Take a look at precision peak detector circuits:
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Peak detection circuits are more susceptible to error from distortion and noise spikes in the waveform as compared to a precision rectifier, which generates the average value.

 

This is about as simple a full-wave precision rectifier rectifier circuit I know of, with an added passive RC single-pole output filter.
Using a quad op amp, you can do two circuits with one op amp package.
Edit: The output is the average value of the input signal for any signal waveform within the op amp's circuit frequency response.

A larger R3C1 time-constant will reduce the signal ripple, if needed, but at the expense of a longer settling time.

 

I see your point but depending on the application, that might not be a problem. Other error sources might far exceed a small resolution loss.

It would take a 5 volt offset to assure that a 0 to 5 volt AC signal did not go negative, and then how is a 0 to 5 volt input range going to work??? I don't think it will. The AC to DC converter is the way to go. And, as often happens, Crutschow has a good solution.

 

It would take a 5 volt offset to assure that a 0 to 5 volt AC signal did not go negative, and then how is a 0 to 5 volt input range going to work??? I don't think it will. The AC to DC converter is the way to go. And, as often happens, Crutschow has a good solution.

A simple two-resistor voltage divider would bring it back into range. Easy peezy.

 

A simple two-resistor voltage divider would bring it back into range. Easy peezy.

How do you do that with only a positive supply voltage available?

 

How do you do that with only a positive supply voltage available?

Maybe I've got it all wrong but I'm picturing the ±5V sensor signal referenced at the +5V supply voltage, an AC signal centered at +5VDC. Take that signal to ground through a ÷2 voltage divider. The resultant signal lies within the 0-5V range for the PLC.

 

I have built the model using the schematic in post #11. I did substitute the LM324 for the LT1006 to see how the simulation compared. It looks really good!!

Question 1: The LM324 has a 4 circuit IC available which would help in reducing my PCB footprint (Instead of using quantity 3 of the LT1006 for each of the 12 channels). Is there any problem or changes to the schematic that would be required to use the single LM324 IC (Digikey part no:LM324ADT) ? If so, do I need to short the unused pins to ground or can I let them float?

FYI: I also replaced the 10K Pot with a 1.2K resistor...........

Thanks again.

 

Question 1: The LM324 has a 4 circuit IC available which would help in reducing my PCB footprint (Instead of using quantity 3 of the LT1006 for each of the 12 channels). Is there any problem or changes to the schematic that would be required to use the single LM324 IC (Digikey part no:LM324ADT) ? If so, do I need to short the unused pins to ground or can I let them float?

No, the LM324 should work as a replacement without any other circuit changes.
The main difference is that the LM324 can have slightly more signal error.

For unused op amps, tie the (+) input to ground and the (-) input to its output.
But if you are building 12 of these circuits with 3 op amps per circuit, you can use the unused one for the next circuit.
That way you would only need 9 op amp packages for the 12 circuits.

If you can get by with a 1-pole filter, the circuit in post #22 uses only two op amps per circuit, for a total of 6 op amp packages for the 12 circuits.

 

No, the LM324 should work as a replacement without any other circuit changes.
The main difference is that the LM324 can slightly more signal error.

For unused op amps, tie the (+) input to ground and the (-) input to its output.
But if you are building 12 of these circuits with 3 op amps per circuit, you can use the unused one for the next circuit.
That way you would only need 9 op amp packages for the 12 circuits.

If you can get by with a 1-pole filter, the circuit in post #22 uses only two op amps per circuit, for a total of 6 op amp packages for the 12 circuits.

I was going to suggest the same thing! With the LM324 you may actually need to do a calibration. If temperature becomes an issue there are also the LM224 and the LM124, which covers the whole millitary temperature range.

 

Thank you guys for all of your help. I appreciate it!! I will circle back on this once I get the board operational.

 

I've built and tested a few of these, based on Texas Instruments circuits from their Burr-Brown IC data books and applications handbooks. Creating an adjustable output is straight-forward using an in-amp or differential amp. Below is an old schematic I did (which hasn't translated into my new software very well) - Left side is input, right side top is signal out (absolute value of input), right side bottom is a "direction" flag. If you want any further info, let me know!

