Repairing a current-to-voltage preamplifier - Page 1
Forum section: electronics / repair

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 25, 2020, 07:10:53 pm
content: I need help with this pre-amplifier. Somehow, the op-amp is broken, and I need to find a replacement for it. The problem is that it's enclosed, and I can't figure out what kind of op-amp it is without breaking it open. It's supposed to be part of a special microscope. I was wondering if anyone here had some experience with this kind of stuff? From what I can tell, these are my three options:Find a replacement module that's enclosed to swap it out for the old one. Does anyone know where I can find a replacement?Break it open and replace the internal op-amp inside.Break it open and recreate the module inside.Options 1 is what I would prefer, but it's hard to find something like this. Options 2 and 3 seem more realistic, but I would have to break it open, and I would prefer not to do that in any case.

author: jdragoset
author_rank: Frequent Contributor
timestamp: June 25, 2020, 11:02:46 pm
content: Please show what surrounding circuit this hybrid works in.If it's not high frequency/precision, the entire circuit can be fabricated with an off-shelf op amp on a test pcb and plugged in.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 26, 2020, 03:51:03 am
content: Thanks for your reply. Let me see if I can't get back to you on that if I can get the surrounding circuit.With regards to your suggestion of fabricating an off-the-shelf op-amp, if we fabricate it ourselves, I'd try and break it open to see what's inside first. Going back to option 1 in my original post, can you buy modules like this as standalone parts? If I can buy it as a standalone part, I can determine whether or not it is cost effective to make my own replacement.Also, are there pre-made off-shelf op-amps assembled on test PCBs like this? Where can I find them?

author: duak
author_rank: Super Contributor
timestamp: June 26, 2020, 05:11:03 pm
content: With monolithic techology the way it is, there should be some off the shelf opamps that could deliver equal or better performance. The problem will be to design and package a replacement. I designed a PDA in the early 2000s and found Graeme's Photodiode Amplifiers book to be useful.https://books.google.ca/books/about/Photodiode_Amplifiers_OP_AMP_Solutions.html?id=sHV0c5hBW4QCI've seen other online material of similar quality here and there. I think Graeme was with Burr-Brown that was later acquired by Texas Instruments.I expect from the feedback resistor value that it's a FET input amplifier. Since it's a hybrid, it might have a matched FET input pair followed by a monolithic op-amp. The other parts are probably separate and attached to a substrate with conductors screened on.To get a feeling for the age of the technology, do you know when the preamp and/or instrument was manufactured or when it was first produced?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 28, 2020, 03:42:41 am
content: Duak, thanks for your reply. The equipment is part of a vacuum amplifier for a scanning tunneling microscope. Sorry, I don't believe I can show any more pictures of the equipment, but I can give some more info about the circuit as best as I can.The input going into the pre-amplifier is a current signal which can range from picoamps to nanoamps. In other words, very small. The pre-amp serves to convert that current to a voltage equivalent signal. The device that is reading the output of the preamplifier is reading it at a rate of 500k samples per second. That's about all I can really say about the device.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 29, 2020, 03:14:51 pm
content: Duak, going back to my original post, if it was a matter of designing/packaging a replacement, I would probably just open up the broken one and examine it to see what components it used. You mentioned about off-the-shelf opamps. In this case, since we don't know what the op-amp is given the schematic, are there 'pre-made' op-amp modules that can function in the same way, as like a 'drop-in' replacement? We keep on talking about off-the-shelf op-amps, but that would require designing a board and getting the necessary components and whatnot. Eventually, we may have to go that route, but I thought that there were modules like this pre-assembled and can just be bought and used.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 29, 2020, 04:05:20 pm
content: There are only very few parts in the circuit. Besides the parts shown, they may be a capacitor at the supplies. The 2 diodes could be one part (e.g. BAV199 or similar).I don't know of any ready made chips with the function, so it would need the diodes, 2 resistors and the OP.For just a single unit one may get away with air wiring or a hand carved PCB.The 100 M resistor may be slightly tricky to find in a small form factor. The parasitic capacitance in parallel to the resistor may be important too, as it effects the bandwidth: to little and the circuit may oscillate, to much and the circuit gets slow. The values depends on the input capacitance and of the OP and the diodes. It may need a few tries to find the right value / OP with suitable speed and capacitance.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 29, 2020, 04:33:25 pm
content: Kleinstein, thank you for your analysis. I kind of figured that it may come to this, building my own, but I wanted to try and at least see if there was a way to find this in the wild somewhere. I found some op-amp evaluation boards, but just looking at them tells me that they are not what I want.Thank you for your comments about the circuit. I never considered those factors before about the diode and parasitic capacitance, so that's worth looking into. I found some 100 Mohm resistors on Digikey, though they are a bit more pricy than other resistors. Regarding what you said about the parasitic capacitance being parallel to the resistance, are you referring to the 100 Mohm resistor? Perhaps an addition of a feedback capacitor in parallel with 100 Mohms will help?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 29, 2020, 05:27:42 pm
content: The critical capacitance is capacitance parallel to the 100 M. Here fractions of a pF can make a difference. So it would not be so much about a discrete part cap, but maybe an isolated wire that may be bend / shorted to right capacity.It is normal that a 100 M resistor is more expensive and the choice is smaller.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 29, 2020, 05:47:45 pm
content: If it comes down to it, I'll have to break open the old broken pre-amp and see what I can work with. The schematics that I posted earlier, it seems to not be set in stone/fixed value, but it seems to 'allow' for a wider range of values. Perhaps the manufacturer intended for this to allow different varieties of op-amps to be used?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 29, 2020, 06:09:13 pm
content: The choice of OP will effect the noise and bandwidth and required capacity parallel to the 100 M. The 10 K resistor at the input should not be critical in value. One could try a simulation including estimated parasitic effects.

