Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

what is the significance of using Inverting Op-amp?

Status
Not open for further replies.

786

Newbie level 6
Joined
Aug 11, 2008
Messages
12
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Location
India
Activity points
1,380
what is the significance of using Inverting Op-amp instead of using Non-Inverting Op-amp?
Does it realy makes any difference in amplification or what else?
In one the design i found that, Inverting op-amp was used: +ve pin was grounded & -ve volatge was provided at -ve pin to get o/p in +ve form.
Here in this, why we can't simply use non-inverting amplifier & GND the -ve pin &
fed +ve volatge at +ve pin to get +ve o/p.

Can any body reply?
Thanks in advance
 

The non inverting amplifier has better input impedence , infact the voltage follower has the highest impedence. . Hence for all sensors the first stage tends to be a voltage follower (generally called the buffer). in my view it is better to use non inverting than inverting.
 

TI has a very good handbook on OmAmp applications, you can get it here.
I think that the better configuration, depends on your particular aplication. You can take a look here to see the diferences betwen each configuration. Also at the end of the page, you can see a list of good references.

Regards.
 

pauloynski said:
Non-inverting circuits have high input impedance while inverting circuits have almost zero input impedance.

The last part of the sentence is not correct.
The input impedance of the inverting opamp is (nearly) identical to the first resistor (between signal input and inverting opamp terminal).

Concerning "significance" of inverting/non-inv. circuits:
In many cases (filters, oscillators, controllers,..) you've got no choice. The application requires either inverting or non-inverting gain.
 

No. As seen by the opamp´s input terminal the impedance is almost zero. If you place a series resistor that´s a choice of yours (or needed by the circuit).
 

pauloynski said:
If you place a series resistor that´s a choice of yours (or needed by the circuit).
That's not just straight true. If you don't place a series resistor you have a transimpedance amplifier, so it's not just a "choice", you're converting an voltage amplifier (inverter) to a current to voltage amplifier (inverter), you're changed completly the circuit behavior. Look here.

Regards.
 

I agree with you but placing a series resistor converts the input voltage to a current. In this way a transimpedance amplifier is not so different (just another way of seeing what happens). For me we are talking about similar things. This pdf shows both circuits (see figure 1b) **broken link removed** In any case, it is the application that will demand an input resistor or not. Discussing differences between an inverter and a transimpedance amplifier is out of the scope of this post, for me it is enough.
Regards
 

pauloynski said:
In this way a transimpedance amplifier is not so different (just another way of seeing what happens). For me we are talking about similar things.
I agree with you, but he/she is asking about the main differences between both configurations, I just want to make to her/she clear, that without this resistor, the circuit is not exactly the same.

Regards.
 

One other aspect is common mode on the op amp input. The inverting has none and so the common mode rejection ratio not being infinite does no harm.
 

With inverting configuration, any transfer function can be readily achieved. That is not the same for non-inverting configuration. A good example is an "ideal" integrator.
Regards
Z
 

zorro said:
With inverting configuration, any transfer function can be readily achieved. That is not the same for non-inverting configuration. A good example is an "ideal" integrator.
Regards
Z

Sorry, but that`s not true.
Of course, there are also non-inverting integrator topologies:
BTC, NIC (DEBOO) and the Phase-lead integrator.
 

LvW said:
zorro said:
With inverting configuration, any transfer function can be readily achieved. That is not the same for non-inverting configuration. A good example is an "ideal" integrator.
Regards
Z

Sorry, but that`s not true.
Of course, there are also non-inverting integrator topologies:
BTC, NIC (DEBOO) and the Phase-lead integrator.

I don't mean that it is not possible to synthesize a transfer function using OA's in noninverting configuration.
What I meant is that, in single-feedback inverting configuration, as the V/V transfer function is just the the quotient of two transfer impedances, the desired transfer function is easily synthesized.
For that task there exist good tables of two-port nets (like those included in Giacoletto's handbook).
With non-inverting configuration the task is not so direct and the topology in general is not so simple. For example, in the case of the Deboo integrator, not only it uses more componentes, but if the balance between the resistor ratios is not exact the pole moves from s=0, making the integrator less "ideal".
Of course, each topology has its pros and cons. And some of the most popular use OA's in non-inverting configuration (e.g. Sallen-Key).
Regards

Z
 

Here are basic formulas for operational amplifiers:
 
Last edited by a moderator:

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top