Bogus parts from Reputable Dealer? - Page 1
Forum section: electronics / beginners

author: Jwillis
author_rank: Super Contributor
timestamp: January 19, 2020, 11:28:25 am
content: I don't want to mention the North American dealer just yet out of respect until I figure what might have gone wrong.I bought 10 pieces of TLV2371IP as suggested for a parallel mosfet linear electronic load. My intention was to parallel 2 mosets with 2 TLV2371IP to control.I was simply wanted to test the circuit design and expand to 4 mosfets when things go correctly with 2 . Seems simple enough right?Now on the first try 1 of the op amps worked correctly, but I couldn't get the second one to work . It seems around 250mV is leaking back from pin 3 and the out puts is the same as the supply 9V on the one that wasn't working right .No voltage is applied to the number 3 pin . I changed 6 of them with similar results with slightly different voltage leakage into pin 3 . The one that worked continues to work. So figuring it may be a bread board issue. I re orientated the bad circuit but got the same results .So I changed out the bad one for another one . Same results .6 out of 10 got changed . Same results. So I switched the two op amps in the two circuits .Well the good op amp works as it should in the other circuit now and bad op amp doesn't work where the good one used to be.So its not the circuit . So I changed the "bad " again (# 7) and low and behold it worked . So I pulled it out to test the last one left out of the 10 and same results as before it won't work. 8 out of the 10 are uncontrollable .Ok so I have 2 out of 10 that work. Now I admit I'm not an expert on op amps but I was being really careful about static and making sure everything was hooked up right before power up. Both circuits are exactly the same Resistors are all 1% . Capacitors are new and measure fairly close. Checked the mosfets and both are functioning .Both circuits receive a parallel voltage at pin 3 . I tried to get the ones that don't work by adding an offset thinking that may be the problem but the offset with a 200k trim pot don't seem to do anything.i tried a voltage divider at pin 3 but that does'nt make any difference either. The voltage is at pin 3 is coming from the chip.Now before I go and make a complete fool out of myself to the dealer I would like to test each failed op amp for leakage and functionality.I don't have really spectacular gear. So what would be the simplest way of doing that so I have evidence to show to the dealer in case they decide to get nasty about it.Now, from I understand when I supply the op amp with 9V there should be no voltage at the inverting and non inverting pins and no voltage at the out put relative to ground correct ? When used in common mode I should only get a voltage at pin 6 when a voltage is applied to pin 3 ,correct?I haven't had this sort of thing happen before with other op amps I've used and I'm out of ideas.

author: nigelwright7557
author_rank: Frequent Contributor
timestamp: January 19, 2020, 01:53:15 pm
content: I would have thought a reputable dealer wouldn't buy rubbish.I have bought IC's off ebay in the past and had bad experiences like you in that some worked and others didnt.On a previous project I bought in some op amp's for RS (reputable dealer.)The circuit behaved very strangely with impossible voltages on op amp pins.Turned out my fast op amp was also bi polar and had quite a bit of input current which messed up my potential divider voltage.Another problem I found with op amps is the bandwidth, which in reality is only small signal bandwidth.If you have a large signal going in the bandwidth is much less.

author: E-Design
author_rank: Regular Contributor
timestamp: January 19, 2020, 02:09:23 pm
content: not at all likely "bogus" parts. Post your circuit.

author: ebastler
author_rank: Super Contributor
timestamp: January 19, 2020, 03:06:28 pm
content: Quote from: E-Design on January 19, 2020, 02:09:23 pmAgree, and could you post a photo of the breadboard setup too please?Something oscillating maybe?

author: ArthurDent
author_rank: Super Contributor
timestamp: January 19, 2020, 03:31:09 pm
content: If you did get bogus parts from a major supply channel, this isn’t new. Twenty plus years ago I worked for an international company and we made some electronic test equipment for utilities and we had open parts orders from major parts suppliers. All of a sudden almost all of this equipment we made was failing test and it was found that one shipment of a particular fairly expensive IC that we had been ordering for years from the same supplier was the problem. After an investigation it was found that all those ICs in this one shipment were relabeled crap. We never found out how this happened but the supplier certainly wouldn’t have done it knowingly and they made good on the repairs we had to make.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 19, 2020, 03:39:22 pm
content: Looking at your level of "documentation", it's most likely that there is something wrong in your circuit design.Please post full schematics, photo of the layouts, and actual measurement results (like scope traces) in a readable way.

author: rstofer
author_rank: Super Contributor
timestamp: January 19, 2020, 04:43:46 pm
content: It sometimes happens that known bad components, those that failed final test, make it into the supply stream by fraudulent schemes. It's a real problem for the factories, not only do they have the losses associated with the failed components, they will double that loss when they make good on the failures. They do everything in their power to prevent this from happening including smashing up TO-3 packages with a jackhammer. After all, they still want to recover the gold.I suppose a factory lot could bypass final test somehow. It would be a screw-up but I guess it could happen. The factory and their distributors will do everything in their power to make it right. Their reputation as a manufacturer/supplier is riding on how they handle screw-ups.It is highly unlikely that this is happening, post your circuit.I'm not a big time hobbyist so I only place a few orders per year but I have never gotten a defective component from DigiKey or Mouser. Could be luck, I suppose.