author: PartialDischarge
author_rank: Super Contributor
timestamp: June 29, 2020, 07:06:19 pm
content: It's relatively easy to build such a transimpedance amplifier. I would use a OPA145 with no input diodes, two 0805 50Mohm resistors in series and no capacitance added next to them. The parasitic capacitance from the resistors to that tight metallic enclosure is enough and will give more than 30KHz in BW.Total BOM cost is less than $5

author: Kleinstein
author_rank: Super Contributor
timestamp: June 30, 2020, 05:54:38 am
content: The OPA145 may be a good choice, at least it is low noise and moderately fast.The capacitance to the tight case has a different effect than capacitance parallel to the 100 M, so it can not replace it. It depends on the details (e.g. input capacitance, BW of the OP) if one can get away with the parasitic capacitance, or may need a little (e.g. 0.1 - 1 pF range) more for stability.

author: KeepItSimpleStupid
author_rank: Regular Contributor
timestamp: June 30, 2020, 08:13:41 am
content: So, with the spec of dust info, your probably dealing with an STM.I put an EBIC system together. Building the stage was the hard part. I built a faraday cup, so that the beam current could be measured.The 100 M resistor is the real pain.Ib is a critical parameter and Ib varies with temperature which affects Vos.See:https://www.analog.com/en/products/ada4530-1.html#product-qualityI used the OP41 for one design. That one is obsolete.https://www.analog.com/media/en/technical-documentation/obsolete-data-sheets/75599329OP41.pdfThe AD549 is a suggested replacement.Usually there is a capacitor across the 100 M resistor . Put the (-) input on a PTFE post.

author: exe
author_rank: Unknown
timestamp: June 30, 2020, 09:49:22 am
content: Quote from: KeepItSimpleStupid link=topic=245046.msg3112306#msg3112306The 100 M resistor is the real pain.[/quote

author: TimFox
author_rank: Super Contributor
timestamp: June 30, 2020, 02:56:29 pm
content: If the linearity of the current-to-voltage amplifier is critical, you should use physically long resistors to reduce the voltage co-efficient of resistance. In my work, with output voltage up to 10V and linearity requirements of 0.1%, we found that the VCR of 50 megohm SMT resistors was important, and that 1206 parts were measurably better than 0805 parts. We also had to choose vendors carefully, and many do not specify VCR. At the physical level, for a given resistive material, the deviation from Ohm’s law depends on the voltage gradient along the resistance path (over any thermal effects due to self-heating).

author: LoveLaika
author_rank: Frequent Contributor
timestamp: June 30, 2020, 06:26:34 pm
content: Thanks for your reply. If I'm dealing with input currents of picoamps to nanoamps, how does the input bias current affect the results? Ideal op-amps have 0 bias current, but in reality, the op-amp will need some input current. Essentially, the smaller the value, the better, but if the value and the input current coming in are on the same scale (like here, picoamps vs. picoamps), how much difference of a voltage will that make?Looking at the comments and feedback I got from this post (and through Google searching other STM papers), assuming I'm going the route to build my own, the op-amps I'm considering are the following:OPA196:https://www.ti.com/lit/ds/symlink/opa196.pdf?ts=1593537201092&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FOPA196OPA145:https://www.ti.com/lit/ds/symlink/opa145.pdf?ts=1593540646198&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FOPA145AD549:https://www.analog.com/media/en/technical-documentation/data-sheets/AD549.pdfOPA196 and OPA145 have input bias currents of 20 pA and 2 pA respectively, but the AD549 has an input bias current of 60 to 250 femtoamps. Such a difference makes me think that this is more suitable for STM applications (not to mention the price tag).

author: Kleinstein
author_rank: Super Contributor
timestamp: June 30, 2020, 06:56:39 pm
content: The bias current will give some offset to the measured current. Often the Bias current is highly temperature dependent. It depends on the application if this really is a problem. The other problem is that input current noise also gets higher with more bias, though very low bias does not guarantee low noise. The bias current usually has the corresponding shot noise, but there can be compensation in the bias - so the noise level can be higher, but hardly lower.Quite often the Johnson noise of the 100 M resistor will dominate anyway. So besides noise and bias the input capacitance can be parameter to look at: less capacitance gives higher speed / less capacitive FB needed.

