I believe there is a mistake in the formula for the voltage at the inverting input Vn.
It's part of a voltage devider so shouldn't R1 be R2, on the upperhalf of the division?

Thank you!

-Floris F.

 

Let's assume that R2 is 9k and R1 is 1K and Vo is 10V.
Try out the values with the given formula and with your suggested formula and see what answers you get. Which seems to give the most sensible result.

 

Thanks!

 

Let's assume that R2 is 9k and R1 is 1K and Vo is 10V.
Try out the values with the given formula and with your suggested formula and see what answers you get. Which seems to give the most sensible result.

I am wondering how this help? R1 and R2 can have any value or feedback factor from 0 to 1.

 

I am wondering how this help? R1 and R2 can have any value or feedback factor from 0 to 1.

Did you do the two calculations?
It should be fairly obvious which is correct if you do.

 

I am wondering how this help? R1 and R2 can have any value or feedback factor from 0 to 1.

The circuit is a non-inverting amplifier. The example provided by AlbertHall is valid.

Select R1 and R2. Select Vs. Compute Output, then compute Vn based on the formula. Then you can decide if the example was valid.

 

View attachment 115670
View attachment 115671
I believe there is a mistake in the formula for the voltage at the inverting input Vn.
It's part of a voltage devider so shouldn't R1 be R2, on the upperhalf of the division?

Thank you!


-Floris F.

Hi,

Just so you know, the lower resistor in the schematic is always on the top in the expression. Since in this case the lower resistor is R1, R1 is in the top of the expression because it is in the bottom of the schematic divider.

 

Did you do the two calculations?
It should be fairly obvious which is correct if you do.

The circuit is a non-inverting amplifier. The example provided by AlbertHall is valid.

Select R1 and R2. Select Vs. Compute Output, then compute Vn based on the formula. Then you can decide if the example was valid.

I think you are suggesting calculate Vn either ways and then compares with Vs.
However, in case we don't know the value of Vs and the feedback is just a voltage divider. I don't think it is a good way to know which one is right.
Just study the voltage divider because it is basic that everyone needs to know.

 

Hi,

Just so you know, the lower resistor in the schematic is always on the top in the expression. Since in this case the lower resistor is R1, R1 is in the top of the expression because it is in the bottom of the schematic divider.

The TS made their mistake by assuming that the resistor labeled R2 is always in the numerator. You seem to be making a similar mistake by assuming that the resistor at a certain location in a drawing is always on the top of the expression. Not all schematics with voltage dividers put the output voltage across the lower resistor. For that matter, not all voltage dividers are with respect to ground.

Far better to understand how to analyze the circuit.

 

I think you are suggesting calculate Vn either ways and then compares with Vs.
However, in case we don't know the value of Vs and the feedback is just a voltage divider. I don't think it is a good way to know which one is right.

The TS proposed a formula found in their textbook was incorrect, so a member gave them some values to "test" that hypothesis. Did they? To date, the TS has not returned to this forum. I expanded on the test to show the TS a way to "decide" for themselves how to confirm it.

I agree that the TS is hung up on the resistor designations and not their function in the circuit.

 

The TS proposed a formula found in their textbook was incorrect, so a member gave them some values to "test" that hypothesis. Did they? To date, the TS has not returned to this forum.

I take post #3 from the TS as saying that the example solved the problem for them.

 

It's part of a voltage devider so shouldn't R1 be R2, on the upperhalf of the division?

it is answering the question from the perspective of the divider.

An ideal opamp in negative feedback has two properties:
1) the potential on the non-inverting input is the same as the potential on the inverting input;
2) the input pins have infinite impedance.

In your case, 1) leads to Vn = Vinput; and 2) leads to Vn = Vo / (R1 + R2) * R1.

From that you can solve for Vo and it should be Vo = (R1 + R2) / R1 * Vin.

 

The TS made their mistake by assuming that the resistor labeled R2 is always in the numerator. You seem to be making a similar mistake by assuming that the resistor at a certain location in a drawing is always on the top of the expression. Not all schematics with voltage dividers put the output voltage across the lower resistor. For that matter, not all voltage dividers are with respect to ground.

Far better to understand how to analyze the circuit.

Hi,

Yes i agree, but with the drawing given, it was clear that the other resistor was on the bottom so keeping it simple i suggested that. What you are suggesting results in a more complex formula for a voltage divider which although would certainly be more general, is not the way we usually show this concept.

We usually have:
Vout=Vin*R2/(R1+R2)

and what you are suggesting is:
Vout-Vref=(Vin-Vref)*R2/(R1+R2)

or something like that, and that we must also have a schematic to go with it to show where R2 really is.
So if you want to go with that, be my guest

However, in light of your post i'll post my example circuit:

Vin o---R1---x----R2----o GND

where Vout is at the node 'x' above.
In this circuit, the output is:
Vout=Vin*R2/(R1+R2)

and R2 here is usually referred to as the 'lower' divider resistor even though it is not lower in the exact schematic, and R1 is usually referred to as the 'upper' divider resistor.

I hope that helps clear this up.