author: wraper
author_rank: Unknown
timestamp: January 19, 2020, 04:58:05 pm
content: Quote from: Jwillis on January 19, 2020, 11:28:25 amNope. And output can be anything. No voltage related to what? And op-amps generally don't have "ground" which can be considered as reference point to begin with, consider it just as negative power supply. Often GND (V-) pin has negative voltage in reference to actual GND of circuit. And operation of inputs/output is referenced to actual GND and has nothing to do with voltage on power supply pins..QuoteWhat you mean by common mode to begin with? If you leave even one of inputs floating or connect them together, output can be anything.

author: ebastler
author_rank: Super Contributor
timestamp: January 19, 2020, 05:04:19 pm
content: Quote from: Jwillis on January 19, 2020, 11:28:25 amAs stated a couple of times above, we would really need to see a schematic. But in the meantime, I was trying to figure out what I can from the text.The first two sentences in the (shortened) quote above seem contradictory. Whatdidyou connect to pin 3? And, reading further to the final paragraph -- did you leave pin 3open, by any means? If so, that won't work!Finally, a comment unrelated to content and technology: You would really do yourself and all readers a favor if you adopt the standard typographic convention of putting a spaceafterpunctuation marks, not before them.EDIT: Ah, looks like wraper has the same suspicion and was quicker than me putting it in words...

author: wraper
author_rank: Unknown
timestamp: January 19, 2020, 05:10:11 pm
content: Here is some info about unused op-amps. It can give you some clue what will happen when you don't connect inputs as for normal operation.

author: jmelson
author_rank: Super Contributor
timestamp: January 19, 2020, 06:23:53 pm
content: Quote from: blueskull on January 19, 2020, 02:00:03 pmI buy mostly from Digi-Key, and also Mouser and Newark. A long time ago, somebody at Digi-Key was stealing parts, and a BUNCH of cut tape orders were coming up short, like half what I ordered. I guess their security people eventually figured out who was doing it and canned them. Otherwise, I have had VERY few errors from any of the above. I did get the wrong regulator once, it was supposed to be fixed 5 V, and I got the adjustable regulator version. One other issue was for a time they were taking FPGAs out of the Xilinx tray and repacking them in little tiny cut tape strips of 5, which I then had to put back in trays to run in my P&P machine. In all that handling, fragile leads could get bent a little.I just reflexively do static-safe handling, grabbing the bench before touching any devices or boards, and keeping all sensitive parts in static-safepackaging of some sort. I almost NEVER have problems like you report. I've probably built gear with 100K ICs over the last 40 years, and cancount on one hand the number of bad parts I have encountered when there was no reason (improperly connected power supply, wrong signal connection, part soldered on in wrong orientation, etc.)I did have one incident when buying a bunch of Analog Devices op amps from a major NON-franchised distributor. After mounting 25 of these chips on a board, half of them blew up. AD had recently changed their marking/logo and put their triangle logo near the pin 5 side of the chip. I suspect another assembler put bunches of chips in their P&P machine backwards, and assembled boards. Then, after massive failures, they scrapped the boards, the chips were picked off and cleaned up, and re-reeled. After talking with AD, I thought to ask, was it possible for a reel of ICs to have varying date codes? They said that was ABSOLUTELY impossible in their process, and clearly indicated the parts were either counterfeit or salvaged. So, I swore to NEVER, EVER, deal with that outfit again.But, other than that, I have had great reliability when buying parts from the major, franchised distributors. I know that in the past, Newark scrapped all returned merchandise, as ourt local surplus outfit bought it by the pallet load. Some of it was bad, like power supplies, but most of it was fine and just "ordered the wrong part". But, refusing to restock anything that has been opened should keep their stock pure of counterfeits and damaged parts.So, I'm guessing you have caused static or electrical overstress damage in your experimenting.Jon

author: Jwillis
author_rank: Super Contributor
timestamp: January 19, 2020, 11:03:47 pm
content: The conceptual design was derived from other similar designs andhttps://www.richardsonrfpd.com/docs/rfpd/Microsemi%20How%20to%20Make%20Linear%20Mode%20Work.pdfThe TLV2371 op amp will operate with single supply and no voltages are exceeding maximums .Common-mode input voltage (VCM) 0 - VDDVhttp://www.ti.com/lit/ds/symlink/tlv2371.pdfAs I stated I'm not as experience as many others here and if I get some terms incorrect I apologize. But the individual circuits are similar to other circuits described by many other people.The two circuits are identical. Two of the 10 op amps purchased work correctly in my circuit where as the other 8 do not.If no voltage is applied to pin 3 there should be no voltage at pin 3 . The 2 working op amps confirm this. The 8 non working op amps have approximately 200mV at pin 3 with no voltage applied to pin 3. Ergo they must not be functioning correctly.

author: wraper
author_rank: Unknown
timestamp: January 19, 2020, 11:21:15 pm
content: What is the meaning of "no voltage to mosfet"? If there is no voltage on the drain, feedback loop won't work at DC, thus opamp voltage should go up to the positive rail voltage if any positive voltage is present on non-inverting pin. Also I don't get about "This circuit works" on the left side of schematic. Does it work with all op-amps or not? Did you run right part of circuit with 2.2k resistor from inverting pin to GND disconnected?EDIT: Also resistance of gate resistors is way too high.

author: rstofer
author_rank: Super Contributor
timestamp: January 19, 2020, 11:27:57 pm
content: Where is the negative supply connection on the 2d op amp? It is clearly shown on the op amp at the left and not shown at the op amp on the right.