author: TimFox
author_rank: Super Contributor
timestamp: July 01, 2020, 02:00:59 am
content: The Johnson noisecurrentof the feedback resistor decreases with increasing resistance, so one should use as high a value as possible with low input currents. The resistor's noise current adds (in quadrature) with the amplifier's input noise current and the amplifier's input noise voltage divided by the total resistance seen at the inverting input.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 02, 2020, 02:50:18 am
content: @KeepItSimpleStupid Thank you for your reply. Yes, I'm indeed working with an STM. Specifically, it's the Scienta Omicron VT STM. I don't know much about it other than the fact that this I-to-V op-amp works with the STM tip in vacuum and converts the current (It) to a voltage where it goes to another pre-amplifier that's outside of the vacuum.Looking at the schematic I posted, TIP is the current from the STM tip (tunneling current), and {It} is the resulting voltage, but what I don't understand is Io. That is supposed to be a reference voltage, but I don't know what the usual values of it are, as it serves to reduce noise. I'm trying to run a basic SPICE simulation against an ideal op-amp to see how it is so I can get an idea of what the behavior will be, but what would the non-inverting terminal be referenced at?

author: Kleinstein
author_rank: Super Contributor
timestamp: July 02, 2020, 08:16:18 am
content: The non inverting terminal would be ground for the simulation. With the STM one may have the tip voltage (and thus a constant voltage) there and have the sample grounded.For the simulation the capacitance at the input (tip + cable) and the OPs input can be important.A larger resistor in feedback can give less noise current, but it would also slow down the amplifier.

author: duak
author_rank: Super Contributor
timestamp: July 02, 2020, 04:45:02 pm
content: This is for an STM? I used Google to look for "STM transresistance amplifier". Here's some links off the first page:https://aip.scitation.org/doi/full/10.1063/5.0011097https://www.researchgate.net/profile/Emiliano_Pallecchi/publication/234889355_Development_of_an_ultralow_current_amplifier_for_scanning_tunneling_microscopy/links/0046351a838873e8f3000000/Development-of-an-ultralow-current-amplifier-for-scanning-tunneling-microscopy.pdf?origin=publication_detailhttps://arxiv.org/pdf/physics/0610280.pdfChoosing the parts shouldn't be that difficult. I haven't worked with pico and femto Amp currents, however I expect the bigger problem will be to design and fabricate an assembly that will work. I would look into design and construction techniques for electrometers. One thing done is to lift the summing node components off the circuit board and support them with the minimum number of PTFE standoffs. Looking at the module in the image above, I see the input is separated from the other pins. Distance is one way to reduce leakage currents.I wonder if the original design was done as a hermetically sealed module not only to control leakage currents, but also to control outgassing under vacuum? I worked on various projects that used multi-watt laser diodes and as we gained experience with building these things found that adhesive outgassing affected long term reliability because it deposited a contaminating layer on the optics. This was at atmospheric pressure and just above the typical room temperature. We went to low outgassing products and increased the curing and outgassing times.Do you know if there are any organic based items in the microscope's vacuum chamber?

author: magic
author_rank: Super Contributor
timestamp: July 02, 2020, 09:01:24 pm
content: Now that sounds like fun30kHz means 33µs period. For such bandwidth, your feedback network's time constant must be no more than 5µs. With a 100MΩ resistor this is at most 0.05pF capacitance. That's 1/20th of a picofarad, less than some resistors have between their ends.You probably don't want to increase that resistor to 1GΩ.Maybe a different question for a changeHow did you test the original amplifier? What really is wrong with it?How will you test the replacement? You may want to verify its frequency response to avoid surprises.

author: KeepItSimpleStupid
author_rank: Regular Contributor
timestamp: July 05, 2020, 03:09:06 am
content: The label looks like Io. The output of the I-V converter is Generally Vbias-Irf. The Vbias is your Io pin. It can be used as a zero adjust.With I-V converters, Vos and Ib are probably the most important parameters. Guess what, Ib likes to vary a lot with temperature. Io would normally go to ground.I think the reason why there isn't a capacitor across the 100M resistor is the speed that's required in your case. I think you have to worry about stray capacitance.Like I said earlier, I put the (-) pin on a Teflon standoff on the latest design I did for in-house use. I primarily cared about AC performance (trapezoid, low frequency generated from a mechanical chopper) and I needed +-100 mA, +-10, +-1 and +-0.1mA full scale at 10V. I added selectable 2 Terminal/4 terminal measurement capability. I added guard. I could do +-50 mA of suppression.If suppression was used I was limited to biasing to +-5V, otherwise +-10V. I added a >10V or <-10V RED/Green led that pulsed for 1s.I was going to make that Io pin and drive it from an D/A converter, but the plug was pulled (you spent enough time). I ended up with a 40pA offset and I had no way of adjusting the zero. The application was a front end to a lock-in amplifier used to do quantum efficiency (wavelength vs efficiency) measurements of solar cells.The idea was to add a resistor that would change the I-V converter to amplify the offset voltage and then use two equations to figure out what voltage to output on the offset terminal.The gotcha was the IEEE-488 D/A converter I used did not output 0V for Zero. I heard that IOTtech's version of the product used a relay for 0V.There was a big gotcha for me. The converter would not work for our calibration cells. I panicked. it turned out I had to isolate the capacitance with an LT1010 and OP-amp.I did use 400 M-ohm resistors to ground for input loads for the differential amplifier across the device. for low currents, you don't measure the voltage.Since it was for low-current measurements, I used mercury wetted relays to switch the ranges.keithley's low level measurement handbook is useful:https://download.tek.com/document/LowLevelHandbook_7Ed.pdfis useful.