 

It is too early to be certain but it looks like this circuit has a differential input and output, and that somehow there is a missing diode. But I may be off on the missing diode as I have not carefully analyzed the operation of the circuit. And it looks like your drawing program is demanding that you define those "X" points as connections. You may need to zoom way in and adjust the wire length.

 

Hey Guys......
Everybody missed the trick question.....
This is a "CURRENT TRANSFORMER" ( "CT" ) output.
A current transformer is very high impedance, and REQUIRES a load resistor to calibrate its output range.
The output voltage of a current transformer will go into outer-space without a proper load on it.
They are known to be DANGEROUS DEVICES for this reason.

On the other hand, this makes solving this problem quite easy,
( that is, if the load resistor is NOT "built-in" to the transformer ).

All you have to do, ( if the load resistor is external ), is to run the transformers output though a bridge rectifier,
and then recalibrate the load resistor using a good known "Amp-Clamp" meter.

The output of the transformer should be shunted with a TVS Diode to prevent accidentally making a lot of "Blue Smoke",
if the output of the bridge should become accidentally unloaded somehow, and spike to several hundred volts.
No load resistor = instant blue smoke !!!
There's usually not much current involved, but the voltage can go EXTREMELY HIGH with no load.

This will give you exactly whatever DC voltage range you adjust the resistor to,
with no additional electronics, and will be very linear and accurate.

Keep in mind, a current spike in the wire being monitored by the current transformer,
causes a corresponding voltage spike in the output,
which may smoke an external circuit,
so over-voltage, and voltage spike protection, is STRONGLY advised.

Also, since this is an AC circuit, a smoothing filter capacitor might be necessary.
Now all you need to know is whether you need RMS or Peak Amps for proper calibration.
.
.

 

All you have to do, ( if the load resistor is external ), is to run the transformers output though a bridge rectifier, and then recalibrate the load resistor using a good known "Amp-Clamp" meter.

When the AC signal voltage is below 1 volt or so, the DC voltage drops to zero due to the voltage drop across the diodes.

ak

 

Hey Guys......
Everybody missed the trick question.....
This is a "CURRENT TRANSFORMER" ( "CT" ) output.
A current transformer is very high impedance, and REQUIRES a load resistor to calibrate its output range.
The output voltage of a current transformer will go into outer-space without a proper load on it.
They are known to be DANGEROUS DEVICES for this reason.

On the other hand, this makes solving this problem quite easy,
( that is, if the load resistor is NOT "built-in" to the transformer ).

All you have to do, ( if the load resistor is external ), is to run the transformers output though a bridge rectifier,
and then recalibrate the load resistor using a good known "Amp-Clamp" meter.

The output of the transformer should be shunted with a TVS Diode to prevent accidentally making a lot of "Blue Smoke",
if the output of the bridge should become accidentally unloaded somehow, and spike to several hundred volts.
No load resistor = instant blue smoke !!!
There's usually not much current involved, but the voltage can go EXTREMELY HIGH with no load.

This will give you exactly whatever DC voltage range you adjust the resistor to,
with no additional electronics, and will be very linear and accurate.

Keep in mind, a current spike in the wire being monitored by the current transformer,
causes a corresponding voltage spike in the output,
which may smoke an external circuit,
so over-voltage, and voltage spike protection, is STRONGLY advised.

Also, since this is an AC circuit, a smoothing filter capacitor might be necessary.
Now all you need to know is whether you need RMS or Peak Amps for proper calibration.
.
.

OK, yes, we missed that indeed. Clearly a case of misdirected mindset. OF COURSE, the output of a current transformer at low currents, and the diode turn on voltages, just MAY provide some non-linearity to things at lower currents. Thus to be really fair and hold to less than 1% erroractoss the entire range. it does get complicated. But probably from one volt up to five volts it will be quite close.

 

Okay I see some confusion here.
The output is from a current transformer, but the TS stated it was a voltage, so I had assumed it was from the transformer burden resistor.
But if that resistor is external, than certainly a bridge rectifier can be put before that resistor to give a rectified output voltage, and that should give a linear output voltage down to low currents.
That can then be filtered to get the average AC voltage.