hybridfree/eevblog-posts huggingface dataset:
Op Amp Clipping - Page 1
Forum section: electronics / beginners

author: cswanson
author_rank: Newbie
timestamp: October 11, 2014, 05:51:38 pm
content: I am confused as to why I am receiving a clipped signal at Vtop 2.76 when I am supplying a voltage of +- 5v to the op amp CA3140E.The signal is a 1Khz sine wave Vpp 800 mV. The circuit inputs the signal to a CA3140E (http://www.jameco.com/Jameco/Products/ProdDS/12204.pdf) with an inverting configuration. Rf is 47K Ohm and Re is 10K Ohm for an A=4.7*signal. This output is sent to another CA3140E with Rf=Re=10K Ohm. However, I am applying a voltage from a V divider of R1=6.8K Ohm and R2=1K Ohm. I have the intention to raise the signal to vary above 0V for input to an ADC. Ideally the sine wave will vary between 0 and 5V.The purpose of this is to 1) Learn about Op Amps and 2) create a circuit to record pH values for a pH sensor.I am using this url as a guide, but I am having problems getting it to function which could be a limitation of my understanding or any other reason:https://www.sparkyswidgets.com/portfolio-item/ph-probe-interface/.What I am confused about is the output signal from the second CA3140E is clipping at 2.76V and I believe the op amp should have enough headroom since it has a supply of +- 5V.What am I overlooking?ps.: I am providing the +-5V supply to the op amps using a 7660C charge pump configured as a supplier of negative voltage.http://www.jameco.com/Jameco/Products/ProdDS/51174.pdf

author: Simon
author_rank: Unknown
timestamp: October 11, 2014, 05:55:31 pm
content: Check that the output can go that close to the supply rails. Some opamps won't go closer than 3V to the power rails

author: SeanB
author_rank: Super Contributor
timestamp: October 11, 2014, 06:09:24 pm
content: And the 3140 is one of them.It will go down to the negative rail with no load, but will only go to 3v below the positive rail. With a 5v supply that means the output will only swing from 0v to 2V max.

author: cswanson
author_rank: Newbie
timestamp: October 11, 2014, 06:12:19 pm
content: Thanks!Can you recommend a different op amp that may be able to deliver what I am looking for?

author: Simon
author_rank: Unknown
timestamp: October 11, 2014, 06:17:11 pm
content: I am told LM358 as i too had the same problem last weekend.

author: SeanB
author_rank: Super Contributor
timestamp: October 11, 2014, 06:20:13 pm
content: Look for rail to rail input and output opamps, probably those with a max supply of 6V are likely to be usable.

author: cswanson
author_rank: Newbie
timestamp: October 11, 2014, 06:23:38 pm
content: Fantastic! Thanks
Op Amp comparator problem - Page 1
Forum section: electronics / beginners

author: dfox416
author_rank: Contributor
timestamp: March 11, 2014, 02:45:06 am
content: Hi,I want to use an op amp comparator to convert a 1V p-p sine wave to a 0-5V square wave. I tried using a standard comparator circuit and I got no output. Then, I put a 10pF capacitor between the negative op amp supply rail and ground and it worked. I am using an LF353. I attached schematics of the two circuits. Does anybody know what is happening?

author: w2aew
author_rank: Super Contributor
timestamp: March 11, 2014, 03:16:56 am
content: Quote from: dfox416 on March 11, 2014, 02:45:06 amThere are a couple of problems here.1) Op amps aren't really optimized for use as a comparator (poor dynamic saturation characteristics, etc.)2) The LF353 is designed primarily for supplies greater than +/-6V (split), which means at least 12V from VCC to VEE. Operation at lower supplies will be sub-par. Proper operation with only 5V is probably not going to work well at all.3) The input common mode range and output swing are not compatible with what you are doing.4) If you're judging this circuit based on simulation - then the simulator is NOT predicting reality. Models used in simulation make the assumption that the device is being biased and used correctly. If you don't use the part correctly, the simulator won't tell you that - it will just lie to you. After all, it is nothing more than a set of equations designed to mimic *proper* operation of the circuit. If the circuit is not setup right, the equations, and results, will be meaningless.So, I'd say to start off using a comparator, bias one input with a DC voltage that is at the mid-point of your sine wave. You may need to add a DC component to your sinewave to balance it around your decision point. Or, bias both inputs, and AC couple your sinewave in...
op amp comparator with retroalimentation? - Page 1
Forum section: electronics / beginners

author: Andres Nino
author_rank: Newbie
timestamp: July 15, 2019, 01:44:39 am
content: Hello guysI would like to know how this op-amp distribution is called? I am trying to search this on google but I don't know how it works and how it is called in order to find more information and be able to analyze this circuit.you'll find the pic attachedThank you in advance for your helpwarm regardsAndres
Op-Amp Confusion - Page 1
Forum section: electronics / beginners

author: logictom
author_rank: Unknown
timestamp: July 24, 2011, 03:20:13 pm
content: Hi, I'm trying to build a circuit which amplifies a small pulse with a large dc offset. I understand the principle of what I need to do but I'm having trouble implementing it.I have found a circuit in an appnote which works in ltspice but I'm unclear on a few of the details.As I understand it, the signal is ac coupled by C4, R2 & R5 are a voltage divider to offset the signal above ground to ~0.6v.I'm unsure about R1, is this to minimise the loading of the voltage divider? Or is it working with C4 as a high pass filter?R4 & R3 are the standard gain setting resistors, C2 is linked to the gain but I'm unsure how or why.I've been reading through theTI single supply opamp circuit collectionwhich is coming in very handy but I'm having trouble relating this circuit to any of those mentioned.It's been some time since I learnt about opamps and I'm very rusty, am I on the right track or am I barking up the wrong tree?