author: magic
author_rank: Super Contributor
timestamp: July 05, 2020, 06:04:28 am
content: Quote from: KeepItSimpleStupid on July 05, 2020, 03:09:06 amFor a one-off it would be sufficient to lift the IN- pin of the opamp and solder the feedback resistor vertically to contact the IN- node in the air. Similar trick was used in the picoammeter thread and it gives sub-pA level leakage essentially for free.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 15, 2020, 04:35:45 pm
content: Quote from: duak on July 02, 2020, 04:45:02 pmThank you for your reply. Sorry I couldn't get back to you sooner. What do you mean by organic-based items? For what it's worth, I don't believe so, but to be fair, I'm not running the STM experiments. Ideally, in the long run, since this thing isn't manufactured by Omicron anymore, we're trying to build a replacement based on the information above.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 15, 2020, 04:43:22 pm
content: Quote from: magic on July 02, 2020, 09:01:24 pmThanks for your reply. Sorry for not responding sooner. Been busy with other tasks.Currently in the lab, we have two IVC preamps. One of them broke, and we're using the second one for the time being. To be fair, I'm not doing the STM measurements myself. According to the professor who is doing it, the one that broke was broken because one of the pins broke off. Thus, he used some epoxy in an atteInmpt to use a piece of metal to connect back to the broken tip. While we're not necessarily trying to fix this per se, we are trying to replicate the functionality of this preamp so we don't have to buy it again (Omicron said they don't make this anymore, so they're out).I honestly haven't tested the preamp myself. I'm just going by what I was given.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 15, 2020, 04:50:11 pm
content: Quote from: TimFox on July 01, 2020, 02:00:59 amThanks for your reply. Sorry for not replying sooner.Looking at the specs, what bothers me is how the op-amp supposedly gets a gain of 10e8. In this inverting gain config, wouldn't you get a gain of 10k given the resistances? How would you get that 10e8 gain with those resistor values?Aside from that, the offset voltage is stated to be 500 uV, so assuming I understand what that means, when pins 1 and 2 are unconnected, there's a 500 uV voltage measured between the two if the op-amp is powered? How do you pick the diodes shown in the schematic?

author: duak
author_rank: Super Contributor
timestamp: July 15, 2020, 05:22:31 pm
content: I mentioned organic compounds above. The original preamp looks like it's built into a metal/ceramic package that might even be hermetically sealed. If I understand how many STMs are put together, the preamp is located within the vacuum chamber and the components are exposed to high vacuum. Many of the components on the replacement are made of organic materials such as epoxy that may absorb contaminants under atmospheric pressure and outgas under vacuum. I'm wondering if this might be a problem?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 15, 2020, 09:37:52 pm
content: I don't think the preamp was within a chamber or the like. When I saw it, it wasn't in a chamber. It was laying on the top, easily accessible with connectors attached to the pins. If it helps, here's a link that might give you a better idea.https://groups.google.com/forum/#!searchin/surfacescience/IVC|sort:date/surfacescience/PpgaSCq90lw/JZdzigi4DgAJIn a nutshell, the current coming from the tip goes into that tiny little preamp which turns it into a voltage. That voltage goes into another preamplifier and another series of amplifiers, which then goes to the computer to be read.If it helps, my system is a VT STM by Omicron.

author: magic
author_rank: Super Contributor
timestamp: July 16, 2020, 05:41:09 am
content: You suggested that this preamp sits in the vacuum chamber by stating that it sends its output to another amplifier outside the chamber.Quote from: LoveLaika on July 15, 2020, 04:50:11 pmYes, closed loop voltage gain is 10⁴, at least at frequencies where the opamp is fast enough to support it. But that's not the spec they give - it doesn't have a gain of 10⁸ but current-to-voltage conversion ratio of 10⁸. And why wouldn't it? Calculate what current must be fed into the TIP input for the output to deflect by 100mVQuote from: LoveLaika on July 15, 2020, 04:50:11 pmThe IO input is high impedance so it needs to be externally driven or else it could drift randomly or slam into one supply rail. Once that is sorted, yes, the TIP will be maintained within ±500µV by cancelling the TIP input current with current through the feedback resistor. That's how those converters work.

author: TimFox
author_rank: Super Contributor
timestamp: July 16, 2020, 01:20:39 pm
content: The gain of a transimpedance amplifier (current-to-voltage) has the dimensions of ohms: this gain would be 108

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 07:33:21 pm
content: Thanks. Sorry, I was thinking of the mindset of V-to-V and not I-to-V. The math makes sense now.If you don't mind me asking, the diodes seem to limit the voltage swing between the two op-amp inputs when there is a current present at TIP, so that way, the two voltages don't have a really big difference between the two inputs. Is this correct?I did some research and while it doesn't say what diodes are being used, I looked at corresponding circuits involving the same STM, and the preamplifier after that uses a comparator op-amp that has diodes like what is shown in the images in my first post. It uses the BAV45, and judging by the diode description, it may be suitable here given its two forms:https://my.centralsemi.com/datasheets/BAV45.PDFhttps://my.centralsemi.com/datasheets/CPD65-BAV45_WPD.PDF

