Op-amp weirdness - Page 1
Forum section: electronics / beginners

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 03:21:32 am
content: Hey,I hope someone can point out what I'm missing... I've been playing around with TLE2022 op-amps, my main goal being to get a reliable DC voltage follower that goes all the way to 0, on single supply.Now I have a 5.1V Zener diode plugged into the amp's output, and from there on I have direct negative feedback to the inverting input. My positive power rail is 18V, negative power rail is 0. That's all of the setup.First I plugged the non-inverting input into the 0V power rail. On the output I got ~0.7V, which dropped across the Zener, and I got actual 0 Volts at the end. Then, I started playing around with various voltage dividers... I gave it 90 millivolts, and it followed quite perfectly. Then 1.8 Volts, no problem at all. Then 9 Volts, and... I got a whopping error of ~0.6V!First, it's weird how it worked at low voltages, but at 50% of the rail I'm getting such an unacceptably high error... Also it somehow makes me wonder that the error is about one diode drop... Any idea?Next up is, I'll try to just divide the input voltage down, and set a gain higher than one to multiply it back up... Maybe it will remove the error.Is it possible that I'm already getting clipping? The Zener drop is 5.1V, the voltage to follow 9V... That's 14.1V, well below the positive rail...

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 03:31:50 am
content: Remember that there's no fixed diode "drop", it's an I(V) curve that goes all the way to zero. Even though it's nominally 0.6V-0.7V, there's a current level that will give 0.4V, and 0.15V, and 100V...(If you want to see a large number, go find the real equation for a diode's I(V) curve and compute I(100V)...)So... at low voltages, where will the current flow to give you the diode drop? Into your negative rail? You don't have one... That's why it changes with voltage.What are you calling the "output" of this circuit? The output of the op amp, or the ass end of the diode?

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 03:40:52 am
content: Quote...Excalibur? Are you f-king kidding me, TI? Time to rope in the marketing guys...Anyway, this is why the diode won't magically get you 0V output. Don't you think they'd include one inside the op amp if it could do that? They're practically free...The current has to flow somewhere.Give it a negative rail or redesign your circuit not to require zero.

author: BravoV
author_rank: Super Contributor
timestamp: August 04, 2013, 03:52:37 am
content: Quote from: Sigmoid on August 04, 2013, 03:21:32 amI stopped reading your next paragraphs when finished the 1st one, your problem is clear, op-amp doesn't go to a perfect zero volt when using a single supply, even they claimed its "rail-to-rail" output capable.As mentioned by c4757p, supply it with a negative rail is the only way.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 03:53:59 am
content: I think I may not have been clear enough.The way I'm wired up is:I+ <- test voltageOut -> Zener cathodeI- <- Zener anodeThen, I grounded the Zener anode with a 1kOhm resistor.My measurements were:Test voltage (I+)Zener anode voltage (I-)0V0V0.018V0.018V0.18V0.18V9V8.4VSo... as you can see I get zero fine. (As I should. The Out pin of the op-amp naturally cannot go to zero, but the Zener diode blocks that voltage. When I apply a small voltage to I+, the op-amp detects the error, and goes as high as it takes to raise I-, which in this case is about 3V, where I get the 0.018V at the Zener anode. This far it's not theory, it practically works.)What's not fine is when I get into higher ranges...

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 03:56:03 am
content: Quote from: BravoV on August 04, 2013, 03:52:37 amThanks for deciding that I'm an idiot without even bothering to read the entire post. Please check my latest post, it clearly IS possible to get a voltage follower that goes to zero. I just have one on my breadboard.My problem is a negative error at higher voltage ranges.

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 04:01:50 am
content: Have a look at figure 16 in the datasheet. Then imagine your op amp isn't quite "typical".

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 04:23:24 am
content: Sorry which figure?We may be looking at different datasheets...http://www.mouser.com/ds/2/405/slos191d-121626.pdfAnyway, I solved the problem, now I'm just curious why this works... What I did was divided the test voltage by two before feeding it into I+, and then put a similar divider into the feedback path, to get a gain of 2x:Test voltage -> Voltage divider 1kOhm/1kOhm -> I+Out -> Zener cathodeZener anode -> Voltage divider 1kOhm/1kOhm -> I-The result is quite magical. XD The error disappeared. (I'm taking the voltage follower measurement from the Zener anode.)The only thing I can think of is that the minute resistance of my test leads and breadboard contacts somehow weirded out the amp with a direct feedback, and this way whatever caused the error is now insignificant.

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 04:27:13 am
content: The one that you zip to if you type ctrl-F "figure 16" - duh!Maximum Peak Output Voltage vs Output Current. At 9V, your 1k pulldown will be drawing 9mA, getting you really close the output swing dropoff.Quote from: Sigmoid on August 04, 2013, 04:23:24 amAh-ha! So I was right! It works because you are halving the output current, putting you back into the safe zone.Just use a 10k resistor instead of 1k.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 07:04:43 pm
content: Quote from: c4757p on August 04, 2013, 04:27:13 amLOL, I managed to figure it out in the meantime... XD Just like a software person to disregard physical limitations...It seems the truly linear safe zone is below 7.5mA at this Vcc+.

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 07:11:08 pm
content: Quote from: Sigmoid on August 04, 2013, 07:04:43 pmThere is a cure for that: get a few oftheseand make them do something useful.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 04, 2013, 09:29:26 pm
content: Question below is REDACTED... It likely wasn't the op-amp that was screwing up, but the meter... If I calculate with an input resistance of 10MOhms for the voltage inputs on the BK Precision, as it's given for the 6.6V range on its datasheet, then it's quite obvious how I got the error when measuring the input voltage.I'll leave it here for posterity though, I guess it's a generic mistake for anyone without enough experience.As I mentioned it works great with input impedances in the 1-10kOhm range. However, when I try to feed it from an input with a 1MOhm impedance - which I guess shouldn't be considered extreme for a voltage follower -, I get this weird error that's several orders of magnitude larger than what I'd expect based on the offset voltage and bias currents in the datasheets.And to top it off, when I directly measure the voltage between I+ and I-, the meter measures 0V.Thinking along the lines of "what to do if measurement fixes it", I tried adding a 1MOhm resistor between I+ and I-, but it didn't affect the error when measuring from ground, and once again it was 0V when measured between the inputs.I+ to Gnd7.35VI- to Gnd7.88VI- to I+0.000V

author: c4757p
author_rank: Super Contributor
timestamp: August 04, 2013, 10:07:10 pm
content: Yup, common mistake, good to leave it as an example.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: August 05, 2013, 01:04:34 am
content: Above a megOhm, EVERYTHING is a voltage divider. XD