author: ejeffrey
author_rank: Super Contributor
timestamp: July 24, 2011, 04:33:00 pm
content: Quote from: logictom on July 24, 2011, 03:20:13 pmYes.QuoteR1 does two things. It sets the input impedance high so that the circuit doesn't load the source, and it forms a high-pass filter with C4 for AC coupling.QuoteC2 essentially AC couples the gain network. At low frequency C2 becomes an open circuit and the op-amp becomes a unity gain follower. The reason it is there is to avoid amplifying the op-amps DC offset.

author: logictom
author_rank: Unknown
timestamp: July 24, 2011, 06:25:17 pm
content: Thanks for the quick reply.In the 'standard' non-inverting setup R3 & R4 are connected to Vcc/2, in this case we connect to ground, through C2, because at high frequencies C2 looks like a short? That's what is still throwing me, why in this case is it ground and not Vcc/2?

author: logictom
author_rank: Unknown
timestamp: July 24, 2011, 06:53:17 pm
content: Just been doing a little more poking around in ltspice, does this amplification work due to a phase delay caused by C2 charging and it's the difference between the phase that the opamp is amplifying? Or am I talking absolute rubbish?

author: ejeffrey
author_rank: Super Contributor
timestamp: July 24, 2011, 10:28:56 pm
content: Quote from: logictom on July 24, 2011, 06:25:17 pmSince it is capacitively coupled the DC voltage doesn't matter -- ground is just as good as Vcc/2. All that matters is that the capacitor goes to a low-impedance node with a constant voltage. What will happen is that the capacitor will charge up to the operating point set by the voltage divider bias. It has to do this because in the steady state C2 makes sure no current can flow through the feedback network, meaning all three terminals of the op-amp must be at the same potential.

author: logictom
author_rank: Unknown
timestamp: July 24, 2011, 10:52:25 pm
content: Thanks for clarifying that, the circuit makes a lot more sense now :)
op amp feedback resistor power rating - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: July 17, 2017, 06:31:25 pm
content: For anon-invertingop amp configuration, is it correct that the output current is RL|| RF? Let's say the output goes to the next stage amp, "infinite resistance", so then we can just consider RFby itself.So say the output is 2.5V DC, and RFis 20k, then the current is V/R = 125µA and power is V*I = 3125µW?

author: hermit
author_rank: Frequent Contributor
timestamp: July 17, 2017, 06:50:39 pm
content: Without doing the math (you seem to know the formulas)..... Opamps are generally voltage devices with minimal current so power ratings on feedback resistors aren't normally a concern. If you are calculating a low current this would be normal.

author: Benta
author_rank: Super Contributor
timestamp: July 17, 2017, 08:11:21 pm
content: For non-inverting, the current drawn by the feedback network is what runs through the feedback resistor in series with the ground resistor.If you redraw the circuit, so the feedback network is a voltage divider from output to ground, it's apparent.

author: LabSpokane
author_rank: Super Contributor
timestamp: July 17, 2017, 09:39:24 pm
content: A drawing will be of help. I'm assuming from your description that RL is between the output and ground, but hate to guess.

author: mikerj
author_rank: Super Contributor
timestamp: July 17, 2017, 09:47:44 pm
content: Quote from: electrolust on July 17, 2017, 06:31:25 pmYou are a factor of ten out:P=V^2/R2.5^2/20000 = 312.5µW

author: Zero999
author_rank: Super Contributor
timestamp: July 17, 2017, 09:53:19 pm
content: Generally there's no point in calculating the power dissipation in op-amp feedback resistors, because the power dissipation is usually far below the maximum rating of even tiny resistors. The only time I'd do it, is if the op-amp had a high voltage buffer, in the loop, although I'd try to keep the resistor values as high as possible, so power dissipation a non-issue.
OP AMP input and output impedance - Page 1
Forum section: electronics / beginners

author: joersam24
author_rank: Contributor
timestamp: June 03, 2012, 07:13:56 am
content: Hi to all,Can anyone explain in details why should OP AMP have very high input impedance and very low output impedance? Thanks...

author: amspire
author_rank: Super Contributor
timestamp: June 03, 2012, 08:09:39 am
content: Quote from: joersam24 on June 03, 2012, 07:13:56 amThe input of an opamp is designed to be high impedance. Either it is a very small geometry transistor running at very low collector current, or a FET/MOSFET input where the only current is leakage currents. So the inputs are very high impedance by design.Output impedance of a typical opamp is low but not very low. An opamp could easily have an output impedance of a few hundred ohms.Everything changes when you add a feedback circuit. The feedback circuit coupled with the opamp loop gain drives the output to the exact voltage that makes the two inputs the same voltage. With feedback, the resultant output impedance becomes very low - less then one ohm.The input impedance of the amplifier depends on the circuit. Non-inverting circuits will have an extremely high input impedance. Inverting amplifier designs have an input resistance equal to the input resistor. If the input resistor is 10K, then the final amplifier's input impedance will be 10K.Richard.

author: joersam24
author_rank: Contributor
timestamp: June 03, 2012, 08:32:10 am
content: Thanks amspire, Does it mean that high input impedance and low output impedance of an op-amp exists only due its physical construction? I mean it exists only unintentionally? Or does it intentionally designed to serve a certain purpose or applications?

author: joersam24
author_rank: Contributor
timestamp: June 03, 2012, 08:39:07 am
content: Quote from: joersam24 on June 03, 2012, 08:32:10 amI mean for solely a single op amp unit only without external components(feedback resistor or input resistor).