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 07:35:59 pm
content: Thanks. Yeah, I made the mistake of thinking of V-to-V when I should have thought of it as I-to-V. The math makes sense now. Sorry about that.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 07:54:15 pm
content: Quote from: Kleinstein on June 29, 2020, 04:05:20 pmI was doing some research regarding this part and the part that it goes to. It's part of an STM. The tip current goes through this IVC and gets converted to a voltage. That voltage goes back into another preamplifier (let's call that one big preamplifier) for more amplification, which then goes back to the computer to be read.Perhaps it may be meaningless, but I was kind of looking at the circuit ahead, and there's an optional op-amp circuit that was being used. Now, in my 'big preamplifier', option E is not present. Rather, it is bypassed. However, assuming that it is there in some options, I can't help but think that some parts used there that may be used in the IVC, specifically the diodes. The BAV45 is a switching diode, but it's used for picoamps. In this configuration, it seems to limit the voltage difference between the two op-amp inputs, so perhaps it might be used here in the IVC.

author: SMdude
author_rank: Frequent Contributor
timestamp: July 17, 2020, 11:56:13 am
content: Diodes on the inputs like that clamp the input voltage which helps not saturate the opamp input/output.If the signals of interest are only tiny then diodes like that work well and block large unwanted signals, allowing the opamp output to settle quickly so the measurement can be taken.It would be interesting to see scope shots of the receiver output from this device.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 17, 2020, 02:13:45 pm
content: Thanks for your reply. Sorry, I don't have scope shots of the device. Only simulation shots. I ran a hypothetical simulation using the OPA602 compared with an ideal op-amp with and without diodes. You can see the results here. Sorry if it's hard to see.Assuming that my current-to-voltage op-amp uses the same diodes as in that option E circuit, I sent a pulse of 50 nA to my three circuits to see how they behave. It's just like you said, the diodes effectively limit the input voltage. Otherwise, via my ideal op-amp at INV1, I get a large difference of 5 volts rather than 10 mV with the diodes. I think that this is what I want if I want to keep the potentials between the two inputs close to each other as in the case of my IVC.

author: TimFox
author_rank: Super Contributor
timestamp: July 17, 2020, 03:07:00 pm
content: Is this a safety circuit to prevent the probe from coming too close to the sample?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 17, 2020, 03:52:51 pm
content: The circuit as a whole? I don't think so, based on the pictures of the manual I posted in the original post. The diodes shown in the original image might be there to make sure that voltage between TIP and I0 don't deviate from one another by a large scale (only by a few millivolts), but I don't know for sure. I0 is supposedly the reference voltage at which TIP is being held against. I believe it is the same with the op-amp. So, from this op-amp's point of view, I0 is effectively 'GROUND'.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 11, 2020, 02:52:28 pm
content: @SMdude Sorry for not replying sooner. Been busy with other projects that took over time. Regarding what you said, I see that diodes just clamp the input voltage, but it seems to have an effect on the output signal. Take a look at my SPICE simulation below using the ideal op-amp model. You can see how the different diodes affect the output signal. In each op-amp circuit, I'm using the same current signal, and the op-amp without diodes is used as a control to see how the diodes affect the output.Doing research, I found that Central Semiconductor's diode, BAV45, is obsolete. Using other common diodes, you can see at the bottom how the output is greatly distorted (at least when compared to the control). Op-amp aside, what diodes would be good replacements for the BAV45? That seems to be the major issue.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 11, 2020, 04:49:59 pm
content: Quote from: KeepItSimpleStupid on June 30, 2020, 08:13:41 amI'm sorry for replying so late again. Unfortunately, other projects got in the way, and now I'm back to researching this problem again. With your suggestion, how would you go about simulating whether or not the component will work? I tried running a simulation, comparing the AD549 against an ideal op-amp model in SPICE. They all use the same current signal. You can see how the ideal op-amp gives a close to perfect square wave when I convert the current square wave signal to voltage. When compared to the AD549, the AD549 performs rather poorly. I know that it won't match up to an ideal op-amp, but is this the best way to determine whether or not the op-amp will work, or am I testing this in the wrong way?

author: magic
author_rank: Super Contributor
timestamp: September 11, 2020, 06:15:38 pm
content: It's probably capacitance which makes the difference between diodes. Try changing CJO to confirm.Has it been established that you need those diodes? IMO there are only two issues here:1. Overvoltage from external circuitry damages the opamp. 10kΩ plus internal protection of most opamps should prevent this to a reasonable degree.2. Weird output from the opamp damages circuitry outside the TIM module. Is this a concern of any sort?

author: aqibi2000
author_rank: Regular Contributor
timestamp: September 11, 2020, 06:32:48 pm
content: How have you verified it is faulty?Inside is most likely a standard SOIC

