Understanding the Behavior of Inverting Op Amp Circuit with Switched Terminals

In summary, the conversation discusses a circuit and its transfer function. It is mentioned that reversing the op amp terminals will change the behavior of the circuit and the transfer function will no longer apply. The conversation also touches on the assumptions made when analyzing the circuit and the possibility of analyzing it without prior knowledge. There is also a discussion about a balanced amplifier and its use as a differential amplifier.
  • #1
david90
312
2
please look at this circuit
http://hyperphysics.phy-astr.gsu.edu/Hbase/Electronic/ietron/inva2.gif

The transfer function for the the circuit above is

Vout = (-Rf/Rs)*Vs

if I were to switch the terminal of the op amp so that Vin is inputted into Vp and Vn =0 instead (everything is the same except that Op amp terminal Vp is now Vn and Vn and is now Vp), how does it change the behavior of the circuit? The transfer function is still the same but I doubt the behavior of the circuit is the same. Or is it?
 
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  • #2
No, it isn't the same and that formula doesn't apply any more.

It is just the formula for an inverting amplifier and assumes the DC conditions and other connections are correct for an inverting amplifier.

If you reverse the inputs to the chip or reverse the power connections, the transfer function does not apply.
 
  • #3
What are the DC assumptions? I still don't understand why if you switch the op amp terminals then the transfer function becomes invalid.

if Vout = (-Rf/Rs)*Vs is not the right transfer function, then what is the new transfer function? Will the circuit work at all?
 
  • #4
You wouldn't expect the transfer function to apply if the opamp had no power would you?
What if you reversed the power supply?

The transfer function only applies if you have the circuit functioning correctly as an inverting amplifier.

If you did reverse the input connections, the transfer function would change. It looks like a Schmitt trigger circuit that way, as you would get positive feedback.


400px-Opampschmitt_xcircuit.svg.png


You might like to read about Schmitt Trigger circuits here:
http://en.wikipedia.org/wiki/Schmitt_trigger
 
Last edited:
  • #5
vk6kro said:
You wouldn't expect the transfer function to apply if the opamp had no power would you?
What if you reversed the power supply?

The transfer function only applies if you have the circuit functioning correctly as an inverting amplifier.

If you did reverse the input connections, the transfer function would change. It looks like a Schmitt trigger circuit that way, as you would get positive feedback.
400px-Opampschmitt_xcircuit.svg.png


You might like to read about Schmitt Trigger circuits here:
http://en.wikipedia.org/wiki/Schmitt_trigger
Say somebody give you the circuit above and you have no idea what it does. How do you know if you can analyze it by writing the transfer function using the assumption vp=vn and In=Ip=0 (op amp terminal current)?
 
  • #6
The transfer function for this circuit would be discontinuous and also depend on the output voltage.

You can analyse it, though. You assign actual values to the resistors and then assume the output is at the negative supply voltage. Suppose you make R2 = 90 K and R1 = 10 K and the supply is +/- 9 volts.

Then you start increasing the input voltage from the negative supply voltage towards the positive supply voltage.

The two resistors form a voltage divider between the input and the output.

At some point, (when the input is greater than +1 volt) the +input will go positive and the output will swing positive. This will swing the +input even more positive.

If you now reduce the input and keep doing it until the + input goes slightly negative, (when the input is less than -1 volt) the output will go negative again.

Try it and you will understand this circuit a bit better.
 
  • #7
david90 said:
Say somebody give you the circuit above and you have no idea what it does. How do you know if you can analyze it by writing the transfer function using the assumption vp=vn and In=Ip=0 (op amp terminal current)?

Unfortunately if they have no idea what it does beforehand one doesn't know for certain that they can. But, as it is the simplest case to analyze (and it is often right), you can just assume that you can. If you arrive at a contradiction (say vo depends on more than just vi, or you get two equations which can't both be right, for instance vo=2vi and vo=3vi) then you know your initial assumption was wrong and you can start over knowing that vp!=vn but In=Ip=0 (for the ideal opamp).

In the schmitt trigger example vp!=vn for all vi.
 
  • #8
vk6kro said:
The transfer function for this circuit would be discontinuous and also depend on the output voltage.

You can analyse it, though. You assign actual values to the resistors and then assume the output is at the negative supply voltage. Suppose you make R2 = 90 K and R1 = 10 K and the supply is +/- 9 volts.

Then you start increasing the input voltage from the negative supply voltage towards the positive supply voltage.

The two resistors form a voltage divider between the input and the output.

At some point, (when the input is greater than +1 volt) the +input will go positive and the output will swing positive. This will swing the +input even more positive.

If you now reduce the input and keep doing it until the + input goes slightly negative, (when the input is less than -1 volt) the output will go negative again.

Try it and you will understand this circuit a bit better.

I have a college textbook on op amp and it seems like circuits with Vo that feedback to Vn can be analyzed using vp=vn, ip=in=0. Is my observation correct?
 
  • #9
Did you understand how the Schmitt trigger worked?
 
  • #10
Here in thumbnail is a balanced amplifier. R2=R3 and R1=R4. You can use either pos or neg input, and ground the other. This circuit minimizes input bias current offsets.

Show that if this amplifier is used as a differential amplifier, the output common-mode signal is zero; i.e., Vout = G(Vpos - Vneg)

Bob S
 

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  • #11
The transfer function for the the circuit above is

Vout = (-Rf/Rs)*Vs



Try this:
Rf = 100 K
Vs = 2 volts
Rs = 1 K

So, Vout = ?

This is a problem with blindly following transfer functions.
 
  • #12
Bob S said:
Here in thumbnail is a balanced amplifier. R2=R3 and R1=R4. You can use either pos or neg input, and ground the other. This circuit minimizes input bias current offsets.

Show that if this amplifier is used as a differential amplifier, the output common-mode signal is zero; i.e., Vout = G(Vpos - Vneg)

Bob S

So what are you trying to say?
 
  • #13
vk6kro said:
Did you understand how the Schmitt trigger worked?

Yes I know how it functions but I haven't try to understand why it behaves the way it does.
 

1. How does the inverting op amp circuit work?

The inverting op amp circuit is a type of electronic amplifier circuit that uses an operational amplifier (op amp) to amplify a signal. It works by taking an input signal, inverting it, and then amplifying it. The output signal is the amplified and inverted version of the input signal.

2. What are the components needed for an inverting op amp circuit?

The main components needed for an inverting op amp circuit are an operational amplifier, resistors, and a power supply. The op amp acts as the amplifier, the resistors are used to set the gain of the circuit, and the power supply provides the necessary voltage for the circuit to operate.

3. How do you calculate the gain of an inverting op amp circuit?

The gain of an inverting op amp circuit can be calculated using the formula: Gain = - (Rf/Rin), where Rf is the feedback resistor and Rin is the input resistor. This means that the gain of the circuit is determined by the ratio of these two resistors. For example, if Rf is 10kΩ and Rin is 1kΩ, the gain would be -10.

4. Can an inverting op amp circuit be used for both amplification and attenuation?

Yes, an inverting op amp circuit can be used for both amplification and attenuation. The gain of the circuit can be adjusted by changing the values of the feedback and input resistors. If the feedback resistor is larger than the input resistor, the circuit will provide amplification. If the feedback resistor is smaller than the input resistor, the circuit will provide attenuation.

5. What are the advantages of using an inverting op amp circuit?

There are several advantages of using an inverting op amp circuit. It has a high input impedance, meaning it does not draw much current from the input signal source. It also has a low output impedance, making it suitable for driving other circuits. Additionally, it has a wide frequency response and can be easily configured for different gain values.

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