LVDT problem.  How to shed a few volts? - Page 1
Forum section: electronics / projects

author: kc1980
author_rank: Regular Contributor
timestamp: May 21, 2010, 09:33:50 pm
content: Preface:This is NOT my design! I'm simply trying to troubleshoot it.Background:I have a tool that employs two LVDTs to measure differential thickness of sheet-type materials. The LVDTs have self-contained internal circuitry for excitation/oscillation and to output a DC voltage that is proportional to its displacement. I have no clue about the internals.Each LVDT has four wires: 6V (nominal) supply, ground, output_high and output_low (the latter two make up the differential output). Each LVDT's output (i.e. difference between output_high and output_low) is highly depending on its load (i.e. open circuit = higher voltage, load resistor = lower voltage). Also, I've found that the LVDT outputs are dependent on the supply voltage (i.e. higher supply = higher voltage between output_high and output_low).There are two LVDTs in the tool. The output_low wires from both LVDTs are tied. The output_high are attached to either end of a 1.3k ohm resistor and the signal is essentially the voltage across the resistor. I do not know the nature of each LVDT's input impedance (it is not constant and not a simple resistance).Problem:Though both LVDTs are sufficiently linear, one of them scales differently than the other. In other words, if I were to displace both LVDTs the same exact distance, the output voltages will be off by a bit. Think: output x some constant. Another way to think of this is that when the individual LVDT outputs are plotted, the line is straight but the slope is off.Proposed Solution:I know that the each individual LVDT's differential output is dependent on its source voltage. I would like to trim the supply voltage on one LVDT. I'd like to avoid messing with the 1.3k load or anything else on that end, as the LVDT outputs are unpredictably dependent on its output load. I need a very simple and elegant way to shed 0.05-0.3V on one LVDT's supply. I was thinking about using a spliced-in, low-voltage-drop diode, but too bad it's not adjustable and the voltage drop might still be too high. The other option is to construct a small circuit, but I'd really like to keep it as bulletproof and idiot-proof as possible. Can you guys think of any other options?THANKS!Ken

author: kc1980
author_rank: Regular Contributor
timestamp: May 21, 2010, 10:06:58 pm
content: Anyone know of any LDO regulators that will get down very very low at a minimum of 100mA? That's another option.

author: Unknown
author_rank: Unknown
timestamp: May 21, 2010, 10:49:24 pm
content: An LDO with a MOSFET pass element, like an MCP1700 or MCP182x, might just barely/almost make 50mV dropout at 100mA. Note that they have a max. input voltage of 6V, and absolute max of 6.5V, so you're quite close to the limit. The max. output voltage is 5.5V, which might be an issue. I think someone else (Analog Devices IIRC) has a device with a slightly higher (7.5V?) absolute max input voltage, not sure about the output voltage.If you have power to burn, a resistive divider might even work. It depends on how constant the load is. The fluctuations in the load should be significantly less than the current through the divider, so if the load actually fluctuated between 0 and 100mA, you'd need 1A (probably not an option). But if the load is almost constant, it might just be good enough with only 10-100mA.A DIY LDO regulator built from discrete components (with MOSFET pass element) would also be an option, but significantly more complex.

author: tecman
author_rank: Frequent Contributor
timestamp: May 24, 2010, 07:48:54 pm
content: The simple answer is a cheap and simple LM1117 LDO. You can make the output adjustable with a simple resistive divider (look at a data sheet,www.national.com). The major problem is that the dropout voltage is about 1.3v~1.5v As a proposed solution, power both LVDTs with each one having its own separate regulator. If you could bump the supply voltage to 7.5-8 volts, then you adjust the first to 6 volts nominal and the second to 6 volts +/- whatever you wish to trim.Paul

author: jimmc
author_rank: Frequent Contributor
timestamp: May 25, 2010, 07:34:02 pm
content: Why not replace 1.3k load resistor with 100R pot in series with 1.2k fixed resistor.Load on LVDT is still fixed at 1.3k so linearity remains the same.High end of pot to output_high, low end of 820R to output_low. Take output between pot wiper and output_low.This allows adjustment of output from present value down to 1200/1300 = 92% of it. i.e. 8% adjustment range.Jim