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 11, 2020, 08:13:53 pm
content: I haven't physically confirmed whether the diodes are necessary. Like you said, they seem to be for overvoltage protection. As for the second concern, I don't think it necessarily is a problem. The output of the IVC goes into another pre-amplifier which converts it to another voltage (with respect to another reference voltage than the one in the IVC). You can see it below. Output goes to LG6 and the reference voltage of the IVC goes to LG7. In my case, option E is not present when the IVC is present. Conversely, when option E is present, the IVC is not.From what others have said, if the diodes are used for just circuit protection, then a 1N4148 will work in lieu of the BAV45, despite the horrid simulation results below. It could, like you said, be an issue with the SPICE model. Honestly, it's been so long, I don't exactly remember where I got the BAV45 SPICE model. You can see how I'm simulating it below. Maybe it's just a matter of me simulating it incorrectly. Sending a square wave current source, you can see how it's being converted into a voltage. Protection diodes slow it down a bit, but the conversion is still clear with some diodes while others distort it.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 11, 2020, 08:17:40 pm
content: I haven't verified that the IVC is faulty. It's just that according to one of my professors, when removing the connector, the leg snapped off of the case, and epoxy was used to 'glue' the leg back on. Due to being unable to test it, and due to the part being really expensive, they want a replacement. Thus, I'm doing my best to try and build an identical circuit with the information provided. I suppose you may be right, being a standard SOIC, but there's no way without cracking it open, and that's not a risk that they are willing to take.

author: magic
author_rank: Super Contributor
timestamp: September 11, 2020, 09:11:54 pm
content: What I say is that a diode with too much capacitance will slow it down.Regarding protection, I asked if the diodes protect the input side from the TIA. Is there any concern about the signal source being damaged if the TIA applies more than 0.7V to it, despite the 10kΩ series protection? If not, maybe just get rid of the diodes.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 12, 2020, 01:19:14 am
content: The input current is coming from the tip of an STM. As that current goes through the IVC going through that 10k ohm resistor, Ohm's law, I0 has a voltage. Correct me of I'm wrong, but that's the voltage you're talking about at the input signal source? With regards to that, the max input voltage that we can put at the tip is +-10 volts, so i think we're okay.Going through some other comments and asking around elsewhere, I kind of realized how dumb I was. Below is a simulation of mine, using the ideal op-amp model and the 1N4184 model (comparing it to the BAV45). Someone mentioned a feedback capacitor, and that seems to magically solve that ringing issue I posted in my previous circuits. Here, I thought it was a bad model, but it was probably the parameters itself. Previously, I didn't add in that feedback capacitor because the original schematic didn't show it, but I guess I need it now if we're switching diodes.Unfortunately, that brings up the issue of how to get a feedback capacitance that small. Lowest I can get is 100 fF. @magic, I really appreciate all the help you've given me, and I apologize for not realizing this sooner, but in your experience, how would one achieve such a low capacitance? (or maybe it might be easier to use a different diode; then again, this is an ideal op-amp, so perhaps things will change when using a real op-amp model)

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 12, 2020, 03:33:33 am
content: @KeepItSimpleStupid Thank you very much for your earlier replies. Sorry, I've been revisiting this issue, and taking into account your advice by adding a feedback cap, it's been very helpful. You can see in my posts with ideal opamps how a feedback capacitor helped when I used different diodes. Now, I'm kind of experimenting with that idea, trying various diodes and op-amps in order to see what would be useful. You can see it in my attached simulation below. Compared to an ideal op-amp model, you can see how the addition of diodes and the feedback capacitor affect the output.Sorry for rambling. I was just excited at the prospect of getting some progress, I wanted to show this off. Thank you very much for your help. If you don't mind me asking, looking at these results, what do you think so far?

author: magic
author_rank: Super Contributor
timestamp: September 12, 2020, 08:21:04 am
content: I have no experience besides building the TIA picoammeter with LMC662 and, more recently, designing a audio bandwidth TIA-like circuit which I haven't even built yetYou can't have more than 50fF feedback capacitance IIRC. I posted about it long ago, it's no rocket science but basic math. From the limited experience I have, I can tell you that you will probably need to connect a bunch of smaller resistors in series just to get their parasitic capacitance under this target. You don't need to shop for 50fF capacitors. Hell, even the opamp will have capacitance between its pins, better use a single rather than a dual for that reason alone.Instead of throwing feedback capacitance at the oscillations, remove as much of input capacitance as possible to prevent them from occurring. You must find some low capacitance, low leakage diode. Sorry, can't help with that, but 1N4148 will likely fail on both counts, forget it. You can connect a few diodes in series if neither the tip nor the opamp needs protection from voltages lower than a few volts, this will reduce capacitance and leakage.Maybe omit the diodes if you feel adventurous, but consider if there is a possibility of 15V or whatever appearing at the output node under some fault conditions and finding its way through the feedback resistor to the tip. And whether it's a problem, conidering all the series resistance along the way.Out of opamps available in LTspice, try LT1169 or the LT179x. Low input capacitance IIRC and low input noise current. You could also run noise analysis, LT includes noise models of their chips. Run AC analysis, run transient analysis with sinewaves.I have no idea how much noise and GBW you need. GBW helps the opamp hold IN- at zero volts and reduce effects of parasitic capacitance on IN-. It also reduces input impedance seen by the tip. Well, there are some very fast JFET opamps like OPA627. It doesn't help that no one knows what was in the original module.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 12, 2020, 02:44:38 pm
content: Thanks for your reply @magic. With the tests I ran at 50 fF, I was working with the ideal op-amp model. Things certainly changed around when I started to use non-ideal op-amp models. You can see that in my other reply to @KeepItSimpleStupid. Using the OPA602 as a start, the feedback capacitance increased to something more realistic at 0.1 pF. Thinking about what you said, I never did account parasitic capacitances that come with resistors. Taking that into account, I may not even need that feedback capacitance depending on the op-amp and diode I choose?I was working with the 1N4148 to use as a 'control' of sorts to see how to work with it and use that as a jumping off point. That diode is very common, so I figured that working with that will give me an idea of how to move on when working with other diodes. The issue with the BAV45 that I believe was used previously is that it is now obsolete. It had leakage current in the pico-amp range (and capacitance of ~1.3 pF), and I can't find that or any alternative part with that low leakage current in any available/in-stock part on Digikey. The idea was to try and work with easily available parts so I don't have to spend a fortune trying to find some part like the BAV45.Believe me, I wish I could see what was in the original module, but opening that up and seeing what's inside is out of my hands.

