operational amplifier circuits etc.

Discussion in 'Mathematics and Physics' started by PG1995, Apr 19, 2014.

1. PG1995Active Member

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Hi

An op-amp tries to make differential voltage across its input terminals zero which also means that no input current flows into the op-amp.

In this case of Figure 12-22, it's easier to understand what is really happening. As Vout = (-Rf/Ri)Vin, therefore when value of Ri is low, the Vout is made more negative so that 'positive charge carriers', i.e. conventional current, are pulled away from the inverting input terminal of the op-amp with greater force and no charge carriers enter into the op-amp.

Q: The circuit on the left uses inverting configuration whose gain is given as Vout/Vin = -Rf/Ri.

In the given case Vin is negative therefore Vout = (Rf/Ri)|Vin|.

I'm assuming that the relation for Vout given above is correct. When the resistance of photoconductive cell decreases (which in this case is Ri), the Vout increases. Conceptually, I don't get why an op-amp would increase its Vout as Ri decreases.

Likewise, when Ri increases, Vout will be decreased. Why?

Regards
PG

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Vin=-3V, Rin=Rout so gain =-1, Vout=+3V Does this work for you?
Vin=-3v, Rin is 1/2 Rout then gain --2, Vout=+6V
Vin= -3V Rin is big then gain is small, Vout is small

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3. NorthGuyWell-Known Member

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IMHO, since current doesn't flow into the op amp, the best way is to look at it as a voltage divider formed by Ri and Rf. The middle point and Vi are fixed, and Vout is allowed to vary. Current flows through it in one direction. Therefore, when Vi is below ground, Vout is above ground. And when Vi is above ground, Vout is below ground.

Less Ri resistance with the same battery voltage means more current through Ri (Ohm's law). Ii = Vi/Ri.

Assuming ideal op-amp, current through Rf is the same as through Ri. If = Ii = Vi/Ri

Since Rf resistance doesn't change, more current through Rf means more voltage drop on Rf (Ohm's law again). Vf = If*Rf = (Vi/Ri)*Rf = (Rf/Ri)*Vi

One end on Rf is fixed at 0V, so Vout is equal to the voltage drop on Rf. The sign will be opposite to Vi. Vout = -Vf = -(Rf/Ri)*Vi

It is the same regardless of the direction of the current.

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5. PG1995Active Member

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Thank you, ronsimpson, NG.

NorthGuy: I'm sorry but it looks like I wasn't able to put my question clearly because, as I see it, you have missed the point of my question. Just to clarify, I was trying to understand the operation of the circuit at the bottom.

Yes, I agree with you. Actually, this is what I also said, "In the given case Vin is negative therefore Vout = (Rf/Ri)|Vin|." As Vin is negative in the given circuit, therefore Vout is going to be positive.

I also agree with you that Vout will increase as Rin decreases; in the given circuit Ri is photoconductive cell.

I request you that please have another look on what I said in previous post. What I'm trying to understand is that why Vout decreases as the resistance of photoconductive cell increases, and why Vout increases as the resistance of photoconductive cell decreases. I'm looking at it in intuitive terms. Thank you.

Regards
PG

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6. NorthGuyWell-Known Member

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I'll try in intuitive terms:

More resistance means weaker current. Weaker current has less ability to develop voltage.

Less resistance means stronger current. This strong current pushes output voltage higher.

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Hi,

First, the input current is not zero. Also, if the input current is very low it is not because the differential voltage goes to zero, although that may help. The input current is dependent on the "bias" current of the particular op amp which can be found on the data sheet and usually varies with temperature. The input current is 'taken' to be zero for simplified theoretical problems like this although it is good to know the real reason why there is any input current, even though it is low.
Remember that although the differential voltage is zero or close to that, the absolute voltage on both input terminals is not always zero but some offset value or some actual DC level set by the circuit. IF the non inverting terminal is at ground, then the inverting input will be 'close' to zero but not actually zero. If it was really zero there would be no output because the differential output amplifies that small difference and that is how we get any output at all.
Also, if both inputs were zero they could be drawing current OUT of the op amp terminals, which means there is negative input current.
If you short the two inputs you'd get zero output on an ideal op amp, but a real op amp may output some voltage because of the input offset of the op amp chip.

The simplified gain can be calculated by knowing that the output feeds the feedback resistor and the input resistor 'shunts' of of that current into the input voltage source. The output will attempt to make the inverting terminal equal to the non inverting terminal in the simplified view, so we reach equilibrium when the output voltage forces the inverting terminal to the same voltage as the non inverting terminal, and to do that we have to look at the voltage divider action of the two resistors viewing the feedback resistor as the upper resistor of the divider and the input resistor as the lower resistor.
Doing that, the voltage across the two resistors is:
Vx=Vout-Vin

which as you can see could not be simpler.
The voltage at the junction referenced to Vin (not to ground) is therefore:
Vj=Vx*Rin/(Rin+Rfb)

and the voltage at the inverting terminal is that voltage with a Vin offset so we have:
Vinv=Vj+Vin

Combining all these, we get:
Vinv=(Vout-Vin)*Rin/(Rin+Rfb)+Vin

which simplifies to:
Vinv=(Rin*Vout+Rfb*Vin)/(Rin+Rfb)

Now we can just solve that for Vout and we get:
Vout=((Rin+Rfb)*Vinv-Rfb*Vin)/Rin

and if we assume that we have the simplified op amp we set Vinv=0 so we get:
Vout=-Vin*Rfb/Rin

and divding both sides by Vin we get:
Vout/Vin=-Rfb/Rin

so we see the gain is negative.