author: amspire
author_rank: Super Contributor
timestamp: June 03, 2012, 11:00:30 am
content: Quote from: joersam24 on June 03, 2012, 08:32:10 amAn opamp is based on a concept that is a very high gain amplifier, with a high input impedance differential pair stage, a single pole gain fall off characteristic with increasing frequency and a wide range output. The concept has been around since before ICs existed, and it still is one of the most successful analog amplifier concepts.There are other amplifiers, like a transconductance amplifier that is very different. An amplifier designed for video use may only need a 75 ohm input impedance. That would not be an opamp. To be an opamp, the amplifier has to match the opamp concept.Richard.

author: Mechatrommer
author_rank: Super Contributor
timestamp: June 03, 2012, 12:15:49 pm
content: Quotethe idea is so the input can suck little energy from source and the output can give high energy.
Op-Amp Input Offset - Page 1
Forum section: electronics / beginners

author: chrisabird
author_rank: Newbie
timestamp: January 08, 2017, 12:50:22 am
content: If someone could point me in the right direction to read more on this I'd be very grateful.I have a LM324 configured as a simple voltage follower, single rail, output connecting back to inverting, and an input to the non-inverting. The output it is fed into an analog input of a ATMega328.What i am noticing is that there is a -200mv discrepancy between the input and the output, i was expecting it follow withing the tolerance of the input offset (2mv).Have i totally misunderstood this, and am i looking in the wrong place for this problem?

author: Ian.M
author_rank: Super Contributor
timestamp: January 08, 2017, 01:00:10 am
content: 200mV seems excessive. Are you sure your +in is within the acceptable common mode range, which does *NOT* extend to the OPAMP's positive supply?

author: chrisabird
author_rank: Newbie
timestamp: January 08, 2017, 01:10:53 am
content: Thanks Ian.M.Yep, that was the problem, my input signal is just outside the common mode range. I forgot to check that when switching parts

author: danadak
author_rank: Super Contributor
timestamp: January 09, 2017, 12:08:11 pm
content: Useful amplifier calculator -http://masteringelectronicsdesign.com/summing-amplifier-calculator-java/Regards, Dana.

author: Brutte
author_rank: Frequent Contributor
timestamp: January 09, 2017, 12:31:16 pm
content: Why buffer? Does the source of the signal has high output impedance and ADC requires a low impedance source?Well, LM324 does not have a bias current compensation, just use similar impedances on both inverting and non-inverting nodes.BTW, this ADC is impossibly slow (15ksps?), unless your source has an impedance of megaohm range, just use a capacitor instead of a buffer. Guaranteed 0.0V offset.
Op Amp Input Protections - Page 1
Forum section: electronics / beginners

author: Kevin.D
author_rank: Frequent Contributor
timestamp: February 05, 2013, 07:35:36 pm
content: Just a quick one on op amp input protection .Does the input voltage limit's given in op amp data sheets apply toun-powered op amp .?In other words if I have an op amp who's input's can take up to 20V max above or below supply rail on it's inputs according to data sheet. Then when it is un-powered and there is a voltage above >20V being put on one of it's inputs ,do I have to protect it .?The thing is , I have this circuit which is designed to handle input's up to 35V with no problem when it is powered on.But there is a possibility the opamp may lose power itself and the inputs to it could still be present.

author: CarlG
author_rank: Regular Contributor
timestamp: February 05, 2013, 07:54:26 pm
content: Check Maximum Voltage Ratings.If it says 20V then it can take 20V.If it says Vdd+0.3V (or something similar) and Vee-0.3V, respectively, (which usually is the case), then you have to limit the current someway. And be aware that if you forward bias the substrate diode you supply your circuit that way, and may lead to unexpected behaviour.

author: Bored@Work
author_rank: Super Contributor
timestamp: February 05, 2013, 08:13:29 pm
content: Some opamps contain internal protection, others usually don't. If your's doesn't, then the typical protection is diodes to the rails and a current limiting series resistor in the input. I once learned that the resistor should limit the current to max 1 mA and things go really south at 10 mA. Later people kept telling me 5 mA is the limit to go for, and that's what I use, esp. because the resistor shouldn't get to large.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: February 05, 2013, 08:58:14 pm
content: Ok thnx .I re read data sheet ,it's input's can go 20V above or below supply rails without damage it says.(op 490)So I think it therefore must have some in built protection like what you said ,diodes tied to power rails internally with current limiting resistors . I think I will just add the extra diode to the supply rail then ,that will sort it . thanx
Op amp input voltage - Page 1
Forum section: electronics / beginners

author: miketan3904
author_rank: Contributor
timestamp: July 01, 2020, 03:49:49 pm
content: Hi. I would like to know if we can apply a voltage higher/ lower than the supply of the op amp to the input of the op amp. For example, if I am powering up the op-amp with a single supply, can i apply a -3V at its input?

author: TimFox
author_rank: Super Contributor
timestamp: July 01, 2020, 03:57:01 pm
content: The input common-mode range is one of the specifications on the op amp data sheet, and varies substantially from part to part.Most op amps will not tolerate a common-mode input voltage 3 V negative with respect to the negative power terminal.

author: David Hess
author_rank: Super Contributor
timestamp: July 02, 2020, 02:10:51 am
content: Common mode range is separate from the absolute maximum input range.Most operational amplifiers will not tolerate input voltages which are higher or lower than the supply voltages however there are some exceptions with parts specifically designed to support higher or lower or both input voltages.Also beware that not all operational amplifiers will tolerate a large difference in voltage between the inputs.