author: magic
author_rank: Super Contributor
timestamp: September 13, 2020, 04:49:01 pm
content: If they allow you to power up the original module on a bench you could run some tests on it.Check output voltage with shorted input to know the offset voltage, necessary for step two.Check output voltage with open circuit input (and maybe some shield around it) to know the internal leakage current.As above, but also measure noise.Feed sinewaves through a 100MΩ resistor (say, 10x10MΩ chain for low cost and low capacitance) to see the input impedance vs frequency.Any DIY build could also be tested that way. In real life, not in simulator.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 15, 2020, 10:00:41 pm
content: Thanks Magic. I'll try and see if we can run tests. It's always worth a try if I can.I apologize for pivoting a bit, but I wanted to ask about one of the parameters. The manual stated that the -3 dB bandwidth specification at the fixed gain needs to be greater than or equal to 30 kHz. Now, I never took a course in op-amps, so the frequency stuff is a bit fuzzy to me, but I was wondering if my logic here is correct.This is an IVC, but if you look at it a different way, you could also see it as an inverting voltage gain op-amp. Looking at it like that, the ideal closed loop gain is 100 M / 100 k, or A = -10k. Converting that to decibels, -20 * log(10000) is -80. Using the parameters of the AD795 (one of the op-amps I'm considering using), the open loop gain is 108 dB, and it's GBW is 1.6 MHz. Following the chart on the AD795's datasheet, looking at figure 21, it's the open-loop gain, but the closed loop resistor feedback configuration reduces it down to 28 dB (108-80 = 28). Following along at 28 dB, it interesects the open loop gain line at around 50 kHz. That would mean that if the AD795 was used, the -3 dB bandwidth would be 50 kHz. Is this correct? After running some simulations in LTspice, I was designing my board and reviewing the specifications, and that got me to thinking about the bandwidth.https://www.analog.com/media/en/technical-documentation/data-sheets/AD795.pdf

author: magic
author_rank: Super Contributor
timestamp: September 16, 2020, 10:54:12 am
content: No, to find the (approximate) -3dB point of closed loop voltage gain you find the point where open loop voltage gain equals closed loop gain, so 80dB. That's likely gonna be around 160Hz.But what's the implication of that? It means that from that point, more voltage swing will be required at the TIP pin for the output to perform the same swing. If the TIP pin is being driven by a current source, the current source will keep cramming the same current into the pin and will increase TIP pin voltage as needed, until the output deflects enough for all the current to flow through the feedback resistor.Therefore open loop gain determines input impedance of the TIA as seen by the current source, not its bandwidth. Bandwidth depends on feedback capacitance and the ability of the current source to force current into the rising input impedance of the TIA.editOkay, it's not that simple. If OLG drops down to something like 10x, 100mV will appear on IN- for 1V on OUT and therefore only about 90% of the voltage across Rf will be visible on the output. So the conclusion is that we start to seriously lose I/V conversion gain as we approach the GBW of the opamp.It would help you to simulate this circuit with ideal opamp, resistors etc. and inspect all the waveforms and see how they change with varying parameters.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 16, 2020, 10:30:43 pm
content: Thanks for the reply Magic. I managed to ask around, and I've managed to get data of a sample waveform used as a simulation. Approximating it to an ideal sinusoidal model, I'm also comparing it against a sinusoidal current source that appears to more or less match up with the sample data (-300 pA offset, amplitude of 30 pA, frequency of 10 Hz). It's just a little easier to look at for this simulation.Working with that, now I see where I was messing up in my simulation. Comparing it with the ideal model, op-amp models that have around 5 pA or less (input bias current) seem to do very well. It's more important now to find an op-amp that works well in tiny currents rather than larger currents.