 

Hi,

Yes i agree, but with the drawing given, it was clear that the other resistor was on the bottom so keeping it simple i suggested that.

My problem with your suggestion is that it was not limited to the drawing given -- it explicitly stated that the bottom resistor in a schematic is ALWAYS on the top of the expression. That's a blanket, unqualified claim that simply isn't true.

What you are suggesting results in a more complex formula for a voltage divider which although would certainly be more general, is not the way we usually show this concept.

How it is usually shown is really not relevant -- do we really want people to learn to just regurgitate equations based on how something is usually shown? That just leads to them regurgitating equations wrong as soon as they encounter something that is not drawn or labeled exactly the way something is normally shown (the OP being a perfect example). I see this time and time again (and we see it here time and time again).

 

I take post #3 from the TS as saying that the example solved the problem for them.

I've seen thanks for posts that consisted of a question to the TS ... and the TS never responded other than thanking the member.

So, my expectations have be lowered considerably since reading them.

 

I've seen thanks for posts that consisted of a question to the TS ... and the TS never responded other than thanking the member.

So, my expectations have be lowered considerably since reading them.

Oh, OK.

 

My problem with your suggestion is that it was not limited to the drawing given -- it explicitly stated that the bottom resistor in a schematic is ALWAYS on the top of the expression. That's a blanket, unqualified claim that simply isn't true.



How it is usually shown is really not relevant -- do we really want people to learn to just regurgitate equations based on how something is usually shown? That just leads to them regurgitating equations wrong as soon as they encounter something that is not drawn or labeled exactly the way something is normally shown (the OP being a perfect example). I see this time and time again (and we see it here time and time again).

Hi,

Ok well what i meant was that in resistor voltage dividers the bottom resistor is always in the top of the equation. Resistor voltage dividers are usually drawn with a top and bottom resistor, but if we have the drawing then we know that the top is connected to the source and the bottom resistor is connected to ground. That's the usual connection.

But i would be interested to hear your rendition of computing a general voltage divider with two resistors where we dont have a drawing or whatever you are saying we dont have.
Of course we could use Nodal, but for two resistor voltage dividers we often give a simplification. This is what the OP was referring to.

So what formula would you suggest?

 

In the general case of an unloaded two-resistor voltage divider in which we don't have a drawing, I would describe the situation as having two resistors, Ra and Rb, in series. The voltage at the end of Ra and Va and the voltage at the end of Rb is Vb. The output, Vo, is the voltage at the junction of Ra and Rb. With this we have all that is needed to analyze the circuit in terms of these quantities (and without resorting to assertions about the resistor on the bottom always being at some place in some regurgitated formula).

The current flowing from Vb to Va is (Vb - Va)/(Ra + Rb). The voltage at Vo is then simply the voltage at one end plus the voltage drop/gain due to that current flowing through the resistance between that end and Vo. Start at whichever end you like, the results will be compatible. Either

Vo = Va + [(Vb - Va)/(Ra + Rb)]·Ra

or

Vo = Vb - [(Vb - Va)/(Ra + Rb)]·Rb

Both of these yield

Vo = (Va·Rb + Vb·Ra)/(Ra + Rb)

Notice the nice symmetry of this result -- not "top" or "bottom". Also note how it reduces trivially to the classic voltage divider equation you are referring to when Va or Vb happens to be zero.

EDIT: Fixed errors pointed out by anhnha -- thanks!

 

In the general case of an unloaded two-resistor voltage divider in which we don't have a drawing, I would describe the situation as having two resistors, Ra and Rb, in series. The voltage at the end of Ra and Va and the voltage at the end of Rb is Vb. The output, Vo, is the voltage at the junction of Ra and Rb. With this we have all that is needed to analyze the circuit in terms of these quantities (and without resorting to assertions about the resistor on the bottom always being at some place in some regurgitated formula).

The current flowing from Vb to Va is (Vb - Va)/(Ra + Rb). The voltage at Vo is then simply the voltage at one end plus the voltage drop/gain due to that current flowing through the resistance between that end and Vo. Start at whichever end you like, the results will be compatible. Either

Vo = Va + [(Vb - Va)/(Ra + Rb)]·Rb

or

Vo = Vb - [(Vb - Va)/(Ra + Rb)]·Ra

Both of these yield

Vo = (Va·Ra + Vb·Rb)/(Ra + Rb)

Notice the nice symmetry of this result -- not "top" or "bottom". Also note how it reduces trivially to the classic voltage divider equation you are referring to when Va or Vb happens to be zero.

You made some mistakes here.

Vo = Va + [(Vb - Va)/(Ra + Rb)]·Ra

or

Vo = Vb - [(Vb - Va)/(Ra + Rb)]·Rb

Both of these yield

Vo = (Va·Rb + Vb·Ra)/(Ra + Rb)

 

You made some mistakes here.

Vo = Va + [(Vb - Va)/(Ra + Rb)]·Ra

or

Vo = Vb - [(Vb - Va)/(Ra + Rb)]·Rb

Both of these yield

Vo = (Va·Rb + Vb·Ra)/(Ra + Rb)

Yep -- thanks for pointing them out. They've been fixed. That's what I get for typing and thinking at the same time.