author: TimFox
author_rank: Super Contributor
timestamp: July 02, 2020, 02:45:34 am
content: Right. Common-mode range is for proper operation, but exceeding the absolute maximum input can irreversibly damage the part. Both are specified on the data sheet.
OP-AMP Input Voltages Slightly Above V+ Rail - Page 1
Forum section: electronics / beginners

author: mban401
author_rank: Newbie
timestamp: October 23, 2024, 08:06:15 pm
content: Hello!I'm looking to use an OP-amp where I tie the V+ rail to GND and the V- rail to approx -10V. I'm expecting between positive 0-0.1V on the inputs. Can OP-amps be run this way?I'm looking at the LM2904A from 3PEAK in particular that states an input range of "(+VS) + 0.3V" in the datasheet, but will this OP-amp work correctly with the V+ tied to ground and inputs 0.1V above V+?Thanks in advance!

author: mban401
author_rank: Newbie
timestamp: October 23, 2024, 08:26:06 pm
content: Never mind, I'm a dummy, it's specified in the datasheets as "Common-mode Input Voltage Range" and depends on whether it uses PNP, NPN, or MOSFETs for the differential amplifier circuit.

author: magic
author_rank: Super Contributor
timestamp: October 23, 2024, 08:28:36 pm
content: There are two separate specs:absolute maximum ratings - if you exceed those, the device may be damagedratings for proper operation - no magic smoke yet, but the device may misbehaveThe Vcc+0.3V is probably absolute maximum. The operating spec you are looking for is "common mode input range". When it comes to operation near the positive rail, people sometimes use and abuse TL072 for this purpose and the internal schematic looks like this should work (possibly with limited performance if taken too far). There are unfortunately not many alternatives, since negative rail operation is usually preferred. I seem to recall that some rail to rail input opamps explicitly advertise ability to work slightly (100mV or so) beyond the rails, but I can't name any example off the top of my head.LM2904 is not one of those chips, unless it's LM2904 in name only, and actually some completely different rail to rail opamp with a familiar name slapped on it to attract (and mislead) potential buyers. Original LM2904/LM358 is designed for operation near the negative rail and it's unable to compare voltages even 1V below the positive rail, let alone above it.

author: Zero999
author_rank: Super Contributor
timestamp: October 23, 2024, 09:40:35 pm
content: The newer TL072 is specified to work to +V + 0.1V.https://www.ti.com/lit/ds/symlink/tl072.pdfYou didn't say anything about the output voltage range though.

author: Peabody
author_rank: Super Contributor
timestamp: October 23, 2024, 09:42:17 pm
content: Take a look at the TLV2401.

author: PGPG
author_rank: Frequent Contributor
timestamp: October 23, 2024, 09:58:51 pm
content: Quote from: mban401 on October 23, 2024, 08:06:15 pmIn 3PEAK LM2904A datasheet I see: "Common mode Input Voltage Range" from (V-) to (V+)-1.5V.

author: Ian.M
author_rank: Super Contributor
timestamp: October 23, 2024, 10:08:46 pm
content: OPAMPs with a common mode input range extending way above V+ are a thing. LT/ADI's offerings can be found by searching for:over-the-top OPAMPAlso seehttps://www.edn.com/ltc-design-note-robust-high-voltage-over-the-top-op-amps-maintain-high-input-impedance-with-inputs-driven-apart-or-when-powered-down/
Op Amp integrator - limiting +Vsat to zero (ground) - Page 1
Forum section: electronics / beginners

author: sspj9
author_rank: Contributor
timestamp: December 21, 2018, 06:54:23 pm
content: Part of the Dual Slope ADC consists of an integrator which integrates the input voltage according to Vin/-Vref .I am using TL071 connected as an inverting integrator.Quick summary of operation:The output of the integrator charges negatively up to a certain count value when Vin is connected.When the count value is reached, the controller switches to -Vref.This causes the integrator to discharge and hence increasing output voltage (note due to inverting output).NOW... Instead of the integrator to continue to integrate towards a positive output voltage while -Vref is connected, I want it to stop at zero voltage so that when Vin is connected as input, the integrator will start integrating negatively instantly.Is this possible by connecting the op-amp supply terminals +Vcc to ground while -Vcc to -ve supply?I tried this but observed clipping in the negative rail too, is it a bad approach?Any other possible solutions?

author: David Hess
author_rank: Super Contributor
timestamp: December 21, 2018, 10:04:58 pm
content: An operational integrator has overshoot due to its limited bandwidth so even if the input is disconnected when zero is reached, time will still be needed to reset.You might want to study run-up and run-down integrators which continuously integrate the input.

author: sspj9
author_rank: Contributor
timestamp: December 21, 2018, 10:27:42 pm
content: Quote from: David Hess on December 21, 2018, 10:04:58 pmI was thinking, would it be possible to run an enable switch (like a mosfet) across the integrator capacitor so that when the integrator reaches zero, the output of the MCU sends a signal to the switch which hence closes it and shorts out the capacitor thus keeping the output voltage fixed to zero till -Vin is connected again? (ie till when MCU is ready for the next reading)

author: David Hess
author_rank: Super Contributor
timestamp: December 22, 2018, 03:13:19 am
content: Shorting out the integration capacitor will certainly work but is only done in lower performance applications.Another way I have seen it done is to have the integrator temporarily enclosed inside the control loop of another integrator which adjusts the input current to force the output to zero.
OP-AMP Maximum ratings - Page 1
Forum section: electronics / beginners

author: kaykan
author_rank: Newbie
timestamp: March 26, 2014, 09:20:06 am
content: Hello,I have a question about OP-AMP maximum ratings,Pick an LM358, that have a maximum voltage of 32V on (VCC - VEE) and 32V on (V+ - V-), now supposing this configuration:Now if I apply 20V as input I know that the result will be 3.3V, but the OP-AMP will be damaged?Is a good Idea use an op-amp to protect (for example) an ADC with a configuration like this?If I apply 40V (over 32V) the op-amp will be damaged?thanks for help!.c

author: han
author_rank: Frequent Contributor
timestamp: March 26, 2014, 09:29:55 am
content: if there is a internal clamp diode, i'm sure it's bye bye for the opamp..

