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8-channel LVDT measurement using AD598

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abicash

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Hi
I am planning to develop a system which will measure 8 channels of 5-wire LVDT signals.
Initially I was planning to use 8 no's AD598 and sync them to eliminate the beat signal but the cost and availability of the IC led me thinking in an other direction. On the AD forum somebody suggested about using ADA2200, but again this was with 8 no's of ADA2200.
I plan to multiplex these 8-channels to a single AD598 and read it on a Microcontroller 12-bit analog channel.
I am looking for guidance here with some reference material or some schematic concept with few recommended parts. If anyone can comment, it will be great.
I also am unable to understand how will the zero offset for each LVDT be handled.
I will poll each LVDT and I want to finish the process in 3 to 4 seconds. It also involves stepper motors to drive the LVDTs on some fixtures.
Please make me understand how I can acheive this.
Please find a scheme attached here, which i was thinking to use (using multiplexers) where I will use a single AD598 , and switch the primary and secondary of the LVDT with the help of Multiplexers.
Will this work?
If not , can some experts guide me in the right direction?

Thanks and regards
 

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Probably your are talking about a linear variable differential transformer.
The price is bad! Have you looked at the AD698? Think about the PGA970. I see some versions are $11.00 usd.
----edited----
What LVFT are you using?
 
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Hi Ron
Thanks for the reply.
Yes you are correct about the LVDT.
AD698 is used for a 4-wire LVDTs and PGA970 is still buggy.
AD598 works for a 5-wire LVDT and is preferred over the AD698.
Please check the schematic concept I posted earlier to comment whether I am in the right direction.
 
I would think about connecting all "sigref" together.
Add a second 74HC4052. Connect A to A, B to B, INH to INH.
119522

What it the voltage range for the signals? Will they be with in the range of the HC4053? Will the signal ever go below ground or above +5V?
 
Hi

I would not be worried about negative going signals because if the concept works, I will use a dual ended mux from AD.
Checked a few which are available.
Signal will be 3V.

Why do you think that the 4053 won't work?
Why do you want to combine the ref Sig directly to AD598?
 
Why do you think that the 4053 won't work?
I see people that want to mux analog signal and don't know the range for the HC4053. There is also a CD4053 that will work up to 15 volts. I just want to know if you understand there is a voltage range and there is some resistance added.
Why do you want to combine the ref Sig directly to AD598?
Because; then you only have to mux two wires not three. Less parts. Only use one type of part(s).
 
Thanks.
I do understand about the voltage range and the series resistance. I will try to check a device with lesser series resistance.

I am worried about the scheme. It seems correct on paper. Although I don't see anybody using such a scheme.
Most suggest to use n ADx98 for n LVDT sensors.
Then there's the beat frequency problem, so they suggest exciting all the n LVDTs from the same signal.

So do you think this will work?
Can you think of why it will not work, if of course you think it won't work?
 
74HC4053 from TI.
When the switch is closed it has a resistance of about 70 ohms when powered from 5 volts. Vee=0V
119523

Resistance (on) verses signal at different voltages. Note; Vee can be -5v to allow switching from -5 to +5V.
I think your signals will be within the supply range. (did not check) I think, on the receiving side extra resistance will not effect anything.
 
Ok I got the idea. Thanks.
Although what I have shown is for only 4 LVDTs , and I will be using 8 LVDTs.
So the addressing will change.

I have only one chance at this.
I will be making a PCB with this scheme and add a microcontroller and other stuff.
I have asked this on the AD forum but I believe no one wants to answer there.
 
I will be using 8 LVDTs
74HC4051 8-1 switch.
----edited----
I would use two 4051s on each primary side.
Use two 4051s on the secondary side. (connect all the common/center taps together)
On the addressing; connect all the A(s) together, all the B(s) together, and all the C(s). Inhibit=on. Very simple interface.
 
I do understand about the voltage range
1)You need to look more at the voltages! The AD598 can output 2 to 24 volts rms. There is a resistor to set how hard the outputs drive. The HC4051 can not do 24V!
2)I think the output can drive above and below ground. With a +/-15V supply the outputs can pull up and down a long way. I think you need to have the HC405x set for +5V ground and -5V. You need to work with a negative signal.
3)You need to find out what the input side looks like. I think it also is above and below ground.
This is why I keep saying watch out for the voltage range.

What is wrong with the 4051? Which LVDT? Part number--it is important?
 
Because the AD598 can drive very hard: series up the transformers and kill the mix.
The IC can drive +/-11 volts. (say 10V) 4 transformers in series = 2.5V each. (4 series) in parallel with (4 in series)
119525

On the IC pin 17 is often at ground. So the signal will swing + and -. So the mux must have a +5V and a -5V supply. I do not know how big the signal will be but I think less than +/-5V.
 
LVDT probe with +/-1mm Model No. 2924
From Pretec Swiss made.
4051 is an 8x1 mux , whereas I need something like 4052 or 4053 as shown by me.

I will definitely check the voltage compatability. I am yet to design it in detail.
I just wanted to check at top level , if it will work.
I have seen very expensive LVDT signal conditioning boards.
Why couldn't they use something simple like what I proposed?
 
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Isn't it the same thing as using 2 nos of 4052?
I am still trying to understand if connecting 4 series LVDT primaries will work. Never seen anyone doing such a thing.
 
I am still trying to understand if connecting 4 series LVDT primaries will work. Never seen anyone doing such a thing.
If the IC can drive large amounts of current you could connect all of the LVDTs in parallel.
I could not find out how much current because the data sheet is missing information.
 
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