author: ebastler
author_rank: Super Contributor
timestamp: January 19, 2020, 11:28:33 pm
content: Thanks for the schematic, JWillis. Could you also post some photos to give us an idea of the layout and wiring?Edit: I don’t see any decoupling caps for the op-amp supply, which reinforces my earlier suspicion that the op-amps may be prone to oscillations.

author: Jwillis
author_rank: Super Contributor
timestamp: January 19, 2020, 11:44:30 pm
content: To add I've worked with several other op amps without complication in other projects . UA741CN , BA4558 AND 4559 , TL081CP , OPA445 .OPA604 to name only a few. I've also worked with many many other sensitive chips without complication.I'm not the type to boast but I'm also not a complete moron when it comes to complying to safety matters concerning the ESD of these components.The only time I have ever destroyed chips is when I applied to much voltage to the supply. I am learning so mishaps do occur and I do recognize where I go wrong .If I seem upset you would be correct . I have bought hundreds of components form Ebay and Banggod and other non recognized dealers across the world and have had non of them fail when used in the correct manner.This is the second time from this "reputable" dealer that I have had premature failure of components. the TLV2371 and a Mosfet FDL100N50F which failed at 5 Watts.10 V and 0.5 A.I don't buy a lot from Digikey because of the method of shipping is not compatible to rural address. It's to much of a pain in the back side to deal with.But when I get a 90% failure rate with Digikey and a near zero failure rate from other unofficial , non recognized dealers , I question the integrity of supposed Authorized Distributors .

author: rstofer
author_rank: Super Contributor
timestamp: January 19, 2020, 11:44:50 pm
content: Quote from: wraper on January 19, 2020, 11:21:15 pmI thought so too but these devices aren't being used as switches. So, we're not interested in dumping a lot of charge per second into the gate. I'd have to research that, I think.Since there is no appreciable gate current, there is no reason not to reduce the gate resistor to 1K. As long as the op amp doesn't latch up at startup, it should be fine. In the linear mode it should be making small changes to gate voltage.In the meantime, I'm more interested in the right hand op amp negative supply and it appears that the 10k works ok on the left hand MOSFET. Well, we don't really KNOW that it works for all corner conditions but it might.

author: rstofer
author_rank: Super Contributor
timestamp: January 19, 2020, 11:48:29 pm
content: Quote from: Jwillis on January 19, 2020, 11:44:30 pmYes, but... What about the negative PS connection to the right hand op amp. Your schematic shows it omitted and that just happens to be the circuit that fails. You have consistent failures with one circuit and no failures with the other circuit and you want to dump it on the op amps. It is something about that right hand circuit that is different than the left hand circuit.

author: Jwillis
author_rank: Super Contributor
timestamp: January 19, 2020, 11:49:12 pm
content: Ah rats sorry I made a mistake drawing the schematic . I.m aware that my mind is defective sometimes .I put in the correct drawing .I'm really sorry for that .

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 12:05:05 am
content: Quote from: rstofer on January 19, 2020, 11:44:50 pmThe voltage to the Mosfets come from the circuit to be tested by the electronic load . So to test the the op amps working (Input Output) no voltage or current is required to pass through the mosfets at this time. I simply want to get the op amps to respond to the control voltage . 2 of the 10 do respond correctly . The 9v supply ground is connected to the control board and is common to the mosfets as well . The control board works as it should with a single Mosfet. the point of the project is to expand the current capabilities by paralleling mosfets .

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 01:10:04 am
content: Quote from: Jwillis on January 20, 2020, 12:05:05 amUnless there is some current through MOSFET, feedback loop does not work. So in absence of control signal output of opamp basically becomes determined by offset voltage of opamp inputs which is different for every particular IC.Quote... with non working feedback loop. You want for incomplete circuit (disabled feedback loop) to work "properly", whatever it means in your opinion.

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 01:14:34 am
content: Connect inverting input to opamp output and watch how output voltage follows input voltage as on circuit below. Or simply short the 0.1 uF capacitor without disconnecting resistor from MOSFET source and connecting to the gate. It will do the same.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 01:15:17 am
content: Quote from: Jwillis on January 20, 2020, 12:05:05 amIt is required if you want it to work. The whole purpose of this circuit is to try and reproduce the scaled control voltage (0-1.3V on your diagram) across the 0.21 ohm resistors by controlling the current through the mosfets. If you aren't supplying a source of current to the mosfets it cannot do that, all that will happen is that the op amp outputs will go into positive saturation (assuming you're supplying some non-zero value of control voltage).Do you understand how this circuit works? Are you trying to modify someone else's design without a fundamental understanding of what's going on?

author: TerminalJack505
author_rank: Super Contributor
timestamp: January 20, 2020, 01:21:47 am
content: I've attached a screenshot of your circuit running in a simulator under the conditions that you have described--so far as I can tell.The simulation results should be pretty similar to what you would see in the real world. What is it you're are seeing that is different?

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 01:58:43 am
content: Quote from: rstofer on January 19, 2020, 11:48:29 pmMy intention is not to be disruptive in anyway. But I suffer from chronic anxiety and panic attacks which cause me to make mistakes.My apologies for posting an incorrect drawing . But I already stated that both circuits are identical and that when the good op amp was swapped from one circuit to the other it worked as expected . I missed drawing the ground on the one circuit but it does exist in the real circuit . I've re-posted the diagram as it is in the breadboard .The OP is a description of the procedures I made to resolve the issue at hand .Quote from: Cerebus on January 20, 2020, 01:15:17 amThe working circuit delivers a voltage from pin 6 to the Mosfet regardless whether theirs a voltage at the drain of the mosfet or not . it makes no difference . With the good op amps installed the whole project works when a voltage and current is supplied to the control board. The bad op amps lock the mosfets full on and can not be controlled.The bad op amps deliver 9v continuously At pin 6 with 0 volts applied to pin 3 . With 0 volts applied to pin 3 on the bad chips there is still 250mV coming from pin 3.Considering that when the good op amps are installed the whole project works as it should . So I think I have a pretty good idea of what I'm doing.