author: aqibi2000
author_rank: Regular Contributor
timestamp: September 17, 2020, 06:21:55 am
content: If only a leg had snapped off and you haven’t confirmed it is faulty,Just use a gold socket holder to give it some durability and strength and then you can fix the broken leg back on without worrying it will off break

author: magic
author_rank: Super Contributor
timestamp: September 17, 2020, 04:34:53 pm
content: By the way, I tested parasitic capacitance between adjacent pins of DIP8 package, using fake chips from China which had some of their corner pins not connected to anything internally. Three different chips gave the same result.From a corner pin to its neighbor it's about 0.5pF.Might be slightly less between the central pins because they are shorter. No idea about SOIC pacakges - smaller pins, but maybe closer together.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 18, 2020, 01:43:19 am
content: Thanks Magic. I was running my original circuit with my recently acquired data, and it turns out, like you said, a feedback capacitor wasn't really necessary. Based on the sample data it wasn't running nearly as fast as I thought it would be, so a feedback capacitor isn't necessarily needed. I continued to test my design amongst various amplifiers and diodes, and I've more or less narrowed it down to the following:OP-AMP: AD795, LT1792, LT1793, AD549, and OPA140.DIODES: CMOD6001, CMDD6001, BAV116W, BAV199, 1N3595Running these op-amps with my sample data with each of the various diodes and comparing it to the ideal op-amp model, the difference in output voltages range in the micro-volts. Given the parts limitations and restrictions, I think that's about as best as I can make it.I was searching around when I found Dan Berard's homemade STM, and it got me thinking about how I need to readjust my board when dealing with pico-amps: using guard loops, voltage bias planes, teflon stand offs, etc. Figuring how to assemble the board appears to be a challenge in itself, moreso than finding components. That low-measurement handbook by Keithley suggested by KeepItSimpleStupid is proving to be useful.https://dberard.com/home-built-stm/electronics/

author: magic
author_rank: Super Contributor
timestamp: September 18, 2020, 05:15:46 am
content: Check also the amplitude of tip input voltage - lower is obviously better. You have no idea what's the output impedance of the tip, huh?Make a board for standard SO8 footprint, find a good resistor and play with some chips. Not sure what would be the advantage of 1792 over 1793, by the way.Regarding input leakage, scroll back in the thread

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 18, 2020, 06:53:21 pm
content: Thanks Magic. That leg lift trick seems great. I think I'm trying to do that, but it seems to work if the input pin is connected directly to the input and not connected to other components, like diodes and resistors.If we had surface mount components soldered onto the board and used small wire to connect to the input in the air, would the SMD components still introduce leakage?Apologies for asking so many questions, but if you don't mind a critique, what do you think of my board as a rough draft so far? Below is the top copper layer of my 4-layer board. All of the components (aside from the input/output posts) are surface mount, and there are no copper planes for the inner and bottom layers yet. I figured I could replicate the top layer onto all other layers to ensure that the same no-fill zones are in the same spots. I figured I would take a page from the ADA4530's book and use via fences and no-fill zones at the input signal. Focusing on the AD795 exclusively, the datasheet mentioned how I can connect the no-connect pins to the common node and form a guard loop that way. The issue is that the vias get in the way. How would you recommend I incorporate the guard loop and the via fence in my design?https://www.analog.com/media/en/technical-documentation/user-guides/ADA4530-1R-EBZ_UG-865.pdf

author: magic
author_rank: Super Contributor
timestamp: September 18, 2020, 07:21:03 pm
content: You know, I'm a lazy ass and crude hobbyist so I would lift the pin and wire in the air. Components can be "tombstoned" under the flying wire.Can't tell you how much effort is required to keep 1pA or so leakage on standard FR4 board and how much is overkill, but a few harmless tweaks come to mind:Did they tell you to put the vias that close? I would put them farther, under the ground pour. The more resistance between IN- and any kind of ground the better, I think. The point of guarding is to bias the other end of the "resistor" made of PCB material to something approximately equal the signal voltage, but not to reduce PCB resistance.Is 4 layers actually of much benefit for leakage over 2 layers? (No idea).Run a guard under the IC, route the output on the right side. One VCC capacitor will need to move.Make space for a few resistors in series, unless you are sure you will find a resistor with suitably low capacitance. You can always solder-bridge unnecessary footprints, it's not a mass production item, meh.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 20, 2020, 03:19:37 am
content: Well, that does certainly feel like the best option at the moment.I was trying to replicate what AD did in their evaluation board, or at the very least try to incorporate some of their techniques into my design. I was just placing the vias along the guard ring to act as a via fence along with the guard ring itself. Once I get the initial guard ring design down, I'll place more vias around whatever free space I can around the board to connect the ground pours.While I understand the idea of adding pads for more resistors to reduce the capacitance, there would be exposed copper pads if left unpopulated. Wouldn't that cause more of an issue by acting as an opening for noise to jump in?

author: magic
author_rank: Super Contributor
timestamp: September 20, 2020, 07:20:20 pm
content: Populate the footprints starting from the input side. Leave any unused and shorted footprints on the output side, where driving impedance is low.Also, a resistor footprint with a solder blob on it is not even that much larger than a straight PCB trace it replaces.