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OP-Amp amplifier circuit

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girish09

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opamp.jpg

https://www.google.co.in/search?hl=...1785.7j3j4.14.0.les;..0.0...1c.1.fft3X9yxtyg#



This is a inverting amplifier, one query came in my mind that, why current will not flow from Rf only because Rf is having low resistance as compair to op amp inut impedence so path should be from Vin-R1-Rf-GND(at the o/p terminal), in this way also path can complete so what is role of opamp then???
 

not-a-moderator

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I fail to understand "path should be from Vin-R1-Rf-GND(at the o/p terminal)"
You can't connect the output of the opamp directly to ground, you have to connect it to a load first.

In the circuit the non inverting input of the operational amplifier is connected to ground. As the gain of the operational amplifier itself is very very high and the output from the amplifier is a matter of a few volts, this means that the difference between the two input terminals is exceedingly small and can be ignored. As the non-inverting input of the operational amplifier is held at ground potential this means that the inverting input must be virtually at earth potential (i.e. a virtual ground).
As the input to the op-amp itself draws no virtually current (high input impedance) this means that the current flowing in the resistors R1 and Rf is virtually the same.
Using Ohms law: Vout/Rf = -Vin/R1. Hence the voltage gain of the circuit Av can be taken as: Av = -Rf/R1
 

pjmelect

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The output of the op-amp will not always be at ground potential. The output of the op-amp will be at the voltage required to bring the inverting input to ground potential, or until the op-amp reaches saturation.
 

LvW

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View attachment 80527

https://www.google.co.in/search?hl=...1785.7j3j4.14.0.les;..0.0...1c.1.fft3X9yxtyg#
This is a inverting amplifier, one query came in my mind that, why current will not flow from Rf only because Rf is having low resistance as compair to op amp inut impedence so path should be from Vin-R1-Rf-GND(at the o/p terminal), in this way also path can complete so what is role of opamp then???

Simple answer: I am afraid, you forgot that the opamp output carries a voltage as well - that means: Two voltage sources determine the current through both resistors.
 

krixen

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I read over your question a few times and i think you might be asking what is the point of an Op-Amp if it just 'ignores' the op-amp and goes right through Rf.

I thought the same thing when i first learned about Op-Amps, and it has a very useful purpose. It will output a voltage that will not be affected by a load. If your load is a 100 Mohm resistor or a 100 ohm resistor, the voltage at the Vout terminal will always be the same. To find that voltage it is, (Rf/Rin)(Ao/Ao+1)(Vin) or simply Vout = Vin*(Rf/Rin)
 

girish09

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I read over your question a few times and i think you might be asking what is the point of an Op-Amp if it just 'ignores' the op-amp and goes right through Rf.

I thought the same thing when i first learned about Op-Amps, and it has a very useful purpose. It will output a voltage that will not be affected by a load. If your load is a 100 Mohm resistor or a 100 ohm resistor, the voltage at the Vout terminal will always be the same. To find that voltage it is, (Rf/Rin)(Ao/Ao+1)(Vin) or simply Vout = Vin*(Rf/Rin)


Thanks.......Exactly.....you got my point.....But i didn't get ur answer. I know gain of inv amp is -Rf/R1 , But if ckt is going to complete through the path vin-R1-Rf-Rload-Gnd then when op-amp is coming in picture. Because if we will use transistor instead of opamp then current will flow thr transistor so accordingly in the opamp ckt current should flow through opamp then & then only we can say that this is inverting amplifier using opamp and we will use formula of opamp.....I am eagrly waiting for your kind reply....
 

LvW

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.....I know gain of inv amp is -Rf/R1 , But if ckt is going to complete through the path vin-R1-Rf-Rload-Gnd then when op-amp is coming in picture. Because if we will use transistor instead of opamp then current will flow thr transistor so accordingly in the opamp ckt current should flow through opamp then & then only we can say that this is inverting amplifier using opamp and we will use formula of opamp.....I am eagrly waiting for your kind reply....

I think, your error is that you think in current terms only. Don't forget that the opamp is considered as a voltage amplifier (best example: FET input stage with no input current).
Start with your assumption that the current goes through the chain R1-Rf-Rload-Gnd and realize that this current develops a voltage at the node between R1 and Rf.
This voltage enters the opamp input and causes an opamp output voltage, which causes a special effect called "negative feedback".
Now, you have - as I have mentioned earlier - two voltages affecting the whole circuit. In steady-state condition these two voltages determine the current through the resistors thereby creating a so called "virtual ground" potential at the inverting input. For a complete understanding of this effect it is necessary to study the feedback principle.

- - - Updated - - -

Here comes one possible "visualization" of the feedback effect - perhaps it helps:

For steady-state conditions there must be an equilibrium as described as follows:
Both voltages (Vin and Vout) create a very small voltage Vinv at the inverting opamp input terminal that must satisfy the amplifier condition Vout=Ao*Vinv. (Ao=open-loop gain).
Example: Ao=1E6 and Vinv=1E-6 volts.

Now you can write down two current equations:

I=(Vin-Vout)/(R1+Rf) and I=(Vinv-Vout)/Rf

Equating both expressions and setting Vinv<<Vout leads to

Vout/Vin=-Rf/R1.

Note: This simple result is valid for an idealized opamp only (Ao approaching infinity, which means: Vinv=0)
 

irfan ahmad

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salam:
first of all you have to understand that output voltage are inverted then input.
the second thing that op amp has approx infinite impedance (resistance) there fore there is no current passing to op amp's ground terminal.
current path is Rin -->Rf-->output voltage.
because output voltage are inverted then input .
For example;
suppose Rin is 1k and Rf is 10k and we apply 1v to input then according to formula output voltage will be -10v
now x --/\/\/\------inv-input------/\/\/\/\/---y
1v 1k opamp input 10k -10v
Rin Rf

here point x is more positive then y so current will flow from point x to y (conventional current theory)
according to ohms low 1v will apeare across 1k and 10v will apeare accross 10k and
central point will behave like ground terminal and at 0 potential.
i hope this will dissolve your confussion.:sad:
 

krixen

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So if you never had an op-amp and you wanted to create a circuit, what ever load you have, if it changes in value then the voltage drop across it will change, the Op-Amp makes it so that no matter what load you attach to the output of the Op-Amp the voltage across that load will never change. that's a very powerful tool because it can act as a buffer so if you have a lot of amplifier circuits, the op-amp can be attached in between and changes in the individual circuits wont affect one another.

the actual inside of an op-amp is a lot different than what it looks like in the picture, it has a controlled voltage source.
 

telecom engineerr

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Can anyone send me the simulation of difference amplifier please? urgently required

- - - Updated - - -

Or any tool where i can implement the simulation online?
 

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