temperature from the AVR!

Anyhow, at a last ditch effort, I used KitCarlson's lower resistance values for my voltage divider and it works!

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

< spam message deleted by Plons >

I need to leave at least one "message" on the forum to get rid of this "member"

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

If I had a 12-bit DAC with, say, a 1V reference then one count is 1/4096 = ~250uV. That would give ~200 counts between 0V and 50mV.

If that is enough resolution then wouldn't an A4 Xmega nearly do it all? (the 0-5V would need to be stepped down)

Lee

You can put lipstick on a pig, but it is still a pig.

I've never met a pig I didn't like, as long as you have some salt and pepper.