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 02:05:55 am
content: Quote from: Jwillis on January 20, 2020, 01:58:43 amFrankly, you have no clue.

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 02:14:43 am
content: Quote from: wraper on January 20, 2020, 02:05:55 amI's that the only solution you can come up with is to attack me .I've given all the information of what is occurring ,posted schematics and photos In as a reasonable way as I can and people attack me?I posted to find a solution to a problem. To find a way to test the suspect chips and all you've done was give out nonsense and then attack me .If you have something personal against me then then spit it out . Other wise why do you even reply to my post.

author: Bud
author_rank: Super Contributor
timestamp: January 20, 2020, 02:43:11 am
content: Do what Wrapper said in reply #22 to test your opamp when no external voltage source is connected to the Mosfets (no current through the Mosfets). Your problem is the inverting input of the opamp is effectively floating / isolated when no current flows through the Mosfet. Alternatively, connect a voltage source to your Mosfet drain to let current flow through the Mosfet, that will create voltage drop on the Source resistors into the inverting input of the opamp and everything should work. Wrapper provided you with a simulation that shows exact symptoms you experiencing when the Mosfets drain is disconnected.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 02:49:23 am
content: Quote from: Jwillis on January 20, 2020, 01:58:43 amCan you please either (1) annotate your schematic with pin numbers or (2) refer to the signal names. Without digging out a datasheet we've no idea what signal is on what pin.To enable sane discussion I've edited your quotes with pin numbers substituted for signal names and obvious typos corrected.Quote from: Jwillis on January 20, 2020, 01:58:43 amYup, the output at positive saturation just as I predicted. The op amp doesn't need to be bust for that to happen, I wouldexpectit as long as there was some input offset voltage present because you've no effective feedback loop with no power applied to the mosfet sources. What's telling here is that you're not telling us what voltage is on pin 2 (the inverting input) at the same time. For this particular op amp Vout= (Vpin3- Vpin2) * 100,000, so a 90 uV difference between the two inputs is enough to drive the output to the rail. The typical input offset voltage is specified as 2 mV, 22 times more than is needed to force saturation.This MUST be a typo:Quote from: Jwillis on January 20, 2020, 01:58:43 amIf you mean theoutputinstead of thenon-inverting inputthen, again, with no effective feedback loop you cannot predict what voltage will be there.Quote from: Jwillis on January 20, 2020, 01:58:43 amWhat constitutes "works as it should"? That is, what are the signal conditions that you are taking as "works"?You keep claiming that you have "a pretty good idea of what I'm doing" or similar but all appearances suggest otherwise and that you're blaming "bad op amps" for a failure to understand how this circuit should work. Folks are trying to help, but we can't if all you do is say 'I know what I'm doing' when offered explanations and advice. Please convince us that you're not clueless and that you understand the basics of this circuit by answering one question: What is the transconductance of this whole circuit (from the control input to the source of the mosfets)?Yes, this is a deliberate shibboleth and you should be able to answer it in 5 seconds if you know what you're about. If you can't thenpleaselisten to what people are saying.

author: TerminalJack505
author_rank: Super Contributor
timestamp: January 20, 2020, 03:50:23 am
content: Quote from: 3roomlab on January 20, 2020, 03:28:08 amThat's the Tina simulator from DesignSoft. You can get a free version of it from TI here:http://www.ti.com/tool/TINA-TI.

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 04:27:32 am
content: The circuit appears to be working as it was intended to do so for now .And it may still need further work.But these are the observations I'm making .With the 2 "good" op amps installed , when I apply 0 volts to the non inverting pin(3) as it is drawn in the schematic . I get 0 volts at the out put pin (6). No current flows through the mosfets. When a variable DC voltage is applied to the non inverting pin (3) I get full variable range of current sink and both Mosfets have an equal temperature. Which means that both Mosfets are sinking current equally .The circuit seems to work as it is intended to do so.With the "bad" op amps installed in the circuit , When 0 volts is applied to the non inverting pin (3) . I get 9 V at the output pin (6) and the Mosfets go full on. and there is 250mV measured at the non inverting pin(3) when it should be 0V.I've done the suggestions made and the observation remain the same as I described .The observation leads me to believe the suspected IC's seem to be leaking internally. I could be wrong I'll admit that .But I want to test them separately out of the circuit to determine if there is indeed a problem with them. This was the intention from the beginning. A simple circuit to test the suspected chips .

author: TerminalJack505
author_rank: Super Contributor
timestamp: January 20, 2020, 05:08:44 am
content: It sounds like they probablyarebad if what you say is true. I could see you having this particular problem if you accidentally exchanged the two op amp inputs or if you were using different components (other than the op amp.) But if you are just swapping the op amps out on the breadboard and not changing anything else then they should all behave the same.

author: ebastler
author_rank: Super Contributor
timestamp: January 20, 2020, 06:03:31 am
content: Quote from: TerminalJack505 on January 20, 2020, 05:08:44 amI am still not convinced. Having seen the wiring photos, the change in circuit behavior may just be due to having moved the wires around a bit when swapping ICs.Jwillis — have you swapped back and forth between a known „good“ and a known „bad“ op-amp a few times to confirm that the problem is really tied to the op-amp? I.e. the problem reproducibly goes away when you put in the “good“ amp, and comes back when you install the “bad“ one?Also, do you have an oscilloscope, and have you used it to inspect the circuit when it behaves badly? Your breadboard setup uses rather long wires and no decoupling on the power supply of the op-amp; so I still think that oscillations may be the root cause. (And that they come and go either with slight differences in the individual amps’ properties, or more likely with moving the wiring around.)

