About opamp.

Thread Starter

q12x

Joined Sep 25, 2015
266
This is a completely parallel issue i have from my original project with the ldr led driven stuff. Again, a reminder, i am not an electronist, but im a hobbyist, im an artist, so you'll have to forgive my way of talking and explaining. Im doing my best with the limited knowledge i have about some things, but im not Very limited.
In my previous post, mister @Audioguru again suggested me to use an opamp and I knew about its existence but not at all familiar with it. So here is my way of understanding it. It became my quest from that moment. Again, a parallel quest with my main objective.
I did my best to enunciate my problem here in this artpage.
If not that well understandable, do point it out.
Thank you.
dBenzin100_ziue_06 copy 1.jpg
 
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hrs

Joined Jun 13, 2014
253
The opamp will do whatever it can to make V- equal to V+. For this it will use the opamp output. At a supply voltage Vcc SPDT1 voltage of 5V the voltage at V+ will be Vcc V_SPDT1 * (10k/(10k+10k) = 2.5V. The opamp will make V- 2.5V too by making the output 2.5V. 2.5V is enough to turn on LED2.

When Vcc the SPDT1 voltage = 10mV V+ will be 5mV and so will be the output. 5mV is not enough to turn on LED2.

It' s easier to discus schematics when all components have designators, like R1, R2 etc.
 
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LvW

Joined Jun 13, 2013
1,086
The opamp will do whatever it can to make V- equal to V+. For this it will use the opamp output.
....under the precondion that we have a negative feedback loop which will bring/keep the opamp output within the linear operating range.
 

dl324

Joined Mar 30, 2015
11,550
If not that well understandable, do point it out.
You don't give us enough information; specifically the opamp part number.

With a voltage follower, the opamp will try to maintain the output voltage at whatever voltage is at the non-inverting input. If you look at the voltages for your case 2, you'll see that this is not the case. That's probably because the voltage at the non-inverting input is violating the common mode input range specification and the output has experienced "inversion" (this happens with some opamps when the input voltage range is out of the allowable range).

If you used an opamp that allowed a 5mV input (like LM358), the LED wouldn't have turned on for either switch setting because 5mV wouldn't be enough to turn on the transistor or LED. But, when you're using such low voltages, you need to be pay attention to input offset voltage specs.

The opamp isn't functioning correctly in cases 2, 4, 5, and 6.

To turn on the LED with a transistor, the output of the opamp can be as low as 0.7V (depending on base resistor value). When driving the LED with a series resistor, the opamp output needs to be above the forward voltage of the LED (it will light dimly before the voltage drops equals Vf because that voltage is specified at a specific LED current).
clipimage.jpg
clipimage.jpg

Older LEDs had lower forward voltages.
 

Audioguru again

Joined Oct 21, 2019
1,926
In your other post, I said to use an opamp's input to reduce the current on the LM3914 pin 7 so that the LEDs will be dimmed.
Then the output of the opamp can drive the 12k resistor ladder in the LM3914 circuit.
 

Thread Starter

q12x

Joined Sep 25, 2015
266
You don't give us enough information; specifically the opamp part number.
I did give a hint though, 1599414397584.png but not very specific, i admit. And i say there, that this is a simulation only. I did choose ua741 inside the simulator from the only 2 opamps it have disponible: LM741 and LM324.