Note that Vout is not the absolute value of Vin times the 'gain', it is exactly as above.

Another view is the 'current flow' view, where the current is said to flow around the op amp. That works out the same really, and we calculate the output voltage as that which appears across the feedback resistor.

To understand the op amp a little better requires using the non ideal gain to be high but not infinite. Say 1000 would be a good plece to start. Then you can write equations based on just a plain old amplifier that amplifies the difference between the two input terminals. If we had an internal gain of 1000 for example and we had a difference voltage of 0.01 volt, we'd get 0.01 times 1000 which equals 10 volts on the output. With a difference voltage of 0.001 volt we'd get 0.001 times 1000 which equals 1 volt on the output.
Writing the equations this way brings out the fact that the op amp difference voltage really isnt zero but depends highly on the internal gain of the op amp.

We can look at more examples if you like and that is how you get a better feel for this.

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For a concept view, the examples are horrible.

To actually to try to understand things stack the voltage source and the resistors on top of one another in your attachment. e.g. Draw them that way. Get rid of RL because it confusing things.

Now look at that input as a voltage divider.

Now look at it again, with a gain of 10 for example and a 0.1 V input.

If the voltage divider gives you 1/10 of the input, then the output has to be x10 to make the difference across the inverting and non-inverting terminal zero.

Another way to look at the OP amp is that Vout = (A-B)*(some big number). That big number is the open-loop gain.

Another important point to remember is that Ib or the input offset voltage has to have somewhere to go. Input offsets in mV or uV dropped across a small resistor (a wire which is .0001 ohms or whatever is still a big number I=V/R can still be quite a big number. Adding a resistor (say 10K) in series with the inputs does a LOT when trying to now short the new input. That new resistor adds noise to the circuit as it's value increases, so you don't get a free lunch. I got burnt by this one.

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9. PG1995Active Member

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Thank you, NG, MrAl, KISS.

KeepItSimpleStupid: Is it possible for you to provide a schematic which shows what you are saying? It's okay if it's not possible because I intend to return to this question again in next couple of days.

Could someone please help me with this query about wheatstone bridge? In wheatstone bridge the ratio of resistances of two legs, R1/R2=R3/R4, where R1 and R2 constitute one leg and R3 and R4 constitute the other leg. I'm more familiar with the configuration shown in circuit #2 where a variable resistor is used in one leg and the unknown resistance in the other leg. But I think any of the circuit configurations, circuit #1 and circuit #2, can be used and value of unknown resistance can be found. Thanks.

Regards
PG

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Hello again,

What exactly is your new question?

I hope i did not confuse anyone in my previous post where i presented two different views of the op amp.
The first view is where we assume that the differential input voltage is zero, and the second view which is how the op amp really works in real life is where we know that the differential voltage is never a perfect zero but is always slightly non zero and that gets amplified to produce the output.

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In Fig 12-21, draw Vin, Ri and Rf vertical and in-line.

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13. PG1995Active Member

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Hi

The given below is my own attempt at the answer. Do I make sense to you? Thank you.

An op amp tends to equalize voltages across its input terminals. Let's say voltage at the green point is 3 V. Then, it means that the op amp would try to adjust voltage at its inverting input (i.e. red point) such that differential voltage becomes zero.

The current flows from the red point towards the negative terminal (i.e. point in blue) of Vin. We should note that in this circuit positive terminal is grounded.

It was said that as the resistance of photoconductive cell (or, Ri) decreases the output of op amp increases. Why? The op amp wants to maintain 3V voltage at the red point. When the resistance dcreases, the potential difference at the red point decreases because more of the voltage gets dropped across Rf. To maintain the voltage at 3V the op amp increases its output voltage so that more current flows which would increase voltage drop across the cell or Ri. As R goes down in V=IR then "I" should go up to maitain the voltage.

Likewise, when the resistance of the cell increases, the op amp's output voltage decreases because when "R" goes up then you need less current to maintain the voltage, V=IR.

Regards
PG

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The only problem I have is "The current flows from the red point". You need to fix that up and you'll be OK. Remember that for the ideal OP amp, the inverting and non inverting terminals do not draw any current.

You make the "red point" sound like a battery.

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15. NorthGuyWell-Known Member

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That is relative to the blue point. The output voltage, however, is measured relative to the green point.

