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Do you know how this circuit functions and the formulas for analyzing it?

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digione

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What is the Op Amp configuration called specifically and what are the formulas for solving what Vout will be?
When V1 is removed and the non inverting input gets fed from a 10K-25K varying resistance connected to a constant current of 100uA the voltage seen on + is equal to 1.25v-2.5v. This OpAmp is then driving the output to swing between .6v - 4.0v So how does it do it and how can it be proven mathematically or theoretically?
 

ronsimpson

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On the left side is your circuit. On the right side is the same thing, with my change.
R1=R11, R2=R22, R5=R55
Look at R11 parallel with R22. Note I added a 1.25 voltage source.
Please tell me why V3, R11, R22 does the same job as V2, R1, R2.
1551644479302.png
Hope this helps you understand.
You understand R11 parallel with R22 = one resistor. This one resistor looks like it connects to a 1.25V source.
Do you know the gain formula?
 

digione

New Member
What I think is going on is that R11||R22 = 1.5K , 1.5K in series with R55 = 4K, so 1.25v/4K is 312.5ua which leads to a 469mv voltage drop across R11||R22 leaving 781mv sitting at the inverting input. The op amp corrects to make both input match and so it outputs the 469mv to bring that inverting input back up to 1.25v. I know this isn't what is happening though but I don't yet understand why or what really is happening. And I do not know the gain formula. So aside from the formula where am I going wrong in my assumption of this amps operation?
 

ronsimpson

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R11||R22 = 1.5K
Gain = 1+2.5K/1.5K Non-Inverting amplifier.
1551648767220.png
What you have is very much like Figure 8.2 bottom left side. But "R2" does not go to ground but to 1.25V.
---------------------- Things you probably know -------------------
A amplifier wants to have the same voltage on the (+) input as the (-) input.
If (+) is higher then the output goes up.
If the (-) is higher then the output goes down.
If (+) = (-) then the output holds, (does not move)
-----------------------------
If Vin = 1 volt then the output will go up. [(-)=0 volts], The output voltage is divided down by R1,R2. So some division of Vout makes it to (-).
Lets say R1 and R2= 1k because that is easy to think about.
(+)=1V=Vin.
Go get (-) to 1V we need to have 2V on the output. R1 & R2 makes a 1/2 divider.
------------in your example----------
Gain = 1+ 2.5/1.5=2.66
Vin=1V so the (+) input = 1V.
The (-) input should = 1V. So we need 2.66V on the output.
Can you see, if you have a voltage divider using 2.5k and 1.5K, it takes 2.66 volts on top of the divider to get 1V out of the divider?
------------------------------
 

AnalogKid

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This smells like homework. The configuration is called a non-inverting amplifier. Beyond that, why don't you show us your analysis and we'll advise.

ak
 

digione

New Member
I see how you are getting to 1 volt in your example. but if my non inverting input were at 2.65v then the output would need to be 6.51 volts to get the inputs matched. this would be over outside of the range of VCC and its not what I was seeing in the circuit. for a non inverting input of 1.25 to 2.5 volts I was getting a linear output of .65 - 4.5 volts. The circuit was suggested to me by another member on this site but I can't figure out why or how it works this way. below is a complete picture of the circuit, I don't have the hardware on me to play with and its frustrating me not understanding its operation.

1551650996027.png
 

digione

New Member
Its not homework. I am trying to create a glove with flexible sensors in them in order to control a robotic hand using an AVR microcontroller. I've been having problems getting the sensors to read the full range of the ADC inputs and so I came looking for assistance.
 

ronsimpson

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It looks like you have spent many hours on this.
You made a 100uA current source.
It looks like you have a sensor that has a resistance (25k to 10k) and you want to translate that to 0.65 to 4.5V.

First thing is the data sheet said that (with a 5V supply) the output can most likely to up to 4.4 volts. (4.0 worst case) That is what I am seeing.
Input voltage 1.05 to 2.625 and the output is 0.73 to ?? (4.4V)
1551655539946.png
There are some choices.
1) Do you have a power supply of 6V that we can run the op-amp from?
2) We could gat a different amp? There are amps that will go from near 0 to near 5V.
3) We could try making the output go from 0.1 to something.
 

ronsimpson

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I changed R6, 2k to 2.8k to lower the output voltage.
Green trace is 2k. Red trace is 2.8k. Blue is the input voltage.
1551656076844.png
How temperature stable do you need to be? Right now things will change with temperature.
 

digione

New Member
I see the limitations. here and I'm beginning to understand how this works. I was intending on using a 7.4 volt battery that I have from a small drone in order to drive a 5 volt regulator. I could increase the voltage in the circuit but maybe looking for a better op amp would be better if I could get one that would actually swing from 0 to 5 volts. This project was supposed to be more of an exercise in programming as I'm teaching myself assembly programming of an AVR micro-controller. After re-writing my code several times for other reasons (learning curve) when I finally got everything functioning and came back to fine tune the ADC inputs I thought maybe it would be easier to adjust the external hardware....everything is a learning experience right now. It seems to me that If I increase my feedback resistor that I could get a slightly lower output voltage but without a better op amp I won't be able to get a larger output. Thanks for the knowledge, like I said this circuit was provided by another member on this site and I was getting confused by the type of circuit it was. just because the inverting input is tied from vcc to gnd through a divider and when I searched the internet for comparable circuits all the non inverting op amps simply show - input dropped to ground but I see what is going on now and knowing the way to figure out the gain of this circuit is helpful.
 

digione

New Member
I changed R6, 2k to 2.8k to lower the output voltage.
Green trace is 2k. Red trace is 2.8k. Blue is the input voltage.
View attachment 116948
How temperature stable do you need to be? Right now things will change with temperature.
What is the application you are using to make these simulations? I'm not an engineer or anything, I'm a simple industrial mechanic so I can't say that I have even thought about temperature stability yet to be honest with you. I am more or less playing around and testing things out. This is a personal project so if I burn up some of my components its not going to affect anyone but me.
 