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 07:23:15 am
content: Quote from: ebastler on January 20, 2020, 06:03:31 amyes I did do that. I checked on the scope for any osculations with the good ones and I get a clean linear line on all T/D between 5 ms and 2us at .2V/DI have not checked the bad ones at this time because it's getting pretty late and I have to figure a way of preventing the mosfets from burning out with them in the circuit. This is why I need to test them out of circuit.I'll try again tomorrow .I.m very sorry if I upset people .It's pretty clean that I have trouble with getting defensive when I think I'm interpreting some comments as insults .I apologize.

author: rstofer
author_rank: Super Contributor
timestamp: January 20, 2020, 05:14:32 pm
content: I think Wraper nailed it in Reply #22 (and #13 and #21). I kept thinking that there would be some feedback caused by voltage drop across the 0.21 Ohm resistor caused by MOSFET current flow. Lacking current flow, the op amp is running open loop, no feedback, with the inverting pin effectively grounded. Any voltage on the non-inverting pin will cause the output to go high and stay there.The idea of shorting out the capacitor as shown in Reply #22 seems like a good test.You could also measure the Vcc current for the left op amp and then for the right op amp. See if there is a difference. Is the op amp on the right side of the schematic getting hot?I think it is going to take some comparative testing of the two circuits to discover the problem. If you bought your parts from an authorized distributor, I think you should give up on the idea that they are defective and start working on differential analysis. Why would one side smoke the chips and the other side not. Eliminate the MOSFETs and add the short proposed in #22 and see how the output follows changes in the input. Measure power supply currents, all the possible tests you can think of.The problem is not likely to be defective components. More likely, you have a couple of op amps that have some internal difference, still within spec, that allows them to deal with whatever the fault really is.If you had 1 op amp fail in a lot of 10, oh well. It MIGHT happen. But you're going through them like popcorn, the problem is not the supplier.Circuits on the Internet often have problems. If they are handouts from a seminar, you can bet the presenter said something like "Oh, by the way, there is an error in this schematic. You need to ...". Anybody who has ever gone to a seminar has heard something similar. Sometimes it isn't a design issue but an operating condition. Like with this circuit! If there is no MOSFET current, there is no feedback. Op amps don't like running open loop.

author: rstofer
author_rank: Super Contributor
timestamp: January 20, 2020, 05:28:26 pm
content: Quote from: blueskull on January 20, 2020, 07:37:31 amI have a grounded mat that I use when I play with motherboards. The rest of the time I am working on a Formica surface, no wrist strap, an ESD safe soldering iron and doing just about everything wrong. Literally a train wreck in terms of ESD protection.I have NEVER damaged a device! I know it's just luck and the fact that I primarily mess with digital chips (Schottky protection) but I think it is the higher relative humidity where I live. Lots of agriculture, just off the Delta (Central Valley of California), something like that.I'm talking NEVER! I know, someday I'm going to get bit! Let's hope it is an inexpensive component.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 05:38:41 pm
content: Quote from: Jwillis on January 20, 2020, 07:23:15 amYou're watching themkissing? There are places you can get arrested for that sort of behaviour.

author: Jwillis
author_rank: Super Contributor
timestamp: January 20, 2020, 06:11:40 pm
content: When the op amps set up with the inverting pin set to ground and no voltage at the inverting pin it acts like a comparitor with a 0V reference . If the non inverting side goes above zero the output goes to Vdd. If the input goes to zero then so should the output go to zero. In an ideal op amp.But because of imperfect manufacturing we don't get perfect op amps. The differential input transistors of real op-amps may not be exactly matched.So This causes the output to be zero at a non-zero value of differential input, called the input offset voltage.For the TVL237X the input offset can be typically 2mV and as high as 4.5mV according to the Data sheethttps://www.ti.com/lit/ds/symlink/tlv2370.pdfSince i don't have the equipment necessary to do an proper offset voltage analysis I did what I could . I set up a test circuit like one of the pair in the schematic . Except i added a negative rail so I could inject a variable negative voltage into the non inverting side.The idea was to see much much negative voltage is required at the non inverting pin to get the out put to zero. I tested all 10 of the op amps in this way.The 2 that I considered as good showed no voltage was required at the non inverting pin to get the output to zero.The remaining 8 op amps showed much different results. 2 showed results of -3.5mV and -5mV at the non inverting pin to zero the output. They may be with in spec i don't know . Since these chip don't have an internal offset I'll have to do it externally.The other remaining 6 showed even worse results ranging from -14mV ,-35mV , -16mV , -52mv -215mv to the worst one at - 618mv. to zero the output . Now I may not be an expert and the test is less than accurate but those figures seem pretty darn high to be with in spec.

author: ebastler
author_rank: Super Contributor
timestamp: January 20, 2020, 06:26:42 pm
content: Did you also replace the GND supply rail with your negative supply? You are not supposed to pull an input below the negative supply rail on these op-amps. The absolute maximum rating (i.e. the potential damage threshold) is -0.2V, but functionality is only guaranteed down to 0V.So I am not sure your large offset errors are meaningful, since they were probably measured in "no-man's land". If you want to measure the offset error, instead of pulling one input to negative voltages, I would suggest to pull the other one positive.

