Instrumenation Amp Question - Page 1
Forum section: electronics / projects

author: cinetron
author_rank: Contributor
timestamp: April 16, 2017, 02:37:56 am
content: I'm working on a project in which I am using a standard difference amp op amp topology to measure the difference between 2 voltages. In the difference amp I am using 4 10k resistors. I am finding the input impedance too low, and I think even using 100K as each resistor won't be enough. I thought of putting a voltage follower on the inputs to raise the impedance and then realized I am actually building an instrumentation amp so thought, why not just buy an instrumentation amp for this application. I am however finding they are quite expensive. I need something that runs single supply at 30 Volts or more. I only need unity gain. Best I have found in terms of price that looks like it would work for me is the INA826https://www.digikey.ca/product-detail/en/texas-instruments/INA826AIDR/296-30238-1-ND/3045446. Just wondering if anyone knows of any lower priced instrumentation amps that work up to 30 volts. I was hoping I could find a dual version (since my circuit needs 2)of a unity gain instrumentation amp in an SOIC-8 package that would fit the bill but seems like such a beast doesn't exist. If it was unity gain the pin count would work for such a package. I may just end up putting a voltage follower in front of the circuit I have now and call it a done deal. The difference amp I am using uses a LM258 opamp so is quite inexpensive. Any thoughts on alternative instrumentation amps?Thanks

author: Rerouter
author_rank: Super Contributor
timestamp: April 16, 2017, 03:24:02 am
content: Does a gain of 2 inconvenience you?If not you can use a circuit like attached,Input impedance is as high as the op amp input (and your input protections allows),if you still need gain of 1, throw a third amp on the output with a gain of 0.5. as this can be done with very cheap op amps.Edit: the batteries and op amp model just come from me reusing an image, but the topology remains the same

author: cinetron
author_rank: Contributor
timestamp: April 16, 2017, 05:14:24 am
content: Thanks Rerouter. That is an interesting configuration. Yes I would need to keep my gain to 1 however I guess if I was adding a 3rd amp it would be the same number of amps as it would take to build an instrumentation amp ie. difference amp preceded by a follower on each input.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: April 16, 2017, 06:24:19 am
content: Quote from: cinetron on April 16, 2017, 02:37:56 amHuh? Half the point of an instrumentation amplifier is the high gain it can apply to differential signals, thus drowning any common-mode signal. Unity gain in-amps aren't useless, but they're rather niche. And as in-amps are typically used to deal with small, sensitive signals, you usually don't want the crosstalk of a dual package anyway, so only a few options exist as duals. You're looking for a doubly rare bird (which might suggest you ought to change your approach).You are going to have to relax at least one of your requirements. My advice would be to just use two 8-pin packages (SO-8 or MS-8). 8-pin in-amps are common enough that you won't have to worry about an abrupt part discontinuation forcing a redesign on you. If you insist, you could drop the 8-pin requirement and use e.g. INA2128, but I probably wouldn't choose that option. (It's also absurdly expensive; you could get a pair of AD8421s for that price!) Constructing your own discrete in-amps is also a reasonable choice, but you'll be using at least four opamps that way (if you use two two-op-amp topology in-amps), so 16 pins. Something's got to give here.

author: fcb
author_rank: Super Contributor
timestamp: April 16, 2017, 09:26:30 am
content: Probably need to know more about what your sensing, and how accurate it needs to be, how much noise you can cope with, what bandwidth you need, etc... Possibly just use a diff amp configuration with a TL072 and 1-10M input resistors would be fine?I have a product that uses a buffered diff amp configuration for the front end (+/-20V), I use a pair of LTC2057HV's as the buffers and then an ordinary opamp as the diff amp.

author: Brutte
author_rank: Frequent Contributor
timestamp: April 16, 2017, 10:14:55 am
content: Quote from: cinetron on April 16, 2017, 02:37:56 amWell you need to know if/how much the output resistance of the source varies with load. If it does not and is known then you can even ignore input resistors (0R).I think you started from back to front. Just pick some CMRR, gain, offset voltage, temperature range and do/show us the math. If your expectations require 0.00001% resistor matching, 3ppm and 17G nominal values then that is most likely not the right route for any differential amplifier.Your LM258 is a bipolar thing. It does not have biasing current compensation (-42nA-48nA typical) and input offset voltage is serious (+-2mV). Do not expect too much from it.