Pommie

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The output of the circuit is being fed into an ADC that requires 0 to 5V. I posted the circuit in this thread and was hoping one of you analogue guys would get involved. The constant current source is also there and will probably not be very temperature stable. However, from my experience with these kind of sensors, they are not very stable or repeatable so I don't think it needs more accuracy.

Mike.
 

ronsimpson

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If you look for a new amp. Look for a low voltage part. Some thing built for 5 volts. Look for "R-R" output. rail to rail
Micro chip has some that are R-R in and R-R out. You really need R-R out. The input side need only to work in the 0 to 3V range.
-----------------
I am using LTSpice. Free program. I can send you my schematic.
 

ronsimpson

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Sorry for the tall picture. The 0.7 volts from B to C of a transistor is not temp stable.
Here I used a second transistor to help remove most of the problem. What ever problems Q1 has Q2 is the same and substracts out.
You don't have to use 10.5k value. I was just using that value so I can do the math in my head. I wanted to get 1.05V so 10.5k and even I can do the math. LOL In real life I would use 10k and change the bottom resistor a little to make up for the 5% change.
1551657270190.png
 

digione

New Member
Hey Mike,
Been getting a lot of help on here. This is a good group of guys. I'm thinking maybe I should stick to programming and tighten up my issues in the code. I should probably spend some time just playing with op amps on their own rather than in a larger circuit.
Sorry for the tall picture. The 0.7 volts from B to C of a transistor is not temp stable.
Here I used a second transistor to help remove most of the problem. What ever problems Q1 has Q2 is the same and substracts out.
You don't have to use 10.5k value. I was just using that value so I can do the math in my head. I wanted to get 1.05V so 10.5k and even I can do the math. LOL In real life I would use 10k and change the bottom resistor a little to make up for the 5% change.
View attachment 116949
Thanks for the advice. I'll look into getting some other op amps and download ltspice. seems like it would help to be able to simulate stuff before spending time laying anything out on a bread board. Really appreciate it.
 

Pommie

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Out of curiosity, am I reading the datasheet for the 2N3906 correctly? Figure 16 seems to suggest that the variation in Vbe is -2mV/°C (Ic = 0.1mA = far left of chart) and so for a temperature range of 0 to 30°C it will vary 60mV from 0.65V. I assume the 0.65V is at 20°C and therefore it will vary from 0.61V to 0.67V which is quite a large variation.

Mike.
 

ronsimpson

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therefore it will vary from 0.61V to 0.67V which is quite a large variation.
Some ICs measure temperature this way.

When making a current source you might have 1V on the emitter resistor and (0.61V to 0.67V) in the B-E voltage. If your reference voltage is fixed at 1.65V then the 60mV change because of temp will appear as 60mV change in the 1V across the resistor. That is only 6% but it can be mostly removed.

What I did was try to make a fixed 1V reference then add one diode drop to it. D=one diode drop. I want 1V+1D. Then the output transistor removes the 1D and we have the reference across the resistor with out much temp variations.
Talking about the circuit in #14.
 

gophert

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Out of curiosity, am I reading the datasheet for the 2N3906 correctly? Figure 16 seems to suggest that the variation in Vbe is -2mV/°C (Ic = 0.1mA = far left of chart) and so for a temperature range of 0 to 30°C it will vary 60mV from 0.65V. I assume the 0.65V is at 20°C and therefore it will vary from 0.61V to 0.67V which is quite a large variation.

Mike.
One of the first things taught about diodes is the -2mV/°C temp coefficient. Same temp coefficient for most every PN junction (base/emitter junction). It is not a "Large variation", it is the expected variation. 4.3 to 4.7V Zeners are pretty stable vs temp because of some competing thermal issues.
 

Pommie

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One of the first things taught about diodes is the -2mV/°C temp coefficient. Same temp coefficient for most every PN junction (base/emitter junction). It is not a "Large variation", it is the expected variation. 4.3 to 4.7V Zeners are pretty stable vs temp because of some competing thermal issues.
I asked because at 100mA collector current it appears to be half that value according to the chart.

Mike.
 

ronsimpson

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Most Helpful Member
I asked because at 100mA collector current it appears to be half that value according to the chart.
It has been many decades, I can not remember but I think we used different current to get slight variations in TC.
With the one transistor current source I see 6% change with temp.
With the two transistor current source I see 1.4% change.
Normally you want the same current in the two transistors. BUT I think I changed the ratio of the two currents until much of the 1.4% was removed.
But …. I am not certain. I studied this when I was 10 years old. At the university I jumped over this because I tested out of the first years.
 
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