Waveform at one opamp terminal is seen at the other terminal? - Page 1
Forum section: electronics / beginners

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 07, 2012, 08:47:44 am
content: Hello there,I am smimulating using op amps I ltspice.I feed a 300khz square wave into one input. I am feeding 1 hz square wave into the other. The problem is that it seems to reject my 1hz square wave as the 300khz signal is measured at both terminals. What is going on?There is no feedback loop in my circuit. This is a comparator.Thanks.M

author: Rerouter
author_rank: Super Contributor
timestamp: August 07, 2012, 08:51:57 am
content: due to op amp feedback, it will do everything possible up to its limits to keep the inverting and non inverting pin at the exact same voltage, as such, i would take a shot in the dark that your feedback network connects your 1Hz source to its output, and the op amp is pulling up and down that input in order to make it match the other input,

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 07, 2012, 08:56:26 am
content: There is no feedback loop in my circuit.M

author: Rerouter
author_rank: Super Contributor
timestamp: August 07, 2012, 09:43:15 am
content: then perhaps post your schematic, or your simulation into pastebin or something similar, without seeing it, its very hard to say

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 07, 2012, 01:14:03 pm
content: Attached is my schematic. Check the inputs of U3. You will find that the squarewave of the voltage source appears at both terminals, which it shouldn't. This is probably one of the limitatons of real opamps. Perhaps the resistance between the inverting and non inverting input is too low.What do you think?Thanks.M

author: ejeffrey
author_rank: Super Contributor
timestamp: August 07, 2012, 01:24:29 pm
content: OK, well your schematic is quite a bit more complicated than you let on. Try reducing it to the key points (say, one opamp and two square wave generators). You say your circuit has no feedback, but it definitely does, so I suspect that you just haven't thought through it all.Edit: one more suggestion: in the future post an image of your schematic along with the simulation that clearly highlights what you are concerned about. I did actually download and simulate your schematic, but got bored waiting for ltspice to plot the results.

author: DRT
author_rank: Regular Contributor
timestamp: August 07, 2012, 01:32:00 pm
content: The problem is D2. It has no load or current path, so the oscillation on the +ve input leaks to the very high impedance node at the -ve input.

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 07, 2012, 01:58:29 pm
content: RDT, I tested 100K, 10k and 1k loads after D2. It didn't make a difference.Ejeffrey, sorry about that. Will post image in the future. But if you assume the square wave at the input terminal is passed through the feedback loop, I think you've made an error.The feedback loop generates the low frequency squarewave.Thanks.M

author: DRT
author_rank: Regular Contributor
timestamp: August 07, 2012, 03:02:42 pm
content: Quote from: king.oslo on August 07, 2012, 01:58:29 pmR9 is not a load. Changing the value won't help. The -ve input is not a virtual-earth because there is no feedback. To put it another way, when the output of U2 is low, the diode is not biased, so the node around your -ve input is, in effect, not connected to anything hence the leakage through the op-amp input stages.

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 07, 2012, 08:03:48 pm
content: No, i didn't change the value of R9. I added a resistor to ground.Now, should I loose R9 for success?M

author: DRT
author_rank: Regular Contributor
timestamp: August 07, 2012, 08:15:00 pm
content: You have several problems to deal with. First, your -ve input is not really under proper control. As I mentioned earlier, when the diode is not forward biased, your -ve input is floating. Second, the op-amp you have chosen is a rather specialist ultra-wide band type. It has input protection in the form of back to back diodes between the inputs - you might want to take a look at the datasheet. Third, using op-amps as comparators is not recommended (especially with this particular one which is explicity designed NOT to cope with large differential voltages on its inputs) - there are proper comparators for good reason.