the voltage at the non-inverting input is violating the common mode input range specification and the output has experienced "inversion" (this happens with some opamps when the input voltage range is out of the allowable range).
Your point is very interesting ! The simulated opamp im using is marked as "real opamp". It also have "ideal opamp" that i didnt used yet. Probably some limitations that are more close to reality for the real opamp, and that was my thought initially. I did a test right now, by changing the type of IC to LM324 and the led is stable at 10mv (5mV on V+). So... type of IC, matters ! Lesson learned.
And about the led current and Vf for each type is true, they are all diferent, thats why i play in this lower margin, to test opamp performance for an easy load as a led. And i understand it have a limited output as low as 50mV,but i cant find it in datasheet, I remember it from a opamp tutorial.
///
- In my mind, it should output good enough current to drive a led, even when is having the slightest current touch in input. So i was testing for that, because in many tutorials says that input is super sensitive (high impedance). And i find it VERY cool and i wanted to see it. But it bugs me why didnt output push the necesary current for the led, why not amplifying as it should. I really do it by looking and not calculating, because i must "feel" it and most of all, understand its basic operations.
For the unlit led2 in case2, I observed there is some voltage drop on the led but not high enough, of 1.44V @ 280uA. It needs 1.7V to fully open and 10mA. Indeed, if i change some parameters, like: 3.82V (1.91V on V+), lower the resistor to 50ohm for led2, the voltage drop on led2 = fix 1.7V, it's open limit with a current of 4.2mA (not very bright but open). See my new modified circuit here:1599419400361.png
So, help me understand this... Why is limiting the current so much? I mean, i find a solution already, by adding a BJT transistor, that solved the issue fine, but why not giving enough current directly trough output? This is the part i want to understand better. In a way, i understand it, but my way, thats why i open this Case, because i want to see your way of seeing it. Its a simple question, it should work as i imagined, but it didnt. Bummer.
Again, thank you so much to all here, for your involvement.
 

hrs

Joined Jun 13, 2014
253
There is 1.9V on the output of the opamp. The LED drops 1.7V. So there's 0.2V over the 50 ohm resistor. I = V/R = 0.2/50 = 4 mA.
 

dl324

Joined Mar 30, 2015
11,550
I did choose ua741 inside the simulator from the only 2 opamps it have disponible: LM741 and LM324.
Someone will be along soon and say how crappy the 741 is. It's a fine opamp; the best thing since sliced bread when it was introduced, but it's not for newbies trying to use a 5V supply. It was designed to use the more typical +/-15V supplies typical of that time.
Why is limiting the current so much?
If we assume that you're using LM324, the output voltage is only guaranteed to be 2V below the positive supply rail. With a 1.9V output, 1.7V Vf, and a 50 ohm resistor, the current would be about 4mA.
\( I = \frac{V}{R} = \frac{1.9V-1.7V}{50\Omega} = 4mA\)

Have no idea why the simulator is giving 4.129mA.
 

Thread Starter

q12x

Joined Sep 25, 2015
266
ill answer both @hrs and @dl324
Thank you both for making the math for me! I really like it when i see it, because ill get more used to doing it. I dont do it that often as you do. Im a bit more...detective than mathematician in my electronic interrogations. I know i should pay more attention to that too, but im more interested in the practical functionality of this device. My curiosity is about It, what can he do.
As an experiment, If i take out 50R, I can lower a bit V+ to 3.4V to obtain that limit of 1.7V for led to be open : 1599422257508.png But that 50R is good to calculate the current, as you already did. That 50R is also limiting the current flow directly into led, but allowing some "overflow" into V-. From what i observed. That's why i keep it. It helps to keep the current balanced in both directions, to the led, and to the V-.
///
- My interest here is to have as low voltage as possible, or as close to 0V as possible, on V+, to test uA741 performance, functionality, and it's sensibility. Actually that is my main reason for this experiment, to understand it's sensibility. - I am wrong? To assume its TOO sensitive? From what i understand so far, it can receive on input +-15V (between +15 to -15 V Range ) so a 0V is valid as an input voltage. So it's 10mV or 1mV or pico or nano or femta Volts. Also -10mV is also true as an input voltage as well. But is a bit tricky to get it without an inverted battery, and from the same source. Yes in the simulator is easy, but im keeping an eye in reality what i can do, too.
1599422635342.png
My reformulated question: the opamp input can receive super low voltages and translate them into it's output, high enough to drive a minimum load as a led, optocuplor or even those super tiny motors big as a pinky finger nail? Or the ONLY way, is to link the output to base of a transistor? This is what i want to get clear.
Or... another way put, why the output is supplying so low current that can not drive the led, having the V+ at 5mV ?
Again, super many thanks for your help. You really do help me a lot. Best guys in the world !
 
