Bridge rectifier not giving an expected output.

[BBROYGBVGW]

Re posting this message.

 

[Harald Kapp]

That's not what I meant. What is your circuit? Of course, you don't have to disclose it, but it might help.

 

[BBROYGBVGW]

This could be an LF347. That part is a quad opamp, although not the most modern part (tha datasheet dates from 1994). Which doesn't mean it is a bad part. It may be well suited to your problem.
Connect the unused inputs as has been said, leave unused inputs open.

Next time you need an opamp (or any other component) I suggest you first evaluate what component you need, then go and buy it. Fitting an existing component (be it freshly over the counter or from your junk bin) may require tweaking the circuit to fit the component. Imho the other way round makes much more sense.

What does your current circuit schematic look like? Care to share?

I'm sure you're right Harald but, as you probably can tell, I'm not up to speed, with OP-Amp theory, sufficiently to be able to make a sensible evaluation. I'm groping in the dark really.

Thanks for the recommendation of the LF347. In my crude method of evaluation, though, wouldn't I be better with a dual Amp? To fit what I need and not have the further task of connecting the unused pins to GND; an added risk of error at my level of skill.

Yes I'm happy to share; I've attached it. It's one you suggested earlier on in the thread, to achieve a good linear transfer from an a.c. signal to a d.c. one.
Thanks for your help.

 

[BBROYGBVGW]

That's not what I meant. What is your circuit? Of course, you don't have to disclose it, but it might help.

There are two applications I'm hoping to use the rectifier for. Both are to convert a signal from a.c to d.c to go into a data logger, which can only read a 0 to 12V d.c., or a 4 -20mA input.

One signal is sourced from a tapping on the primary winding of a high voltage transformer. This is to measure, proportionally, the voltage of the output. The variable output will be very high and impractical to measure direct. The voltage range, which can be determined by at what stage the tapping is made in the progress of the winding, is envisaged to be in the order of 0 to 10V, a.c. of course.

The second signal will be proportional to the primary current and will be sourced from a current transformer in the output circuit. The very low current output will be amplified and converted to a voltage by another little op-amp circuit that I have up and running. The voltage output is 0 to 5V a.c.

 

[BBROYGBVGW]

That's not what I meant. What is your circuit? Of course, you don't have to disclose it, but it might help.

See attached.
I've just drawn up a diagram of how I think the two LF347 in the full rectifier circuit should connect into the current amplifier circuit I’ve already got working to produce a 0 to, 5V a.c. output. The input to this is a low a.c. current from the current transformer as I've explained earlier.
I hope I’ve got the +ve and –ve invertions correct; I do struggle to understand it fully. Any comments would be welcome.

Thanks

 

[BBROYGBVGW]

Corrected version of schematic

I believe I got the inputs to the first stage of the LF347 amp wrong, in my last posted schematic. I’ve attached a revised version.
Hope this one is correct.

 

[Harald Kapp]

I can't find the difference between the two schematics. The even have the same date code.

IC1B is there to generate a virtual ground. That part looks o.k.
But IC1A is connected as a current to voltage converter. From your previous posts I gained the impression that you want to measure the output voltage of the transformer.
If you connect the OpAmp this way, the poor thing (IC 1A) will try to bring the input voltage difference to 0V. Since the output of the IC is connected to the input via a 100kOhm resistor the IC will go into saturation. You may see a square wave output - if you see any output at all.

If you want to use an inverting amplifier, you need to add a series resistor to the input.
In this application you may also use a non-inveting OpAmp.

 

[BBROYGBVGW]

I can't find the difference between the two schematics. The even have the same date code.

IC1B is there to generate a virtual ground. That part looks o.k.
But IC1A is connected as a current to voltage converter. From your previous posts I gained the impression that you want to measure the output voltage of the transformer.
If you connect the OpAmp this way, the poor thing (IC 1A) will try to bring the input voltage difference to 0V. Since the output of the IC is connected to the input via a 100kOhm resistor the IC will go into saturation. You may see a square wave output - if you see any output at all.

If you want to use an inverting amplifier, you need to add a series resistor to the input.
In this application you may also use a non-inveting OpAmp.

I’m sorry if I have caused some confusion Harald. I have two applications for the circuit, as I have explained three postings back. One is for the transformers voltage output, which would just be the rectifier circuit as it stands in the original dig., with a simple voltage divider, as you have recommended. The other is a low current source for which I have used the IC1A to amplify and convert to an a.c. voltage, as you have quite rightly identified. This will then need to be rectified to a d.c. voltage with the circuit you suggested. This interface is a little trickier and that’s why I’ve posted that schematic.
The difference with the first and second schematics, made on consecutive dates, is that I had, in the first, the output from the IC1A connected to the –ve connection on the LF347 and the GND to the +ve connection. I believe the GND of the first part of the circuit should be common with the GND on the second part. So I have swapped the +ve and –ve inputs to the LF347 in the second schematic and commoned the GND.

Hope this explains things.

 

[Harald Kapp]

O.K., I see now.
Version 2 should work. Go ahead, build a prototype...

 

[BBROYGBVGW]

Thanks Harald; I'll keep you posted.

 

[BBROYGBVGW]

Well it seems to work, after a fashion. Perhaps I might be expecting too much but the results are not quite as solid as I had hoped.
First of all, the output is not very stable below 1V; perhaps that could be the forward volt drop (Vf) of the diodes I’m using. Would diodes with a lower Vf produce a better result I wonder?
Secondly I was expecting the output to be the peak to peak of the a.c. voltage but it seems to be closer to the RMS value; I wonder if this is what the circuit is supposed to produce?
Thirdly the voltage does not conform exactly to the a.c input voltage, from the first stage of the circuit. If one extrapolates a plotted line of a.c. against d.c. volts, it does not intersect at zero; it intersects the d.c. axis at around 0.5 Volts.
The a.c output of first stage of the circuit is very good and follows extremely close, in a voltage value, to the current input value. But then the OPA2277 OpAmp is a precision Amp. I wonder if it would help to make things more accurate if I used the same type of Amp for the full wave rectifier.

 

[Harald Kapp]

What you observe is called offset voltage. Yes, that should improve by using a suitable low offset amplifier.
The voltage drop across the diodes should not be an issue. t is compnesated by the OpAmps gain.
The output is nt RMS. The transfer function is Vout=abs(vin). You will need an oscilloscope or a voltmeter with peak measuring function to see the peak value. The "normal" DC range of a voltmeter will average the signal, so what you see is a result nearer to RMS than to peak, If you have a good voltmeter it will even measure true RMS, (almost) regardless of the waveform.

 

[BBROYGBVGW]

Hi Harald, I'm sorry I was away for so long; I was trying a RMS to dc. converter. It was very much worst :-( So I’m back to this circuit.

Ok, thanks for all the info, all noted. Does using a suitable low offset amplifier, mean the same as using a pair of precision Op Amps, or is there some other meaning there?
Thanks.