author: poorchava
author_rank: Super Contributor
timestamp: March 26, 2014, 11:20:54 am
content: Refer to the datasheet to get input voltage range. It will usually be expressed as VDD+x if there are clamping diodes or as an absolute value of there are none.What you are trying to do is not good anyway. I mean you shouldn't drive an ADC with 0...32V signal. Divide it down with a high-impedance divider (several meg, so that source resistance does not affect the measurement too much), then buffer with an opamp that has low bias current and high input impedance (i'd try MCP6051) and use that to drive the ADC.

author: kaykan
author_rank: Newbie
timestamp: March 28, 2014, 08:56:46 am
content: Thank you for support!in the LM385 the maximum ratings is 0V to 32V for VCC, and 0V to 32V for V+/V-, so I suppose that if I set VCC to 3.3V and V+ to 30V the op-amp will not damaged, otherwise I expected a limit for V+/V- to (for example) VCC + 0,5V, right?.Really thanks for help!
OP AMP mic preamp question - Page 1
Forum section: electronics / beginners

author: Proxy638
author_rank: Newbie
timestamp: August 07, 2014, 11:33:17 am
content: HiI'm trying to use some electret mics for the purposes of taking reasonably precise measurementsfor reflection coefficient calculations in an impedance tube.currently we're getting weird errors in my tube and im slowly trying to isolate down all the potential sources of errors.This is a picture of the mic op amp which is a sparkfun LM358 breakout board ( don't know too much about electronics and Im trying to slowly learn)Basically a 770hz sine wave was passed through the amplifier and a fast fourier transform used to breakdown the signal into its frequency compositions?( am I correct in my understanding?) its definetly not the signal generator as it gives one single peak but in the case of this breakout board it's giving multiple peaks so I'm assuming there is some distortion and noise (this is the question, am I correct in this assumption?) is it possible to tell if its more to do with noise or distortion and how would I test?so what would i need to do to get it to as close as a single peak with as little noise or distortion as possible? im interested in the frequency range of 200-3000

author: mij59
author_rank: Frequent Contributor
timestamp: August 07, 2014, 11:46:38 am
content: Hi,What's the output level of the signal generator ?What's the gain of the amplifier ?For low noise and distortion you'll need special microphone amplifier.

author: XFDDesign
author_rank: Frequent Contributor
timestamp: August 07, 2014, 01:14:22 pm
content: The schematic would be useful.From that image you shared, you're not outputting a sinewave, but a square wave. If you have a truly pure sine going into the circuit, then you're likely overdriving the opamp, which is clipping/limiting. With square waves, you'll have strong harmonics on the 3rd, 5th, 7th, ... etc multiples of the fundamental, which is what you're showing on the scope. How much gain are you using?
OP Amp min voltage and current? - Page 1
Forum section: electronics / beginners

author: Pulseit
author_rank: Contributor
timestamp: May 31, 2024, 11:07:07 am
content: At the risk of asking a retarded question.From the Op Amps details,Supply Voltage - Max: 40 VSupply Voltage - Min: 4.5 VOperating Supply Current: 10 mAOutput Current per Channel: 375 mAIs the current output relative to the voltage at which its powered at?I'm looking to power the gate of a logic MOSFET with 5V input as I'd like it to work over the full range of the 0-50v power supply, and it's much easier to just buck the power than buck-boost it.0-50V power supply -> 4.5-50V to 5V regulator -> Op Amp -> MOSFET gate

author: David Hess
author_rank: Super Contributor
timestamp: May 31, 2024, 03:51:09 pm
content: Quote from: Pulseit on May 31, 2024, 11:07:07 amGenerally no, the output range is limited by the supply voltages, but the output current should stay the same at low supply voltages.

author: Terry Bites
author_rank: Super Contributor
timestamp: June 03, 2024, 01:46:22 pm
content: It doesnt mean that you can have all the volts at the max output current all the time.Take the device's maximum power dissipation into account. Po=Io*Vo.It might just emit the magic smoke with all channels on full tilt.
Op Amp Monostable multi vibrator (mild panic) - Page 1
Forum section: electronics / beginners

author: gildasd
author_rank: Frequent Contributor
timestamp: January 05, 2016, 05:45:02 pm
content: This is one of the subjects we might have in a test tomorrow, I disagree with my whole class on the results and method...Remark: we get extra points if we use realistic parts values (as if we were to breadboard the thing)This is how it was drawn on the black board and next to it, redrawn in a more "classical" manner.The diode D clamps Voltage Vd at -0.8 VI need to send a triggering pulse that is bigger to start the circuit.The Pulse with "T" is calculated as T=0.693*R*CT is given as 2 seconds, I choose 47µF thus R=T/(0.963*C)= 2/(0.693*47*10^-6)=61404 rounded to 62000 +-1%Voltage of capacitor (voltage where Vout drops back to -Vcc):/li]Vcap= +Vcc*R1/(R1+R2)=15*10000/(10000+10000)=7.5VAll of this is then regrouped in the following graph;They get C=200µF and R=10000... That just seems wrong.

author: MatthewEveritt
author_rank: Unknown
timestamp: January 05, 2016, 06:36:28 pm
content: I have to agree with your answer.Assuming, as you seem to have done, the R1 = R2 you need to charge to 1/2, which takes ln(2) time constants (0.693).C=200µF and R=10000 gives a time constant of 2 seconds, so doesn't work with R1 = R2, but does work with a different ratio, so I can't disagree with their answer either.