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 07:01:16 pm
content: Please measure input offset voltage properly. First of all, it's as ebaster said. Secondly, "good opamps" most likely simply have input offset voltage of opposite polarity.

author: rstofer
author_rank: Super Contributor
timestamp: January 20, 2020, 08:49:02 pm
content: Once again, Wraper has it spot on. Input offset voltage is to be measured with Vin = Vdd/2 and Vout should also be Vdd/2 (+- offset voltage) according to the datasheet.With 9V for Vdd and the ground pin at 0V, Wraper's circuit is exactly right.Of course, the resistors could be replaced with a pot but I would add 100 Ohm resistors to each end such that the output voltage (input to non-inverting pin) could not reach either 9V or 0V. In any event, the output should track the input as selected by the pot.

author: rstofer
author_rank: Super Contributor
timestamp: January 20, 2020, 09:01:07 pm
content: Quote from: Jwillis on January 20, 2020, 06:11:40 pmIt's probably just my warped way of thinking but, to me, there is a huge difference between 0V and 'no voltage'. I'm thinking that 'no voltage' means the pin isn't connected to anything. 0V, OTOH, is the value of a voltage. Sure, it's zero but it is defined. It's right there between a positive and a negative value for voltage.To be clear, you got 0V output with no connection at the non-inverting pin? Given the existence of an offset voltage, that would mean that the offset voltage is 0V and would be 'just luck'.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 09:05:00 pm
content: For measuring offset the best bet is a circuit like this:It has the advantage of having gain that pulls what is going to be a small or very small voltage into a range that is easy to measure, it has the disadvantage of needing good matching between resistors for optimal common-mode rejection ratio.

author: ebastler
author_rank: Super Contributor
timestamp: January 20, 2020, 09:06:10 pm
content: Quote from: rstofer on January 20, 2020, 09:01:07 pmI had the same thought, but then recalled the input voltage divider in the schematic Jwillis had posted earlier. I do hope that the voltage divider was present, and hence a pulldown to 0V.

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 09:13:29 pm
content: Quote from: Cerebus on January 20, 2020, 09:05:00 pmFor sloppy op amp like one in question (4.5mV max offset @ 25oC), it does not matter. Unless you only have multimeter which does not have mV range. Even crappy 830 series multimeter with 200 mV range would be enough for evaluating if it's in spec using follower circuit I posted.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 10:32:03 pm
content: Quote from: wraper on January 20, 2020, 09:13:29 pmWell exactly, there's the problem, 4.5mVmax/200mV = 2.25%maxof a 200 mV range which isn't a great place to be measuring something. Following the usual don't measure below 10% of range rule you'd change range there if you could. A more reliable measurement for the cost of two extra resistors (perhaps 4 extra resistors if you don't have a split supply handy) seems cheap to me. You don't need to go the whole gain of 1001 - indeed in the case in point (9V rail, 4.5mV offset) that'd get you into trouble - but a gain of 11, 101 or 'whatever pairs of resistors I have on hand' is still an improvement.

author: wraper
author_rank: Unknown
timestamp: January 20, 2020, 10:48:03 pm
content: Quote from: Cerebus on January 20, 2020, 10:32:03 pmIt's for evaluating if opamp is within spec. Not precisely characterizing it, in which case you won't use cheapest multimeter anyway. Even if measurement is 20% off, it does not matter in this case. Also with particular resistor values gain is so high that output will clip to power rails at max offset spec.QuoteThere is nothing wrong with measuring at the bottom of the range with digital instruments as long as you understand specifications and limitations.EDIT, and BTW, this particular circuit with meter in series with input is not suitable for CMOS op amps like TLV2371 due to very high input impedance. You are basically connecting voltmeter in series with tiny capacitor.

author: Cerebus
author_rank: Super Contributor
timestamp: January 20, 2020, 11:37:45 pm
content: Quote from: wraper on January 20, 2020, 10:48:03 pmErm, that's a voltage source symbol, symbolising the offset voltage. Why would you even believe that was a meter? There would be no point of adding gain to then just go and read the unamplified signal at the input.

author: rstofer
author_rank: Super Contributor
timestamp: January 20, 2020, 11:39:48 pm
content: Measuring the offset voltage may be an interesting experiment but at this point I think it would be more helpful to know if the op amp did anything at all. For that, the simple voltage follower with the fixed resistors replaced by a potentiometer (plus the end resistors I mentioned above) would be a more useful test.I suppose if the device is destroyed, the offset voltage might be out of range but that's only one manifestation of a smoked op amp and perhaps it will be apparent that the op amp is toast without having to get into details.I'm going to vote for the simple follower... I can measure the offset voltage at any point in the range of the potentiometer and I can watch the output track the input.

author: Jwillis
author_rank: Super Contributor
timestamp: January 21, 2020, 12:19:29 am
content: Quote from: wraper on January 20, 2020, 07:01:16 pmBefore I get started on this experiment . Do I need a split supply , single supply or no supply .? I'm guessing VDD must be 9V but I want to get this right . I want to be sure as to probe orientation . The voltage at pin 3 must be exactly 4.5V ? I'll try to find exact or as close as I can for the resistors.I would like to do this right so I give you the right measurements.

