AD698 problem

That's not distorted, it is just full-wave rectified.

Read up a bit on what LVDT's do. It's actually a transformer bridge, which the AD698 excites with a more or less sinusoidal waveform.

At the centre of the range, the bridge is balanced, an you don't get any output. As the core moves away from the centre you do get an output whose amplitude is proportional to the displacement.

The AD698 rectifies this for you, but you have to filter out the ripple.

Il 01/08/2011 03:09, Bill Sloman ha scritto:

What I see on my oscilloscope is just a frequency-variable distorted wave. Ok it seems a full-wave rectified but with a lot of distortion.

Anyway the exciting signal is a good sine wave. So there is a mistake somewhere in my design.

I'm not agree. I know how LVTD works, but the datasheet od AD698 says:

"With the addition of a few external passive components to set frequency and gain, the AD698 converts the raw LVDT output to a scaled *dc* signal. "

"The AD698 energizes the LVDT coil, senses the LVDT output voltages and produces a *dc* output voltage proportional to core position. The AD698 has a sine wave oscillator and power am- plifier to drive the LVDT. Two synchronous demodulation stages are available for decoding the primary and secondary voltages. A decoder determines the ratio of the output signal voltage to the input drive voltage (A/B). A filter stage and out- put amplifier are used to scale the resulting output."

In fact, if you watch at the block schematic there is the output filter. Furthermore fig. 10 is pretty clear about the output.

Of course I could be wrong again, in that case please tell me!

Marco

That is the top level description

With AC.

as AC and full wave rectifies the output to

But don't forget that you have to choose - and fit - real capacitors at C2,C3 and C4 to attenuate the ripple as much as you need to give you the signal bandwidth you want, while not letting more ripple through than you can live with.

Since the AD698 includes a full wave demodulator the biggest component of the ripple is at twice the excitation frequency, but you've got progressively smaller components at higher multiples of the excitation frequency, which your low pass filter attenuates progressively more effectively.

The AD698 is set up to make it easy for you to filter the rectified output, but you have to hook up the capacitors (C2,C3 and C4) to make it work, and if they aren't big enough you'll see lots of ripple.

Figure 10 is talking about the DC component of the output after rectification. You are complaining about AC components sitting on top of the DC signal.

You aren't actually wrong, merely failing to properly understand what you are seeing at the output.

Il 01/08/2011 10:23, Bill Sloman ha scritto:

Ideally I want no ripple at all, otherwise how can I measure the displacement? The bandwidth of my system is very low, less than some tens of Hz.

Ok, but at the output I see a "ripple" with a frequency that goes from zero Hz to at 2.5 kHz changing the displacement (w/ 5 kHz excitation). Shouldn't I have a constant frequency ripple with variable amplitude, should I?

You saw my schematic: do you think 1u is too less?

mmm, nope. I don't think so. I'll take a photo of the output signal: it couldn't be described as a DC component with an AC on top.

I will post the link to the image asap. Thank you again

Marco

Il 01/08/2011 10:35, Marco Trapanese ha scritto:

Here a couple of pictures at different displacement. It's hard to speak about ripple. Do you agree?

Marco

There's nothing to stop the core vibrating inside the transformer at at any frequency it feels like. That wouldn't be ripple, it would be signal.

When I built my LVDT-based pressure gauge back in 1967, that was exactly the problem I had. Happily, the signal-processing electronics were linear, and the AC signal coming from the mechanical vibration could be filtered out along with the ripple from the demodulator.

The optical lever that the LVDT replaced hadn't been as forgiving.

You seem to be seeing more ripple than you want, so 1uF would seem to be less than you need.

Anything can be described that way. The interesting question is the frequency of the AC component. If it isn't an exact multiple of the excitation frequency, it isn't ripple coming out of the demodulator.

Il 01/08/2011 14:47, Bill Sloman ha scritto:

Did you see the images I posted? A full-wave rectified ripple doesn't appear like them. And as I said the frequency varies from 0 Hz (DC at constant level, about 5.8 V) to about

For these reasons I don't like to call them "AC-ripples".

-- Marco

I have now seen the images. You need to look at the waveform that is actually being applied to the primary winding of your linear variable differential transformer, and the signal that is coming out of the secondary - before it gets into the AD698.

One possible explanation of the demodulator output you are seeing is large switching spike - probably not coming from the AD698 - which doesn't have much effect on the output of the demodulator when it isn't close to the switching edges, but saturates the output when its positive and negative spikes occur immediately before and after the polarity switch.