author: gildasd
author_rank: Frequent Contributor
timestamp: January 05, 2016, 06:42:20 pm
content: Quote from: MatthewEveritt on January 05, 2016, 06:36:28 pmThat's what I was thinking, the professor did the demonstration on the astable, why should he do it again and T=RC does not make sense.C=200µF and R=10000 can work, but it needs more steps to get there.Thanks!

author: MatthewEveritt
author_rank: Unknown
timestamp: January 05, 2016, 06:47:17 pm
content: Sorry, bit of a brain fart - their answer does make sense if the ratio of R1 and R2 is set to compensate. I've altered my post to reflect this. R2 = (e-1) * R1 should do the trick.

author: gildasd
author_rank: Frequent Contributor
timestamp: January 05, 2016, 06:48:52 pm
content: Quote from: MatthewEveritt on January 05, 2016, 06:47:17 pmNo problem,but it does seem to be the hard way of doing it...

author: gildasd
author_rank: Frequent Contributor
timestamp: February 02, 2016, 11:18:39 pm
content: Crushed the exam... Thanks to all!
Op amp offset measurement/digital compensation - Page 1
Forum section: electronics / beginners

author: daqq
author_rank: Super Contributor
timestamp: December 03, 2013, 11:10:01 am
content: Hi guys,Assuming you have a simple non inverting amplifier and a source with a large enough impedance (10kOhm+) and low voltage output (<1mV, such as a infra thermocouple) that doesn't mind short circuits, could you measure the offset voltage by shorting the source periodically with a MOSFET transistor? In theory you'd only get the offset voltage amplified on the output, which you could measure with your ADC and then simply subtract it from the normal measurement. If you did this often enough, you'd compensate against both the actual offset voltage and its drift.In theory and simulation it works, how about in practice? Would this be a bad idea?Thanks,David

author: Harvs
author_rank: Super Contributor
timestamp: December 04, 2013, 12:58:39 am
content: This is sort of commonly done in DMMs. Often at some point in the analog signal chain, an analog mux will switch from the signal source to GND, then the perform a conversion cycle and subtract it from the result.For auto-zeroing bench meters like the HP 3456A, this is done every conversion cycle by default, or you can switch it off and double your sample rate.As to whether you should short your sensor out, about the only issue I can think of is the leakage current through the MOSFET. You'd need to check the specs of the FET's off-state leakage and see how much error this would introduce to your measurement. Be aware this is unlikely to be linear.
Op-Amp OPA604 max. input voltage - Page 1
Forum section: electronics / beginners

author: RonLat
author_rank: Newbie
timestamp: January 29, 2025, 07:05:24 am
content: In the datasheet of the OPA604 underAbsolute Maximum Ratingsthere is Input Voltage (V-) -1V and (V+) +1V. However on the next page underInput Voltage Rangethere is Common Mode Input Range +/- 13V.1. Why there is a difference between max. input voltage +/- 1V and typical +/- 13V regarding common mode input range?2. What voltage can I finally apply to the inverting and non-inverting pins of the OPA604 without destroying the IC?Link to the datasheethttps://www.ti.com/product/de-de/OPA604

author: magic
author_rank: Super Contributor
timestamp: January 29, 2025, 07:31:38 am
content: 1. That's the range for correct operation, including meeting the guaranteed input offset voltage.2. No more than 1V beyond the supplies, apparently, although I find it surprisingly wide - many ICs specify 0.3~0.5V due to protection diodes.

author: ledtester
author_rank: Super Contributor
timestamp: January 29, 2025, 08:26:35 am
content: The +/- 13V for the common-mode input range is based on the assumption that the supply voltage is +/- 15V.The OPA604 can operate on a range of supply voltages from +/- 4.5V up to +/- 24V.Is it safe to assume that the common-mode input range is 2V from the rails, e.g. V- + 2V to V+ - 2V when using something other than a +/- 15V supply?

author: TimFox
author_rank: Super Contributor
timestamp: January 29, 2025, 11:32:16 am
content: Absolute maximum voltage rating is to avoid damage to the part.Input voltage range is for proper operation of the part.In both cases, carefully understand the circumstances for the rating, especially operating power supply voltages, as discussed by ledtester.
OP-amp output filtering need? - Page 1
Forum section: electronics / beginners

author: veeti
author_rank: Contributor
timestamp: December 19, 2020, 12:34:43 pm
content: HiI'm designing 10A constant current source for milliohm meter.I'm wondering would it need filtering for output of opamp section?Its really stable at simulation without any filters, but in reality?

author: palpurul
author_rank: Regular Contributor
timestamp: December 19, 2020, 02:27:04 pm
content: Maybe you should run a noise simulation to see if the low pass filter at the output of the opamp makes a difference.Isn't 10 amps is a little too much for measuring milliohms? Even 1 amp produces 1mV across a 1 miliohm resistor which is not terribly difficult to measure.