author: danmcb
author_rank: Regular Contributor
timestamp: April 16, 2017, 10:32:22 am
content: what exactly is the application, and your design requirements? what frequency? what kind of CMR? there are many ways to do this, an I-amp is not necessarily the best, as others have suggested.

author: danadak
author_rank: Super Contributor
timestamp: April 16, 2017, 11:08:47 am
content: CMR highly dependent on R accuracy, see attached sim.Reason for using IAs from the likes of Analog devices that are laser trimmed.Regards, Dana.

author: cinetron
author_rank: Contributor
timestamp: April 16, 2017, 07:25:31 pm
content: I should have given more details. It is a low frequency application. I am monitoring the drain voltage of a FET and comparing that to the voltage powering a string of LEDs. Attached is a schematic of the relevant part of the circuit. The LEDs in the schematic are just being representative of the string here. My issue is that when the FET is off there is enough current flowing to the negative input of the op amp to turn on the LEDs slightly, this occurs even if the resistors are 100K . As I mentioned I could easily remedy this by putting a voltage follower in front of the negative input. Just thought I would see if there was an inexpensive instrumentation amp that would have all the resistors, voltage follower etc. in one package.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: April 16, 2017, 08:46:17 pm
content: Four questions for you to consider:Why do you think that bias current into the op-amp is your problem?What's the voltage at the op-amp output when the FET is open?Why a JFET?How do you drive the JFET on and off?

author: danmcb
author_rank: Regular Contributor
timestamp: April 16, 2017, 09:19:52 pm
content: well, maybe I'm being thick, but I still just cannot figure what you are trying to do here.The + input of the op amp sits at +12V, more or less. The op amp tries to keep the negative input the same. This means that current will flow through the LED's and the 10k resistor, so the op amp will simply swing its output towards ground to give that current somewhere to go. A tiny amount (typically nA, I didn't look up this op amp) will actually go into the ive pin, but that is negligible. Most current will simply flow into the op of the op amp, which will cause the LEDs to come on a bit, depending on the number of LED's. Even if the FET is off.When the FET is on, the o/p of the opamp swings high because the FET drain goes pretty much to ground. But - how is the current for the LED's set? is there some feedback to the FET gate? as far as I can see, if you switch the FET hard on, you will probably burn out either the FET of a LED.Are you trying to make some kind of fancy current source? or sense if the FET is on? or control brightness? what exactly?Maybe there is method in this that I am not seeing, but my gut feeling here is that whatever you are trying to do, there is a better way.(edit : as you have drawn it, with the FET off, I would expect to see roughly 10V across the 4 LED's, which means the 10k resistor has about 2V across, pulling about 0.2mA through the LEDs, which will therefore be dimly on. The output of the op amp will sit at about 10V or so.)

author: tszaboo
author_rank: Super Contributor
timestamp: April 16, 2017, 10:41:30 pm
content: Quote from: exmadscientist on April 16, 2017, 06:24:19 amThey are really useful to do precision analog A-B+C. A typical INA has built in resistors that are orders of magnitude better than doing the same thing with opamps + reistors.Quote from: cinetron on April 16, 2017, 02:37:56 amI did a thorough research on INAs some two years ago, and the INA826 is indeed one of the cheapest.But in this case, just place a voltage follower before the diff amp.

author: cinetron
author_rank: Contributor
timestamp: April 16, 2017, 11:59:35 pm
content: I should probably have mentioned why I am monitoring the voltage difference. I am just monitoring to watch out for the drain being shorted to the +24 Volts. So just a safeguard. It has nothing to do with driving the LEDs. As I mentioned the schematic just shows a representation of the LED string which is a 24 Volt LED string. The circuit works really well except for the small current going through the 10K at the -ve input of the op amp when the FET is off so ya a voltage follower should fix that.

author: danmcb
author_rank: Regular Contributor
timestamp: April 17, 2017, 08:38:23 am
content: why not use a comparator?

author: danmcb
author_rank: Regular Contributor
timestamp: April 17, 2017, 09:35:56 am
content: or even a simple PNP transistor that gets turned off if the FET drain is within 0.6V of +24V?