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 08, 2012, 11:27:57 am
content: Thanks. Yes, I saw that in the datasheet, but I didn't think much of it because it was an opamp. Wikipedia told me it should have really high impedance between the terminals.Are you sure my D2 doesn't become forward biased when I add a 10k load after the diode?Thanks.M

author: free_electron
author_rank: Super Contributor
timestamp: August 08, 2012, 04:11:45 pm
content: Quote from: king.oslo on August 08, 2012, 11:27:57 amah, Wikipedia: the ultimate collection of big fat lies, half truths and stuff that is just plain wrong ...

author: DRT
author_rank: Regular Contributor
timestamp: August 08, 2012, 08:14:36 pm
content: I've said this all already, but I'll try again.Please consider the situation when the diode D2 is driven by a low output from the op-amp driving it, i.e. the anode is at 0V. Since the cathode side of the diode has no load whatsoever, your -ve input on the op-amp becomes an un-controlled ultra-high impedance node. You might as well have a disconnected -ve input in this situation. The op-amp input protection diodes will therefore cause this point to follow the +ve input. What do you think D2 is doing in your circuit?Also you are attempting to use these op-amps as comparators, with large potential differences between the inputs. These op-amps cannot be used this way.

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 08, 2012, 08:24:35 pm
content: Quote from: DRT on August 08, 2012, 08:14:36 pmDear DRT,Thanks for your patienceI just want to clarify why I am still talking about this: Yesterday, I made it clear that I have added 1K/10K/100K resistors to my circuit. It is connected from the cathode of D2 to GND. Doesn't this resistor qualify as a load?Thanks againKind regards,Marius

author: DRT
author_rank: Regular Contributor
timestamp: August 08, 2012, 08:42:38 pm
content: Adding a resistor to ground provides a load, but the protection diodes in the op-amp input will still drive the -ve input to match the +ve (less the diode drop). The question of what you think D2 is doing remains. However, the fundamental problem is the use of an ill-suited op-amp as a comparator.

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 08, 2012, 08:49:42 pm
content: D2 was added for testing something. I don't recall for what.Thanks:) M

author: DRT
author_rank: Regular Contributor
timestamp: August 08, 2012, 09:06:11 pm
content: With respect, you shouldn't be asking for assistance with your non-working circuit when random components have been "added for testing something. I don't recall for what."

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 08, 2012, 09:16:05 pm
content: Quote from: DRT on August 08, 2012, 09:06:11 pmYou've got to explain that.M

author: DRT
author_rank: Regular Contributor
timestamp: August 08, 2012, 09:27:10 pm
content: Sorry - I give up. You've been told several times over precisely what the issues are with your circuit, but you don't seem to be taking any notice of it, and then you admit to having added random components for reasons you can't remember.

author: olsenn
author_rank: Frequent Contributor
timestamp: August 08, 2012, 09:35:06 pm
content: I didn't even bother to read your origonal question, but it does make sense for your waveform to be identical at both input terminals of your op-amp. The whole basic operating principle of the op-amp is to try to keep both input terminals equal by varying the voltage/signal at the output. As long the required output voltage to do so falls within the bounds of the supply voltages (+VCC and -VCC) then it will succeed (within the limitations of the op-amp of course).

author: king.oslo
author_rank: Frequent Contributor
timestamp: August 08, 2012, 09:44:48 pm
content: Quote from: DRT on August 08, 2012, 09:27:10 pmDRT, thanks for your patience. You helped me resolved my issue a few hours ago. I don't want to know why the circuit isn't working. I wanted to learn why the 300KHz signal is at both inputs, and why D2 was causing a problem. Thanks to you, I understand that the 'less than ideal properties' of the op amp was causing the problem.Therefore, the reason D2 ended up in my schematic is irrelevant. I only wanted to learn how diodes and opamps function. Do you understand now?Quote from: king.oslo on August 08, 2012, 08:49:42 pmThis post was meant to signify that I wanted to terminate the discussion because I got my answer.Quote from: olsenn on August 08, 2012, 09:35:06 pmOlsenn, as far as I understand the opamp tries to keep both inputs the samethrough the feedback loop. I've learnt that one of the characteristics of the ideal opamp is infinite input impedance. Is this incorrect? I use a feedback loop in my circuit, but the 300KHz signal does not propagate through it. So I dont think this could be causing my problem.Kind regards,Marius

author: DRT
author_rank: Regular Contributor
timestamp: August 08, 2012, 09:46:14 pm
content: Quote from: olsenn on August 08, 2012, 09:35:06 pmNo, he has his op-amps configured open-loop (no feedback) in an attempt to make a comparator. His inputs track because the op-amp inputs are protected against large differential voltages by back to back diode. The 'op-amps always try to keep their inputs equal' concept doesn't apply when there is no feedback.