author: wraper
author_rank: Unknown
timestamp: January 21, 2020, 12:27:32 am
content: Quote from: Cerebus on January 20, 2020, 11:37:45 pmOK, I had a brainfuck.

author: wraper
author_rank: Unknown
timestamp: January 21, 2020, 12:30:46 am
content: Quote from: Jwillis on January 21, 2020, 12:19:29 amWith circuit I posted you don't need split power. 2 resistor divider create a reference point at half of the power source voltage, which you can also consider as virtual ground. With split power, you don't need those resistors and can simply tie non-inverting input to ground.

author: rstofer
author_rank: Super Contributor
timestamp: January 21, 2020, 12:56:38 am
content: Quote from: Jwillis on January 21, 2020, 12:19:29 amAs Wraper's circuit is drawn, you need +9V at the Vdd pin of the op amp and 0V (the other side of the power supply) at Vee but we will call that "Ground".Then you need to get a voltage APPROXIMATELY equal to Vdd / 2 (ie 4.5V) with 2 resistors. Close is good enough. Remember, I am advocating for a potentiometer with end resistors. How well matched will that be when you twiddle the knob from end to end?We're trying to get Vos but, really, we're looking for op amp functionality. With the follower circuit, the output of the op amp should match the non-inverting input, more or less, due to offset voltage. Again, close is good enough. We know that Vos should be < 4.5 mV when the non-inverting input is almost exactly Vdd/2 but it shouldn't be wildly different anywhere else.

author: Jwillis
author_rank: Super Contributor
timestamp: January 21, 2020, 02:26:35 am
content: I noticed that you made a mark from pin 3 to pin 6 so I took 2 measurements with opposite polarities.Vdd 9.06V Resistors are 4.707K and 4.706K Voltage at divider 9.06 V 4.536V Pin #3Chip # 1 and # 2 are the ones that work in the circuitMeasurements in VoltsChip# Output to grnd Negative of probe at pin#6 across to pin # 3 Negative of probe at Pin#3 across to pin # 61 4.535 0.033 -0.0462 4.535 0.033 -0.0463 4.535 0.041 -0.0494 4.535 0.039 -0.0485 4.535 0.040 -0.0476 4.536 0.038 -0.0487 4.535 0.036 -0.0498 4.535 0.040 -0.0469 4.536 0.038 -0.04910 4.535 0.036 -0.048Not sure what to make of this but chip 1 and 2 have identical measurements .

author: wraper
author_rank: Unknown
timestamp: January 21, 2020, 02:58:45 am
content: Quote from: Jwillis on January 21, 2020, 02:26:35 amWhat are you using for measurements? It's really strange that you are getting different reading by just switching the leads. Not to say that simply measuring output voltage shows only 1mV variation but measuring between non-inverting input and output 10+mV, or did you miss one zero? Also why it's in a volts range. Doesn't your meter have millivolt range?

author: rstofer
author_rank: Super Contributor
timestamp: January 21, 2020, 03:26:49 am
content: As pointed out, all measurements between pin 2 and pin 6 seem to be off by an order of magnitude. If so, all of the offset voltages are in spec. The simple output voltage test leads me to believe all of the op amps are operating correctly.Later you can try a 1k load to Vdd and then again to Vee and do the simple voltage measurement again. This will show whether the output stage can actually drive a load.All of the devices look good to me - assuming there is a decimal point error in the offset voltages.

author: Jwillis
author_rank: Super Contributor
timestamp: January 21, 2020, 04:04:47 am
content: I don't know whats going on . The meter is auto ranging . I only have meters that go down that far .

author: Jwillis
author_rank: Super Contributor
timestamp: January 21, 2020, 04:36:54 am
content: So if can I compensate for the offset externally with a single supply or do I still need a negative rail to adjust the offset ?

author: ebastler
author_rank: Super Contributor
timestamp: January 21, 2020, 05:46:35 am
content: Quote from: Jwillis on January 21, 2020, 04:04:47 amYour offset readings (measurements between pins 3 and 6) look implausible, especially in view of the very precise and reproducible output voltage (pin 6 to GND). As you have separately measured pin 3 and pin 6 to be at 4.535 .. 4.536 V, stable to within a mV, why would you see more than a mV of difference between them?Please have another look at your multimeter(s). Most auto-ranging meters have a manual range selection pushbutton which you can use to select the most sensitive voltage range, and avoid the potentially confusing auto-ranging.Also, could you connect a scope in parallel to your voltmeter when you take those readings between pins 3 and 6? Just to check that no AC noise gets introduced during this measurement. You are connecting some nice long "antennas" there, with the multimeter's probe leads. [Edit: And no oscillations due to an unwanted positive feedback loop, coupled via the probe leads.]

author: wraper
author_rank: Unknown
timestamp: January 21, 2020, 09:08:47 am
content: In addition to said above, try attaching some 0.1-1uF capacitor in parallel to one of 4.7k resistors (does not matter which) in voltage divider and measure offset again.