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Thread Starter

q12x

Joined Sep 25, 2015
266
I might have an (intuitive) answer. Only in this (Voltage folower) configuration is valid, because if Vout is not conected to V-, then the opamp will not try to equalize the 2 inputs.
Because the rule V+= V- ; Vout will try to equalize V+ input (5mV) by sending as Low current and voltage as it can, to V-. So thats why i get no practical output current for the led to be driven.
I made another test, to see if V- will be = to V+, by taking out any kind of load from Vout. Now, Vout is used only for feedback.
Why are not equal? What the hell? It should be 5mV on V- !
1599425525609.png
 
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Audioguru again

Joined Oct 21, 2019
1,926
It looks like your opamp that has no part number has an output that cannot go lower than +1.5V in that circuit powered from +5V.
The datasheet for a 741 opamp shows that some of them have inputs and outputs that cannot go within 3V of a supply voltage. But a 741 opamp does not work from a 5V supply.

The datasheet for an LM324 quad or LM358 dual opamp shows that its inputs can go from 0V in your circuit to 1.5V less than the +5V supply and an output that can go from almost 0V in your circuit up to 1.2V to 2V less than the +5V supply.

You show 10k input resistors but an opamp has a very sensitive low current input, try 1M resistors instead.
Your opamp is missing the 2 resistors that would give it some voltage gain.
 

dl324

Joined Mar 30, 2015
11,550
My interest here is to have as low voltage as possible, or as close to 0V as possible, on V+, to test uA741 performance
1599440955359.png
1599440983556.png
uA741 can't be operated from a single 5V supply. The typical input voltage range with a 5V supply is 2-3V. Worst case, it won't operate from a 5V supply.

To have 0V be a valid input voltage, you need the negative supply to be at least -3V.

If you look at the LM324 schematic, you can see that the inputs are PNP transistors. That's why the input can go down to the negative rail (which is usually ground).
1599441358437.png

If you look at the schematic for 741, you'll see that the inputs can't get within about 4 diode drops of the negative supply.
1599441329398.png

My reformulated question: the opamp input can receive super low voltages and translate them into it's output, high enough to drive a minimum load as a led, optocuplor or even those super tiny motors big as a pinky finger nail? Or the ONLY way, is to link the output to base of a transistor? This is what i want to get clear.
Most opamps will only sink or source about 25mA. If you need more current, you need to buffer the output or use a special purpose opamp with a higher output current.
 
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Audioguru again

Joined Oct 21, 2019
1,926
With a +9V supply, the input of an LM324 or LM358 opamp can have a super-low voltage of 10mV (0.01V) and with a voltage gain of 600 times can produce an output voltage of +6V at 20mA.
 

Thread Starter

q12x

Joined Sep 25, 2015
266
Fantastic answer mister @dl324 , love it. Because you argument with the images of internal circuit of both IC's + explanations. Now I understand a bit better (not fully, but better).
I will say with my words, exactly what you just said, to filter it through my knowledge and experience. So... because of the internal structure, those 4 transistors(npn) junctions + R in uA741 to 0 rail, and only 2 more direct (pnp) junctions for LM324 to 0 rail, I will get the sensibility of input for each IC. Excellent answer, really. It makes more sense.

Most opamps will only sink or source about 25mA. If you need more current, you need to buffer the output or use a special purpose opamp with a higher output current.
Yes, this is what I was after. This answer clears the clouds from my sky. Thank you, now is way more clear.
So in some IC's, I have to add a transistor , as you said, to buffer the output, because it's internal limitations.
I'm super happy I clear this problem out.
Thank you so much ! 1-for understanding my problem (i know it was not easy to read all that logical jumbo i had to make use) and 2- for a direct to the hearth answer.
Really, really happy now. You are awesome.
But also mister @Audioguru again is very awesome for participating and shooting his arrows, "again" as usual. Really, thank you. :)

Phiu, that was an intense problem for me.
I didnt made the real test yet ! Again, I did this only in simulator, and I trust them with a drop of salt (not that much). But i will do this test shortly and come with the results, to confirm or not these simulations we did here. I hope it will stick,and remain the same results. Very curious now.
 