Finding such a spike - and where it might be coming from - if it exists

- requires poking around with a dual channel oscilloscope.

Fixing it probably mean finding some defect in the grounding and shielding arrangements.

It certainly isn't a helpful description, but interference - which is the explanation I'm suggesting (as a vaguely plausible hypothesis, rather than any kind of authoritative opinion) is almost always AC-coupled.

Are you saturating the LVDT by 67% overdriving it? Art

Il 02/08/2011 03:19, Bill Sloman ha scritto:

I've already checked both. They are good sine waves, with no evident distortion. The coming out signal changes its amplitude with displacement. Of course it reaches zero at null.

Thanks for the hint. Currently I'm powering the board with a linear power supply. There is an MCU on-board but it's not powered at the moment.

Yes, you're right. I'll try to find out the source(s) of the spikes.

Marco

Il 02/08/2011 03:30, Artemus ha scritto:

The documentation says I can drive the LVDT up to 7V. I'm slightly overdriving it (respect the nominal value) to get more output signal.

Anyway the LVTD is not saturated: I checked the output waveform with the oscilloscope and is a very good sine wave.

Marco

Il 02/08/2011 09:10, Marco Trapanese ha scritto:

There was a big mistake in the schematic. The resistor divider that provide the reference voltage gave a 12V. Instead it should be 2,5V (mid-scale ADC range). So I changed R15 to 49k9 and R19 to 5k6.

After this I was able to see a "standard" AC ripple at exactly the double of the exciting wave (due to the full-wave rectifier). Honestly there is also a high-frequency noise on top: small spikes at 69 kHz.

Increasing R18 to 10k reduced the ripple a lot. So I can play with R18 and C23 to minimize the ripple as requested.

Now the last issue is: the output signal doesn't change with displacement. Only the ripple goes away at null. I designed R21 to get a 5V range (from 0V to 5V).

Marco

Err... the use of an LVDT is often with a moving-needle meter to show displacement. Why would a bit of 5 kHz AC be a problem for a d'Arsonval meter movement? It's not necessary to treat the ripple at all, in that case. Your 'filter' is the mass of the meter's rotating parts and the associated meter resistor.

Il 02/08/2011 18:25, whit3rd ha scritto:

Often, not always.

I have no such meter. Instead there is a nice 24-bit sigma-delta ADC and an MCU that acquires the position and sends the data to a PLC. I need to read the position as accurate as possible with the highest resolution allowed by the noise.

Marco

So something inside the AD698's signal processing chain was saturating? It can be hard to find these sorts of drop-offs - the design equivalents typographical errors in text - where someone has transposed a digit or the like.

My favourite disaster of that kind involved a draughtsman putting TIP29 on the production drawings - and the parts list - when the TIP29A (with a higher breakdown voltage) had been intended. The - very good - design engineer who had specified the TIP29A on his draft circuit diagram was horrified when he found out.

So the demodulator is isn't working right. Are the switching edges appearing at the 0V points on the sine wave - zero and 180 degrees? Or at the peak voltages - 90 and 270 degrees? It ought to be a silly question, but it is something that could conceivably have gone wrong.

Il 03/08/2011 01:56, Bill Sloman ha scritto:

Unfortunately, nothing appears at BFilter (primary voltage) nor at output filter. Just a rough and small "square" (coff....) wave at AFilter. I'm going to mount another PCB. Perhaps I broke something.

Marco

I don't know how you designed your circuit how ever, I'll assume you're using a LVDT induction unit? If this unit has some lead of wire between the coil and input of the AD698, did you put a load R across the input to reduce reflections?

I only ask this because I don't know the freq you have it tailored for and it does make a difference.

We made a LVDT unit once but it wasn't using this chip that you're using how ever, due to the long run we had between the actual coil and circuit the reflection was influencing the final output.

Jamie

Il 04/08/2011 01:47, Jamie ha scritto:

Following the design procedure reported in the datasheet of the AD698.

Yes of course. It's a series-opposed LVDT.

No, I didn't. This R should be placed across the input of LVDT ( = primary) or the input of the demodulator ( = secondary) ?

Thanks for sharing your experience. I didn't think about reflections because I assumed if the datasheet of the LVDT suggests 5 kHz its impedance should match.

Marco

input of the demodulator at the final end of the lead wire. In our circuit we used a 180 ohm, yours may differ depending on the ratio of your coil.

Jamie