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16. PG1995Active Member

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Thank you, KISS, NG.

I would like to clear another point here.

Do you think my connections are alright? Kindly let me know.

I'm assuming the connections are alright. The current can flow along two different paths from the yellow point, either toward the red point or toward the ground (or, black point). I think that the current gets divided to flow along two different paths at any time. In other words, some of the current flows from yellow point toward the black point and some of the current flows from yellow point toward the red point (or blue point). Do I have it correct? Thank you for the help.

Regards
PG

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Hi,

This circuit is easier to understand based on current balancing and the concept of 'virtual ground'.

The inverting terminal is always very close to zero volts because the non inverting terminal is connected to zero volts. The op amp, in linear operation, tends to try to keep the relationship true for the entire operation. We assume it stays in linear operation which keeps the inverting terminal very near to zero volts and we call it zero for short.

With zero at the inverting terminal, we calculate the input current Iin. We then calculate the output current through the feedback resistor Iout, which is the same as Ifeedback because the left side of the feedback resistor is connected to zero volts (as above).

To balance the current we need Iin=Ifb and this is the same as Iout with no additional load, and because it's in linear operation, this equality Iin=Ifb=Iout is always true. All we need to do now then is to convert that Iout to a voltage to get Vout.

Since Iin=Iout and since Iout flows through the feedback resistor Rf the current though Rf can also be called Iin, so the voltage across Rf is simply Vf=Iin*Rf. To get the polarity it's only a matter of examining the circuit to see what polarity would balance the polarity of the input current to exactly subtract to zero or else just what causes zero volts at the "voltage divider" junction of the input resistor and the feedback resistor.

Example:
The Vin battery is 1v, the feedback resistor is 1k.
At a particular temperature, the resistance of the sensor is 500 ohms. What is the output voltage?
With -1v in and 500 ohms the input current is:
Iin=-1/500 amps.
This same current flows in Rf which is 1k, so we have Vout as:
Vout=-Iin*Rf=1*1000/500
which equals 2 volts.
Note i used a minus sign in front of Iin because it's an inverting amplifier.
Checking:
The voltage divider ratio is:
Rin/(Rin+Rf)=500/(1500)=1/3

The output voltage is 2v, the input is -1v, that's a total difference of 3v. The inverting input terminal is at a voltage 1/3 of 3 which is 1v, and 1v up from -1v (-1v is the input battery) is zero volts, so the inverting terminal is indeed at 0v so the result checks out ok.

And yes the circuit looks connected ok.

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It seems correct to me. The output of the OPAMP will try to provide any current that is demanded, but keep in mind that there is a current limit to a real opamp.

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19. PG1995Active Member

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Hi

I believe this is the way I would need to power up a dual polarity op-amp. Is this correct? In other words, a dual-polarity power supply can be made using two 7805 regulators.

If I'm not dealing with negative voltages then can I simply connect the negative supply pin of op-amp to the ground? I have also been told that there also exist single supply op-amps which do not require dual polarity supply. The following links are helpful here, especially page #2 of link #1.

1: http://www.ece.ucsb.edu/Faculty/rodwell/Classes/ece2c/labs/an-116.pdf
2: http://www.linear.com/product/LT1006

In this post, I was told that there exist op-amps which are called rail-to-rail op-amps and they should be used whenever possible. These parts numbers were given: MCP6001, MCP6002, MCP6004, MCP6L01, MCO6L02, MCP6241, MCP631. I wasn't able to find any of these in local store. Could you please let me know some parts which are popular and generally available like LM741.

Moreover, LM741 is considered to be an old horse now. What are newer models which can be used in its stead? At several places TL071 is recommended but do you think its operation is similar to LM741 with better performance?

I'm trying to design a current sensor just for the purpose of using it in a simulation because for hardware I have an current sensor IC. Is this circuit okay? Do I really need R1 because op-amp in itself is high impedance device?

Thank you for the help.

Regards
PG

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That configuration will work, but it is a bit clunky.
Have a look at the 7905 regulator, which is the same as a 7805 but is made for negative supplies like this.

Depending on your application, regulated supplies may not be necessary, look at the data sheets for "Power Supply Rejection Ratio".
Having said that, stabilised supplies are usually used because it is easy to do and you often want the stabilised supply for something else in the circuit.

Also be aware that limiting your supplies to +/-5volts will limit the output of the opamp to (slightly) less than +/-5volts, even with rail to rail opamps.

I recently used an MCP6042 low voltage rail to rail opamp because I wanted to make the most of my 0 to 5v supply, and have a 0 to 5v output, without messing about finding higher voltage supplies from my existing circuit which I was modifying.
Otherwise I would have used a TL071/TL072 if I had more supply volts readily available.

JimB

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Hi,

Just a quick note for the picture "sensor222.jpg".

A1 should be connected as a differential amplifier. It's not really right the way it is now.
A2 should be connected as a buffer or not used at all. It's not connected right the way it is now.

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