author: Zero999
author_rank: Super Contributor
timestamp: December 19, 2020, 03:51:41 pm
content: There is no need for filtering on the output of the op-amp.I think the original poster is confusing filtering with frequency compensation, which is different to filtering. The answer is yes, it might be required, but the Darlington pair doesn't add much delay, so it should be fairly stable, unless a fast op-amp is used.
Op-amp output voltage  - Page 1
Forum section: electronics / beginners

author: crazyjim30
author_rank: Contributor
timestamp: March 13, 2016, 12:08:37 am
content: I came across this question. Our resources suck so badly that I can't even guess at the proper formula. Nothing in my notes, our slides, or the crappy ebook has been any help at all.  An Op-amp has an open-loop gain of 90,000. Vsat = +/- 13V. A differential voltage of 0.1 V p-p is applied between the inputs. The output voltage is what?Sent from my SM-G900V using Tapatalk

author: retrolefty
author_rank: Super Contributor
timestamp: March 13, 2016, 12:15:53 am
content: Quote from: crazyjim30 on March 13, 2016, 12:08:37 amIf differential input voltage is DC then .1 X 90,000 = 9,000 volts but Vsat limits max exertion to Vsat so = 13V

author: crazyjim30
author_rank: Contributor
timestamp: March 13, 2016, 02:03:13 am
content: Ah ha! Now it makes sense. Thanks.Sent from my SM-G900V using Tapatalk

author: danadak
author_rank: Super Contributor
timestamp: March 13, 2016, 11:35:58 am
content: One normally does not use an OpAmp open loop, because its Aolvaries dramatically one part to another, very temp sensitive, andnoise considerations when dealing with very high G circuits.The beauty of all that Aol is using it in a - feedback loop to setwhat is termed closed loop gain. There is a Rome church sizedlibrary worth of research and application notes written on OpAmps,just google OpAmps or Operational Amplifiers. Go to TI, Analog Devices,Linear Technology websites, reams of info.Simulate one in Mathcad or alike and look at what happens withvarious stimuli and feedback factors. Lots of fun. OpAmps are likeResistors and Capacitors in usefulness, its their G and differentialmode, that create so many usefull circuits.Regards, Dana.
Op-amp part replacement comparison - Page 1
Forum section: electronics / beginners

author: Freshman
author_rank: Regular Contributor
timestamp: September 26, 2024, 05:59:06 am
content: I have a question.Can I use this part (https://ww1.microchip.com/downloads/aemDocuments/documents/MSLD/ProductDocuments/DataSheets/MIC6270-IttyBitty-Comparator-DS20006294A.pdf) in place of this part (https://www.ti.com/lit/ds/symlink/lmc7101.pdf?ts=1727329844036&ref_url=https%253A%252F%252Fwww.google.com%252F).The application is for ZCD where the input to the op-amp will be coming from an opto-isolator and given to the non-inverting terminal.U21 in the attached image.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 26, 2024, 06:28:53 am
content: The circuit has no high demands on the comparator.The MIC6270 has an open collector output and would thus need an extra resistor to Vcc at the putput side (may be forther down the circuit).The circuit has quite some filtering action and will thus have a delay and may not trigger at zero crossing, unless there is some oopsite phase shift at the drive side (e.g. capacitive dropper). Ideally one would like additional hysteresis (feedback to the + input) to avoid douple triggering from noise.

author: Freshman
author_rank: Regular Contributor
timestamp: September 26, 2024, 06:59:18 am
content: Thank you for your answer @Kleinstein.Would MIC6270 be sufficient for the application? If so, just adding a pull-up (any suggestion on the value?) at the output would be fine?Could you share some schematic drawing circuit that you suggest for this circuit?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 26, 2024, 07:52:36 am
content: I got confused with the use as ZCD (zero corssing detector) and thought of a comparator. The circuit as shown is using the OP-amp as a follower, not a comparator. For this use a compartor would not be the right choice. One would want a real OP-amp, something like an MCP6001 / MCP6006 / LMV321 or any low voltage (usually CMOS) slow OP-amp. It depends on the later use of the signal if one needs the buffer at all.A ZCD could use a comparator too, but this would be a different circuit and it dempends on what is done with the signal.

author: MathWizard
author_rank: Super Contributor
timestamp: September 26, 2024, 09:04:26 pm
content: Whats the input to the opto from , what sort of frequency ? Is it more for just logic high/low, or is any of the signal meant to be in the linear region for the transistor ?
OP Amp pinout - Page 1
Forum section: electronics / beginners

author: Leuven
author_rank: Regular Contributor
timestamp: June 15, 2015, 02:12:35 pm
content: How do you physically identify pin 1 on a dual OP Amp if there are no markings on it? Is it something to do with one of the edges being chamfered, and that would point out to pin 1/2/3/4 on that side?

author: mikerj
author_rank: Super Contributor
timestamp: June 15, 2015, 02:14:55 pm
content: What package style are you talking about?

author: Leuven
author_rank: Regular Contributor
timestamp: June 15, 2015, 02:16:10 pm
content: TS912AIDT (micropackage)http://www.st.com/web/en/resource/technical/document/datasheet/CD00000501.pdf

author: mikerj
author_rank: Super Contributor
timestamp: June 15, 2015, 02:29:45 pm
content: The drawing for the SO-8 package in that datasheet (page 21) shows the position of pin 1 with respect to the chamfered edge.Looking at the top of the package with the chamfered edge to the left, pin one is at the top left hand side.

author: Leuven
author_rank: Regular Contributor
timestamp: June 15, 2015, 02:37:19 pm
content: Fantastic. Mounted it the other way and already supplied 10V to it. Now it's out of the circuit and measures MegaOhm between ground and positive, and also MegaOhm between inputs and outputs. So this one is fried, right?

author: mikerj
author_rank: Super Contributor
timestamp: June 15, 2015, 03:52:01 pm
content: It's very probably dead, but unless there's an obvious short you can't really tell just by measuring resistance between pins.
OP-AMP Power Supply Query - Page 1
Forum section: electronics / beginners

author: HBates
author_rank: Newbie
timestamp: January 30, 2020, 01:59:39 am
content: Hey All,I have a question regarding the use of a single supply to an operational amplifier. My device is controlled by a Teensy 3.6 and I am planning on using a boost circuit (Figure 1) to boost the 5V pin from the Teensy to 16V and then regulate that voltage using a 12V linear regulator to ensure the supply voltage is as clean as possible.The OP-AMP is setup as a trans-impedance amplifier for my photo-diode. As the Teensy 3.6 can only receive 3.3V MAX on the analog pins I am using a voltage divider on the output to transform the MAX 12V output to 3.3V (Figure 2). I don't want to use the 5V pin directly as I would like to get a higher measuring range from the photo-diode without saturating the detector (hence the 12V reg).What I am wondering is whether or not what I am doing is good practice/correct or if there is a better way of getting a higher amount of measuring range from the Photodiode.Thank you