Thread Starter

q12x

Joined Sep 25, 2015
266
Update:
I did the real experiment and is more weird than expected. But in general lines, it worked as in the simulation, pretty well, which is good. I used what I only have: UA741.
I had to supply with 200mV from my adjustable power supply [S1], it was the lowest voltage possible i could get from it. When is around 100mV is fluctuating very badly. At 200mV is stable. So I had to re-adjust the voltage divider values to 1k and 18k to obtain a precise value(measured) of 10.1mV on the V+(pin3) for my UA741.
I used [S2] a separate 5.1V 2A comercial little electronic transformer to power the IC itself, on pin7 and pin4. This source will also power the output leds.
I also changed the leds resistors to 1k each, because brightness disturbance. I dim them down to a convenient light.
Its the same circuit, really, but with these adapted new values.
The good part is that all the values from the last simulated circuit, worked excelent in my real setup. I was very impressed when i see it working from the first try.
///
Now, the weird part I mentioned:
S2(5V) open all the time.
Led1 remains ON all the time, when S1(10mA) is on or off.

If I cut off the power from S1(10mV), Led2 will get brighter! And viceversa, If I power up S1, led2 will get dimmer.
(this phenomenon is now solved)
If 1 of + or - contacts from the power supply is touching the circuit for V+, led2 gets ON full bright, when the power from S1 is OFF.
If I touch with my skin, ANY power contacts (+ or -) for the circuit for V+, led2 gets ON full bright, when the power from S1 is OFF.
In a way, i think i get my super amplification, but not in the correct way.

This weirdness is evident when using 200R for led2 and 300R for led1.
If I put 1k&10k, led2 just remain open and brighter, whenever I switch S1.
If I interrupt the base for transistor, led1 gets OFF. So clearly is a leak from pin7(+5V) to pin6(Vout). It shouldn't !

In conclusion:
If I cut down S1(10mV), Led 1or2 remain ON !
(led2 is on because V+ is hipersensitive to any kind of touch).
(maybe base of the transistor is sensitive because internally, the IC is "touching" something? )

--image is sharpened now--
IMG_20200907_113905 - Copy copy 1.jpg

I just made this modification in my simulator, and it does present the same symptoms when I remove S1. This is very good to check the realism of my simulator. I removed S1 from the simulated circuit and it really does not affect the output, so led2 remain open. Oau. I didnt think of that before, because i was concentrated on the amplification side of the problem. Damnation, I say!
My interest is to actually see this amplification! So far... i dont see it. The hell continues.
Another idea is to get from you a safe margin in which ua741 works for this Voltage folower, wihout any other weirdness.
I did further test, unlinking Vout from V-. and the more evident effect is on led2, both in simulation and in reality. When unlinked, led2 is at full bright; when linked,led2 is poorly lit. Led1 is not that evident.
///
Another inconvenient, when i click on your images, mister @dl324, they dont get bigger !
They appear as little as you last resize them !
When i click on any of MY images, they do get bigger.
Im wondering if you see my images at their original rezolution, getting bigger when clicked on them?
Or they stay as little as I resize them here in the post?
If they stay as little as i resize them here in the post, then... its very bad.
 
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Audioguru again

Joined Oct 21, 2019
1,926
Your uA741 opamp is too old to operate with only a +5V supply. Also, it MUST have an additional negative supply for its input to be at almost 0V. Since your supply and input voltages are wrong then its output voltage is also completely wrong, the LED 2 should be off.

An opamp as a follower and that has proper voltages has an output voltage that is exactly the same voltage as its input voltage, unless its output is overloaded.

You showed a photo that is so blurry that we cannot see its details.

You talk about "amplification" but do not say if it is for voltage or for current. A follower has a voltage gain of only 1 so it has no voltage amplification. Two resistors allow an opamp to have voltage gain.
The input of an opamp uses a very low current but its output can produce much more current, which is current gain.

