OP-AMP based voltage follower

Hi,
While reading Microelectronics Circuits by Sedra Smith i had the following query
the formula we get for non inverting op amp is 1 + R2/R1, now to turn it into voltage follower it suggests that make R2= 0 and R1 = infinite, i was wondering if R2 = 0 is sufficient and we could keep R1 in the circuit and it should still be a voltage follower, would it work?

Regards,

Hamza

Yes, it'll act as a voltage follower. Basically, the R₁ will then become the load across which the output voltage will be acting.

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Here are two schematics to better explain. They both represent the same circuit i.e. your case in which R₂ is zero and R₁ is any voltage which in this case is 1 k.OHM

hshah8970 said:

Here are two schematics to better explain. They both represent the same circuit i.e. your case in which R₂ is zero and R₁ is any voltage which in this case is 1 k.OHM

Click to expand...

To be exact:
* They both represent not the "same circuit" but the same function.
* R1 is any value (instead any "voltage")
* There is a third alternative with R1 infinite and R2 any value.

^ Yup. Thanks for the correction.

thanks for the replies, ok now keeping to the figure 2, how would the KCL be applied? because of the virtual short there is V1 at node 2, so there is current flowing through R1, ignoring current through branch connected to negative terminal the same current should be flowing in feedback path but o/p terminal and node 2 being at same potential there should n't be any current in feebback path? what am i ignoring?
@ LvW, can u explain how physically V0 = Vin happens for third alternative if we ignore mathematical expression, am i correct in stating it is just down to voltage division rule, because R1 is so large more voltage appears across R1, and approximately that equals V0 for large ratio of R1 with respect to R2?

Regards

I didn't quite understand what you're asking. Current in the feedback path (wire connecting the output to the negative opamp terminal) is zero since both ends are on the same potential. Also because of the virtual short, in this case all pins of the opamp are at the same potential.

The current flow is like this.



The current also flows through the opamp's power supply, which is not shown in the picture.

No current flows through the voltage source marked V1.

@ godfreyl , Sir but why is there current in feedback path, isn't o/p at same potential as the negative terminal?

if the R2 is kept non zero and R1 is kept infinite then what impact it will create .....I mean on the voltage will get followed or it will be useful for sensing low currents ....etc I am asking about practice cases ...in theory it is fine but I am interested in real time cases...

Regards,

milind

hamzahumayun said:

why is there current in feedback path, isn't o/p at same potential as the negative terminal?

Click to expand...

Yes, they're at the same potential (assuming the wire connecting them has zero resistance). That doesn't mean no current can flow.

There's 20V across the resistor, so current must be flowing through it. You just need to figure out where that current comes from. It can't be from the opamp's inverting input because that's a high impedance input that can't source current. So the current must come from the opamp output.

Think about this for a moment:
Any piece of wire carrying a current has no potential difference across it. That's exactly why we use it to conduct current.

Example:
In the picture below, there's current flowing through the resistor.
The top of the battery is at the same potential as the top of the resistor, but current is flowing through the wire that connects them.

Now if you go back to the two circuits in post 2, you'll see that they are equivalent. In both circuits the opamp's inverting input, the opamp's output, and the top of the resistor are all connected together.

godfreyl said:

Think about this for a moment:
Any piece of wire carrying a current has no potential difference across it. That's exactly why we use it to conduct current.

Click to expand...

The reverse is true: any conductor carrying current most certainly does have a potential difference.

(Ignoring reactive elements for the moment,) Any conductor will have a resistance R.

The potential difference across that conductor (e.g. piece of wire) is given by Ohm's Law: V = IR

godfreyl said:

Example:
In the picture below, there's current flowing through the resistor.
The top of the battery is at the same potential as the top of the resistor, but current is flowing through the wire that connects them.

Click to expand...

The total voltage drop is shared across the battery terminals, top wire, resistor and bottom wire. If the current is large, these will be significant values. If the circuit is sensitive to this order of currents, these will be significant values.

Sorry but we can't define a conductor to be something that carries current but has no potential difference across it, Ohm wouldn't agree.

TonyM said:

The reverse is true: any conductor carrying current most certainly does have a potential difference.

Click to expand...

Yes, but for simplicity I was ignoring wire resistance because it didn't seem relevant to Hamza's questions.

TonyM said:

The total voltage drop is shared across the battery terminals, top wire, resistor and bottom wire. If the current is large, these will be significant values.

Click to expand...

All the circuits shown in this thread use a 1K resistor. The extra resistance added by any sensible wiring and/or PCB tracks will be completely negligible.

I appreciate your line of thinking. But surely...

Any piece of wire carrying a current has no potential difference across it.

...should never be part of any explanation? That will just confuse someone further. Especially the answer to his question will be an Ohm's law one.

The reverse is true: any conductor carrying current most certainly does have a potential difference.

Click to expand...

The "wire" in a circuit diagram is assumed to have exactly zero resistance, all connected circuit terminals form a node in terms of network analysis and thus, by definition, don't involve any potential difference.

It's up to you to decide if it's a case of idealization or just expresses the fact, that the real resistance can be ignored related to the circuit of interest. Otherwise, you'll add a resistor symbol representing the real conductor resistance.

FvM said:

The "wire" in a circuit diagram is assumed to have exactly zero resistance, all connected circuit terminals form a node in terms of network analysis and thus, by definition, don't involve any potential difference.

Click to expand...

If you like but not really In these circuits, it's all about Ohm's Law and statement I queried opposed the explanation we're to lead to. I queried for the benefit of the user asking the question and it's good for us all to stay away from bickering or insistences of being right despite the evidence. I'm glad that's the case here and we can just stick with the user's question in hand...

thanks alot i really appreciate it, so following picture from one of the posts answers my query
.

Regards