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: January 30, 2020, 02:51:52 am
content: you could use a log amp like the LOG114 with the photodiode.http://www.ti.com/lit/ds/symlink/log114.pdfThis would boost the dynamic range by a large amount.And it can work at 5V.

author: David Hess
author_rank: Super Contributor
timestamp: January 31, 2020, 08:35:56 pm
content: Quote from: HBates on January 30, 2020, 01:59:39 amThere may be reasons to use a higher supply voltage but extending the measurement range is not one of them if you are just attenuating the output. Your circuit could run just as well on 5 volts, or even lower, with a suitable operational amplifier.EEEnthusiast's suggestion to use a log amplifier to extend the dynamic range is a good one. Another way to extend the measurement range is to directly digitize the photodiode output with a wide dynamic range current-to-frequency converter.
OP AMP power supply using Thailand modules   +- 12 vdc   from  +19 vdc  for $28  - Page 1
Forum section: electronics / beginners

author: ez24
author_rank: Super Contributor
timestamp: July 29, 2015, 07:34:35 am
content: I am looking to put together an op amp power supply from a 19 vdc laptop power supply. I want to learn op amps so I do not know my current requirements but I know I need a +- supply. The neg rail that is causing me problems.Any opinions on this dc convertor from Thailand, ($28) and +- 12 vdc and 1 amp:http://www.ebay.com/itm/MINI-PWR-DUAL-12-Step-Down-Voltage-Regulator-Power-Supply-Output-12Vdc-1A-/251352773668?pt=LH_DefaultDomain_0&hash=item3a85caf424Here is a one chip device for $32 and according to the Chinese datasheet has +- 15 vdc and 200 ma:http://www.aliexpress.com/item/DC-DC-Isolated-Power-Module-24V-dc-buck-converter-turn-negative-15V6W-Manufacturers-selling-special/1746272285.htmlThis is rated at 66 ma ($26):http://www.ebay.com/itm/No-88M0755-Xp-Power-Ih1215S-Dc-Dc-Conv-Iso-Pol-2-O-P-2W-15V-15V-/271634473259?pt=LH_DefaultDomain_0&hash=item3f3ead212bI am leaning towards the Thailand one IF +- 12 vdc is ok for op amp experimentingTIA

author: Zero999
author_rank: Super Contributor
timestamp: July 29, 2015, 08:04:48 am
content: Or you could just use a rail splitter, also known as a virtual earth/ground, to give you +/-9.5V.
OP-AMP question - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: November 28, 2013, 01:06:02 pm
content: In a motor controller circuit i found this op-amp and i have some doubts on how it works.The IN of the op-amp is from a phototransistor used with an infrared led to count the number of holes on an encoder disk.From i what i have understand the op-amp takes the non digital phototransitor output and converts into a square wave.I'm only at first year of university and i have only see the ideal op-amp, so V+ = V- and the current that enter the two input is 0A.Using this two things and the fact that the two resistor in the output forms a voltage dieider i know that V- = V+ = Vout * 10K (10K + 100 K) so Vout = (more or less) 10 Vin.What is the purpose of the 1k on the + terminal? it looks like a pull-down resistor.Also why there is a 10 k in series with the output? To limit the current that goes in the uC?Thank you all for the help.

author: Andy Watson
author_rank: Super Contributor
timestamp: November 28, 2013, 04:43:03 pm
content: Quote from: exapod on November 28, 2013, 01:06:02 pm11 Vin, would be closer.QuoteIt probably is. You would need to know more about the opto-device - does the other end of the detector return to the positive supply?QuoteYes, probably. There could be several reasons for including the resistor, but limiting the current would be most likely. Does the op-amp operate from the same power supply as the µC? Don't overlook the possibility that the designer might have blindly copied the circuit from another piece of working kit - the resistor might not serve any function in this design!

author: Unknown
author_rank: Unknown
timestamp: November 28, 2013, 08:17:48 pm
content: The uC has the same power supply.The opto-device ishttp://www.ottawarobotics.org/datasheets/CNB1302.pdfand it is powered from the same source of the uC and op-amp (3.3v).

author: jeroen74
author_rank: Frequent Contributor
timestamp: November 28, 2013, 08:58:26 pm
content: The collector is likely connected to 3v3 and the emitter to the pull down resistor and the opamp input.A current will flow when light hits the phototransistor, and the current causes a voltage drop over the resistor which is amplified by the opamp.With a If (LED current) of 12mA about 600uA can flow, so the voltage over the resistor is 600mV; instead of the opamp a 10K load resistor probably would have worked too. A Schmitt-trigger would have been better.edit:I just studied the datasheet a little closer and I see the current transfer ratio can vary wildly, from 90uA to 880uA for a If of 10mA. So after all, amplifying it 11 times is not that bad an idea after all(I still would use an Schmitt-trigger)

author: Unknown
author_rank: Unknown
timestamp: November 29, 2013, 02:36:08 pm
content: Thanks for the answer.If i use thishttps://www.sparkfun.com/datasheets/Robotics/QR_QRE1113.GR.pdfinstead of the one used in the schematic can i use the same op-amp circuit? Is better to use a Schmitt-trigger? Can you point me one (Schmitt-trigger) that will work with this sensor?