author: Jwillis
author_rank: Super Contributor
timestamp: January 22, 2020, 04:21:56 am
content: Good news. I concluded that the op amps are fine. No need to freak out to Digikey. Why 2 would work in the old schematic and the others didn't I want to find out and will continue to to learn why.Any way what I did do was make a couple changes to the circuit that work with all the op amps now. First of was to fix the open loop. so I added a 10K from the inverting input to the output. But like before that didn't make a difference until I lowered the voltage to the non inverting input. So I added an adjustable voltage divider to the signal wire that goes to my circuit to lower that voltage to prevent saturation .Now I have to figure out how to balance the op amps so every mosfet is taking the same load.I don't want to go through the agony of reverse engineering the control board right now . But when the load dial is turned the voltage to the gate immediately goes to 7.3 Volts without load . With load a can't determine what the voltage would be unless I put the mosfet back into the board . Then as the dial is turned that voltage rises by a few millivots which opens up the mosfet more allowing more current to be sunk to ground. Why the control board is configured this way I don't know. I should have observed what the control board was doing a lot better.This video help me understand what was happening.

author: nigelwright7557
author_rank: Frequent Contributor
timestamp: February 10, 2020, 01:20:34 am
content: I use RS components and Farnell.I know with them an IR part will be from IR and not an unknown/unreliable source.The only component I have had a slight problem with was Chinese valves from RS.Some of them were microphonic but still worked OK.

author: Jwillis
author_rank: Super Contributor
timestamp: February 11, 2020, 09:22:00 am
content: Quote from: nigelwright7557 on February 10, 2020, 01:20:34 amI was still having problems with these units so I took them to a communications tech and he tested them for me. 4 function correctly are with in tolerance, 2 exploded with 12V supply ,5 leak voltage to the output. 1 leaked voltage back to the non inverting side. They are indeed faulty . I contacted digikey and they are sending replacements .

author: exe
author_rank: Unknown
timestamp: February 11, 2020, 09:57:57 am
content: Quote from: Cerebus on January 20, 2020, 09:05:00 pmDoes it require matched resistors?

author: Cerebus
author_rank: Super Contributor
timestamp: February 11, 2020, 01:42:10 pm
content: Quote from: exe on February 11, 2020, 09:57:57 amYup, ideally.General rule for alldifferentialconfigurations of op amps is that resistors that share the same rôle want to be matched. So in this case you want the 10R pair and the 10k pairs matched. If they're not matched it lowers the common mode rejection ratio. If you've got an theoretically perfect op amp with infinite common mode rejection ratio a mismatch of 0.1% will reduce the CMRR to 60dB, a 1% mismatch to 40dB.What matters most here is the ratio of input to feedback resistor. If you use 1% resistors you'll get a 1% uncertainty in gain, so a 1% basic uncertainty in your measurement. For this particular application the contribution from common mode errors is less important than the contribution from gain errors - you know that you have no common mode signal on the input (it's grounded), so any common mode errors will be from bias current effects on the input resistors which will be negligible in the face of the likely range of input offset voltages. For a  10uA bias current you'll have a common mode input voltage of 100 uV reduced to 1 uV by the 40dB CMMR.If you're dealing with ultra-precision op amps with single digit microvolt offset voltages then of course you're going to need to move to much better resistor matching but if you're paying for those kind of op amps you can afford to pay Vishay for some precision matched resistor dividers.I'd say if you're dealing with >100uV input offset voltages and are happy with results in the region of 1% accuracy then 1% resistors will do you fine. If you're after 0.1% accuracy or dealing in smaller offset voltages then you probably want to go down the route of hand matching some sets of resistors with a good meter (Or you can pay Vishay or someone similar lots of money for top notch matched resistors with low tempcos).

author: nigelwright7557
author_rank: Frequent Contributor
timestamp: February 19, 2020, 12:24:34 am
content: 10K is a high value for gate resistor.I tend to use 390R.

author: Jwillis
author_rank: Super Contributor
timestamp: February 19, 2020, 12:55:22 am
content: Quote from: nigelwright7557 on February 19, 2020, 12:24:34 amThe current draw at the gate of a mosfet is negligible . In the nanoamps. The Voltage drop across 10K resistor would have little effect to the voltage at the gate.

author: rstofer
author_rank: Super Contributor
timestamp: February 19, 2020, 01:54:55 am
content: But it could have an effect on rise/fall times and, thus, heating. If it matters...For power applications like H-bridges, we want to drive Amps into the gate to overcome the capacitance.

author: Jwillis
author_rank: Super Contributor
timestamp: February 19, 2020, 06:36:57 am
content: Quote from: rstofer on February 19, 2020, 01:54:55 amYes if your running a pulse.But not when running in linear. The way its used here in in linear mode.

author: exe
author_rank: Unknown
timestamp: February 19, 2020, 09:32:23 am
content: In a linear application the gate resistor becomes a part of feedback loop and need to be chosen wisely, imho. Although, I can't say if too big value can be a problem or not. Probably not as it seems to be pushing "first pole" to lower frequencies (as together with gate capacitance it makes a low-pass filter), making the whole loop more stable, but I advice use a simulation to confirm this.

author: Jwillis
author_rank: Super Contributor
timestamp: February 19, 2020, 10:21:09 am
content: Quote from: exe on February 19, 2020, 09:32:23 amThe whole circuit has been changed from the original design anyway.https://www.eevblog.com/forum/beginners/protection-for-electronic-load-circuit/. The 10K was replaced with 10 Ohm. But it had no effect on the performance of the circuit . The whole problem turned out to be faulty op amps.Since replacing the components the circuit has been functioning properly at 500 Watts for extended periods of time so far.