When you disconnect the negative feedback from the output to the inverting input then the datasheet shows a graph of its typical voltage gain:
 

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dl324

Joined Mar 30, 2015
11,550
My interest is to actually see this amplification!
By definition, a voltage follower doesn't provide amplification.
Another inconvenient, when i click on your images, mister @dl324, they dont get bigger !
Download them and you can magnify to your heart's content. Both of the schematics I posted are from datasheets.

A uA741 will not operate from a 5V supply. If your circuit happens to work, it's an outlier. When we design circuits, we design them to work under worst case conditions.

Why are you using multiple power supplies? A single supply with a resistor divider and/or pot would be sufficient.
 
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Thread Starter

q12x

Joined Sep 25, 2015
266
Mister @Audioguru again and mister @dl324
Thank you for seeing it. For some reason, I combined all the functionalities of an opamp into my circuit, thinking to something, then thinking on other things, I lost some important aspects like being a folower and not an amplifier.
In my defense, it's my first time when I seriously attacking this IC, which is having multiple utilization/functions. My mindset is a bit fixed about IC's, in general, for me they are like a function (in programming language) that it does 1 thing, like a minifactory that produces only iron, but not copper or brass ingots. I am a rookie with this IC, but not for long, thanks to you ! I see my mistake now. Oau. I do stupid mistakes from time to time, so do you know. :] It completely fly from my mind. Very good that you spot it mister Audioguru and dl324.
This is it with this circuit. I hope.

An opamp as a follower and that has proper voltages has an output voltage that is exactly the same voltage as its input voltage, unless its output is overloaded.
....
A follower has a voltage gain of only 1 so it has no voltage amplification.
The input of an opamp uses a very low current but its output can produce much more current, which is current gain.
Perfect explanation and very good to put it from other perspective multiple times! It really atract my attention and clarify it faster. I read all that you say here before in tutorials, but for some reason, i forget it complete when i was working in my circuit. Also, you put it in a much simpler light, and is easier for me to understand it. So many rules :] But im so new to this also. Really, excellent explanation !
I didnt think of this at all, but you are super right, what i got from it is a current amplification ! Excellent point again. I mean, i am aware it amplifies both voltage and current but I was only concentrated on voltage amplification all the time. Sorry to not mentioning earlier, but from my perspective there are so many rules, and i remember what i can. I will do better in the future.
I've also look on your first image graphic (and put it in my opamp folder in my pc) and is again super well explained, i really understand it now. As long as the frequency increases, the amplification decreases, and viceversa as long as the frequency decreases, the amplification increases. The amplification or Gain. But is a bit more advanced for me this frequency thing, because i never used it before and i dont know if or how to use it in the future. Its good information if you give your attention to it, so thats why i put it in my folder. But for immediate time, i certainly not use this information too much.

You both mentioning repeatedly that is not working from 5V. Really?
Here is what they say:
1599506186432.png
So you are saying it must have Only 22V, or else will fail? Is true, I dont see a range of operating voltages there, so I imagine you are referring as being a fixed voltage? That sucks. Or, if is a voltage range, what is it?

In conclusion,
You both totally solved my problem, by spotting its a voltage follower and not an amplifier. I totally miss it. I totally concentrated in other directions.
Thank you both for excellent answers.
 

Thread Starter

q12x

Joined Sep 25, 2015
266
Then... when is the [voltage follower] used? In what situation? If it's not amplifying anything? If it's only passing the voltage from its inputs to its output? I imagine is a teaching circuit, but not a practical one. I mean, what is a practical usage of it? I think this was my original question when i started playing with this circuit. I was intrigued.
Im thinking now, a possibility might be a 8 bit circuit, like this:
1599507668555.png
and also a voltage inverter (from - to +) as also seen in the same circuit here.
Definitely very new concepts for me, I really never used them before.
Im not sure if it counts as a voltage folower if it's having that Vout to V- Resistor in between.
 
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