Just making sure that I'm understanding this op-amp circuit correctly - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 02:11:56 am
content: I'm reading a schematic, and I think I understand how this circuit works, but I wanted to ask to make sure. This part that I'm trying to understand is one part of a larger circuit, but it's an optional part.From what I can tell, it's a comparator circuit, in that it amplifies the voltage difference between the two inputs. It is not an IVC (current-to-voltage converter/transimpedance amplifier). The 10k ohm resistor is just there to limit the current going into the op-amp and (ideally) shouldn't have any effect on the output voltage. Is my assumption correct?Regarding the amplifier in question, the OPA602, even though it's used in a comparator circuit configuration, the OPA602 is a general-purpose op-amp and can be used in different configurations (such as an IVC). Am I right in assuming this, or is this wrong?

author: unitedatoms
author_rank: Frequent Contributor
timestamp: July 16, 2020, 02:16:29 am
content: The circuit is nonsense. Remove all parts and you have a pair of wires for input and output.

author: TimFox
author_rank: Super Contributor
timestamp: July 16, 2020, 02:19:37 am
content: The trick is in the asterisks: ** is a wire-around option, with the amplifier IC not stuffed, but * is the amplifier/comparator option.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 02:43:55 am
content: Yes, that's the idea. When the amplifier/comparator gain is not present, there's a 0-ohm resistor that shorts the input and output accordingly. I know it's optional, but I'm trying to make sure that my understanding of this circuit is correct (assuming that the circuit is indeed present and not optional).

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 02:45:56 am
content: Yes, I realize that it's optional. When it's optional, in the actual circuit, the op-amp is removed and a 0-ohm resistor acts as a short accordingly to the schematic. In this case, assuming that the circuit is not optional and is present in the overall design, is my understanding of it correct?

author: magic
author_rank: Super Contributor
timestamp: July 16, 2020, 05:57:10 am
content: Yes, it amplifies the input with full open loop gain of the OPA602, whatever that is. Mind that open loop gain at high frequencies is much lower than in a dedicated comparator, due to internal compensation.The 10kΩ resistor is not in the feedback path (there is no feedback path) so it's not a TIA.Besides limiting fault current through the protection diodes, it has a secondary effect of forming an RC lowpass with the input capacitance of the opamp (and diodes) and limiting the bandwidth to maybe a megahertz or so.

author: westfw
author_rank: Super Contributor
timestamp: July 16, 2020, 07:18:13 am
content: Clips the inputs, overdrives the outputs. Is it an Electric Guitar or Synthesizer circuit? They routinely do all sorts of strange things with semiconductors, outside their normal operational range, because they like the way it sounds...

author: magic
author_rank: Super Contributor
timestamp: July 16, 2020, 08:17:26 am
content: My guess would be an electron microscopeI assume D4 connection to ground is a mistake, it should only go to that weird symbol which is the positive power supply.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 03:01:13 pm
content: Thanks @Magic. I saw the design and thought that it looked similar to a comparator that I worked with. That helps a lot.This is actually part of a pre-amplifier circuit for a scanning tunneling microscope (STM). In this case, H is a reference voltage. Since it's going into the non-inverting terminal, depending on the voltage difference between the two (IN+ minus IN-), the output will be driven to either the positive or negative rail in a nutshell?

author: daddylonglegs
author_rank: Contributor
timestamp: July 16, 2020, 03:35:19 pm
content: Perhaps the operation of the overall circuit and machine servos the voltage input to zero?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 16, 2020, 06:56:00 pm
content: If you don't mind me asking, what is the purpose of the diodes there? I searched online, and I saw some op-amps with diodes in a similar fashion to prevent overvoltage.https://www.analog.com/en/technical-articles/key-benefits-of-input-overvoltage-protected-op-amps-in-systems.htmlHowever, with the diodes the way they are in my schematic, what is their purpose? Since they're connected to the inputs, do they just make sure that the voltage at the two terminals remain the same?

author: daddylonglegs
author_rank: Contributor
timestamp: July 16, 2020, 08:10:47 pm
content: The diodes are for protection. They limit the potential difference between the two inputs of the op-amp to a low value ( <=0.7V). They do not hold this potential to 0V. At small voltages ( <<0.6V) the diodes essentially don't conduct and so the small signals that the opamp operates on are still there.The opamp has an open loop gain at DC of about 100,000 and a maximum output of about 15V so it will saturate at about 0.15mV input (plus or minus an offset of up to 3mV). At that input voltage these diodes will be conducting picoamps or less.This will work fine if the op-amp is being used as a comparator though there might a slight turn off delay when the diodes are conducting. I think it is more likely that the system servos the input to zero.

author: magic
author_rank: Super Contributor
timestamp: July 16, 2020, 09:15:03 pm
content: Strictly speaking, I think it's just about preventing the inputs from exceeding the rails. This is a JFET opamp, it doesn't need differential input clamping.If the opamp is part of some larger loop, removal of option E has to remove the whole loop because simply bypassing the opamp and leaving the rest in place changes inverting gain into noninverting and negative feedback into positive.

author: jmelson
author_rank: Super Contributor
timestamp: July 16, 2020, 11:18:05 pm
content: The crazy thing about the options is that the circuit is non-inverting with the wire jumper, but inverting with the OPA602 in the circuit.Seems like that might really alter the function of the circuit.Jon

author: sahko123
author_rank: Frequent Contributor
timestamp: July 17, 2020, 06:19:44 am
content: It looks to me like a simple 0 cross detection circuit but I might be wrong

author: daddylonglegs
author_rank: Contributor
timestamp: July 17, 2020, 08:53:24 am
content: That's a good point about needing another inversion somewhere in the signal chain. Unless the system is only interested in the frequency of zero crossings or the frequency/ amplitude of an AC signal, not the polarity / phase.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 17, 2020, 02:00:35 pm
content: How does the circuit in option E inverts the signal? If I is the input signal and I_0 is the reference, then how does the op-amp invert the signal?In the product I'm working with, Option E is bypassed. All the components are there except for the op-amp. The signal doesn't even go through the 10k ohm resistor before the rest of the circuit.Before option E in my case, there is a current-to-voltage transimpedance amplifier. It takes a current signal and converts it into a voltage while inverting it. I believe that it is also referenced to I_0, the same as option E.

author: magic
author_rank: Super Contributor
timestamp: July 17, 2020, 07:52:44 pm
content: The signal kinda goes to the inverting input, right?So the output goes up as the signal goes down and vice versa. But if you bypass the opamp, suddenly that's no longer true.Chances are high that there was more to that option E than just this single opamp.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: July 17, 2020, 08:26:41 pm
content: Right, sorry. I forgot the formula went like this more or less:VOUT= AOPEN-LOOP* (VIN+-VIN-)Assuming that I is a current, passing it through the 10k resistor will create a voltage at the input, so that gets compared with the reference, H, and amplified.When option E is not present, this circuit is by passed, but there is a current-to-voltage transimpedance amplifier (connected externally) that converts the current I into a voltage. Option E does not have this current-to-voltage transimpedance amplifier, so it's assumed that it converts it here. However, that doesn't make much sense given that there's no feedback loop.

author: magic
author_rank: Super Contributor
timestamp: July 17, 2020, 08:30:48 pm
content: Current has nowhere to go through that resistor because it ends up facing a very high impedance JFET input and two diodes. The diodes don't conduct until you apply many millivolts across them, at which point the opamp's output is firmly at the opposite rail. It makes no sense, there has to be something more to it than this chip.