author: cinetron
author_rank: Contributor
timestamp: April 18, 2017, 03:20:50 am
content: Thanks for the suggestions. I guess what I like about the difference amp is it gives me the opportunity to choose a window of voltage difference between the drain and positive supply voltage at which I can trigger a fault. In my design the difference amp is followed by a comparator. But probably overkill. I really like the simplicity of the PNP solution though I still have to be concerned about the impedance to ground through the base. I'll do some experiments. Thanks again for the suggestion.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 18, 2017, 03:07:44 pm
content: A diode between the output and the non-inverting input to clamp the output just below 12v might work. Adjust the following comparator trip voltage accordingly. This only works if your 24v diode string is really made up of many diodes (not as shown in your diagram).

author: tszaboo
author_rank: Super Contributor
timestamp: April 18, 2017, 03:28:36 pm
content: Quote from: cinetron on April 18, 2017, 03:20:50 amI'm not sure why would you use a JFET in the first place. This is what MOSFETs are made for. Place a 100K resistor from the drain, clamp it to VCC with a diode, and connect it to a GPIO. I dont know, why are you making it harder than it is.

author: cinetron
author_rank: Contributor
timestamp: April 18, 2017, 09:19:27 pm
content: @NANDBlog sorry I guess I drew the schematic rather quickly. I am using a MOSFET not a JFET. Unfortunately a simple resistor from drain won't work because there is always a voltage on the drain. The voltage sits around 15 volts when the FET is off. I like @danmcb suggestion to use a PNP . I tried it using the attached schematic. Unfortunately the voltage at the drain floats up to about 23 volts so the output to the micro remains low even if the +24 Volt is not shorted to the drain. I'm thinking maybe if I replace the PNP with a low threshold small signal FET it may work as planned ie. the output to the micro only goes low if the 24 Volt rail is shorted to the drain. I don't have any on hand to test but was thinking of replacing the PNP with this FEThttp://www.aosmd.com/res/data_sheets/AO3403.pdf.

author: cinetron
author_rank: Contributor
timestamp: April 19, 2017, 05:48:12 am
content: So I realized after thinking about the PNP solution a bit more and messing with Spice that all I needed to keep the transistor fully on was to add a pull down resistor of 10M ohms to the base as shown in this schematic. Now to connect everything up to the micro and see how it works in practice. The transistor also turns off if I short the 24 volt rail to the drain.

author: danmcb
author_rank: Regular Contributor
timestamp: April 19, 2017, 06:52:31 am
content: ? you want to detect whether the bottom of the LED chain is sorted to +24 right?so just a resistor of say 10k from base to +24, and another one to the bottom of the chain. The transistor will then be on whenever the LED's are on. You can adjust the threshold by setting the value of the second resistor - but unless the number of LEDs varies hugely it shouldn't be critical.

author: cinetron
author_rank: Contributor
timestamp: April 19, 2017, 08:42:58 pm
content: @danmcb , thanks for the suggestion . This would work well if I was only wanting to detect the short when the FET is on but I also want to detect it when it is off . The problem with using a 10K to +24V and 10K to the drain (end of the LED chain) is that when the FET is off the PNP turns off . I'll think about it. Maybe your suggestion along with a resistor pulling down the base when the FET is off as I did previously could work in this configuration? I'll do some tests in Spice.

author: danmcb
author_rank: Regular Contributor
timestamp: April 20, 2017, 06:23:22 am
content: that doesn't really make sense. The way you have drawn it - the FET is toast the moment the short happens, even if you detect it. So what's the point?

author: cinetron
author_rank: Contributor
timestamp: April 21, 2017, 01:08:02 am
content: Actually I have done a lot of tests and if I can detect the short quickly enough and shut off the FET it survives. I also have a high side overcurrent sensing circuit as a backup monitoring system. So far after many short circuit tests I have yet to blow a single FET.

author: danmcb
author_rank: Regular Contributor
timestamp: April 21, 2017, 06:58:53 am
content: fair enough. But if I knew I had exceeded a FET's max ratings for, say, 100ns, but it still worked, I would want to change it, because there might be internal damage that would show up later. But perhaps your current limit takes care of that.For a design like this, I would prefer to drive the LEDs with a switchable current source - then you really don't have to worry about shorts, number of LEDs, any of that stuff. It's fundamentally better behaved. But I don't know your design parameters, application, whatever.

author: tszaboo
author_rank: Super Contributor
timestamp: April 21, 2017, 08:18:33 am
content: Quote from: cinetron on April 21, 2017, 01:08:02 amUse this circuit element, called "the fuse". Its not like there is going to be a short one after another. Its a failure mode, treat it like that.