author: kc1980
author_rank: Regular Contributor
timestamp: May 27, 2010, 04:26:06 am
content: Quote from: jimmc on May 25, 2010, 07:34:02 pmI attached a schematic to give you a better picture. Just pretend that the two LT ICs are the LVDTs. Anytime I make a slight adjustment to the load resistor, the LVDT behaves very differently.The battery's actual voltage is 6.3V (fully charged), which throws the LVDT outputs way way off. I don't understand how this tool was supposed to stay calibrated for a whole year. Again, the LVDT output voltages are dependent on the supply voltage. Therefore, I felt that I should adjust the one LVDT that was scaling/tracking differently than the other. But, that means that I need to shave off something like 0.01 to 0.1 volts.Since the nominal LVDT voltage is 6.0V and the battery voltage will drop over time, I thought that maybe I should just get a switching step-up voltage regulator IC and raise it to say 7Vdc. Then, I could use a linear regulator or op-amp with a pass transistor to set the two voltages independently. Of course, this makes for a complex solution that will require a PCB, which I'll need to mount somehow. Ughh......I was hoping to just splice in a few discrete in-line components into the wiring, but that doesn't seem feasible.

author: ModernRonin
author_rank: Contributor
timestamp: May 27, 2010, 05:07:26 am
content: First, I'd bite the bullet and put a 5V LDO reg in there.Second, I'd put an op-amp (differential amp config, unity gain) on the differential output of each DT. I don't like the way the lows are tied together right now...Once you have the op-amps on, use their "offset null" pin to null out any minor voltage offsets.

author: kc1980
author_rank: Regular Contributor
timestamp: May 27, 2010, 05:42:36 pm
content: Quote from: ModernRonin on May 27, 2010, 05:07:26 amThe LVDT manufacturer's documentation calls for a supply voltage of 6.0Vdc +/- 1%. Currently, it is running off of a battery which happens to be at 6.3Vdc (fully charged) -- I have no idea how low it drops in the field between chargings. Seeing that it's been running fine at 6.3, maybe I can take it down to 5.5Vdc without any adverse effects -- an adjustable LDO should do the trick here. The LVDT output increases with battery voltage and load. I think I'll take your advice on the opamps. That way, the LVDTs will see high impedance and will have minimal influence on each other.Also, I don't think the offset null will work. The problem isn't with offset. One LVDT has a higher output versus linear displacement slope. An adjustment to the opamp gain should take care of that.It's unfortunate that I'll have to mount this all onto a PCB an figure out where to stash the PCB itself. It seems that there's no other way around this problem. Thanks for your input!

author: ModernRonin
author_rank: Contributor
timestamp: May 27, 2010, 09:35:51 pm
content: Quote from: kc1980 on May 27, 2010, 05:42:36 pmWow, no kidding. Never mind the 5V reg advice then...

author: kc1980
author_rank: Regular Contributor
timestamp: May 28, 2010, 12:34:46 am
content: Yeah, it sucks babysitting old design. And this sucker isn't even that old; it was designed in 1983. They could've done WAAY better. But hey, I guess that's what's keeping me busy right now so it can't be all that bad. Thanks for the input though.

author: EEVblog
author_rank: Unknown
timestamp: May 28, 2010, 12:47:16 am
content: Quote from: kc1980 on May 28, 2010, 12:34:46 amI know the 80's seems like yesterday, but 1983 was 27 years ago!Half the forum members probably weren't even born then!Does sound like a shit design though.If you need 6V +/-1% from a 6.3V source and you can't find a suitable LDO for the job, how about a DC-DC converter to up the voltage to overcome dropout and then any linear reg/reference system you chose to get your 6V. Is temp an issue?, if so you might need to roll your own linear reg with a more stable 6V reference than you get in standard regs.Dave.