I have read a few books befor I decided to post here.

Virtual Ground:
Most of the OP AMP circuits mention of a virtual ground. I am still not able to figure out what this means...

Some show OPAMPs to have an Ri connected between the inverting and the non inverting terminals(Digital Integrated Electronics - Taub & Schilling). While some have the two terminals grounded via resistors.

What should I consider virtual ground to be?

Feedback:
What I understand by feedback is to have the output supplying current to the input. But all the opamp circuits I have read about have a source connected to the non inverting terminal and a feedback resistor from output to input. and current flowing through this feedback resistor from the input side to the output side. The explaination is that since op-amps draw little current, the current flows to the output side via the feedback resistor.

But if the amplified output is higher than the input, should current not flow from output to the input.

Sorry, I have no diagrammes at the instance. I tried to describe the situation as clearly as possible.

Math Operation :
Are Op-Amp used in digital computers for maths operations? I have been told at college that most maths in the computers like division and multiplication are done by addition and subtraction. Is this done descretely via registers or by op-Amps? And what about complex operations like log, ln, dy/dx, integration, sine, cosine etc.??? Are data tables used?

Thanks in advance.

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

Right OPAMP theory basically states that an OPAMP will try to keep the voltage differnce bewteen the POS and NEG terminal be zero volts.
IF you go and ground one of the pins (normally the POS pin in a inverting OPAMP), by the OPAMP rules the NEG pin wants to exist at zero volts as well. Thus it is a "Virtual earth" since it really isn't connected to earth but he OPAMP will try to make that potential 0volts

Yes current will flow from the output to the input. Think of it like a voltage divider from the output to the source voltage. The output potential will chance to make sure the voltage at the joining of the resistors equals teh other terminal - most cases zer-volts

NO!!! Multiplication is just shifting left. This produces multiplication of two each step. IF you shift by one and add the original you get a multipication of 3.

No analogue in a CPU

If you shift the other way you get a divide.

As far as Trig goes two ways of doing it
1) CORDIC function (more DSP and FPGA code) or use the expanded formula of what SIN is, it is just a series of X^2/2 + X^3/3 .... kind of thing

differencdiation is just taking the difference between samples (not true mathemticaly diff)

Intergration is just summation over a sample periond, not true integration

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

Opamps are analogue, not digital - so you obviously wouldn't use them in a digital computer. In actual fact, opamps were invented for use in analogue computers - an audio virtual earth mixer is actually just an opamp 'adder' from an analogue computer.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Okay you need to be careful when speaking of Virtual Grounds and Virtual Shorts. The concept of this falls out of some assumptions we make about ideal opamps. The first assumption we make is that the input impedence is infinite (and in most practical op amps its on the order of 10^12 ohms). The second assumption we make is that the output impedence is zero. Again in most practical cases its very low.

Now with out going through the math with you (I will show it to you if it'll help you understand further), I'll try and explain the idea of virtual grounds and virtual shorts.

An op amp like someone said above tries to equalize the (+) and (-) terminals IE a virtual short, since they are the same. The virtual ground is a special case of the virtual short, it occurs when you ground the (+) terminal of the op amp and use a negative feedback configuration.

That is more of a hand waving arguement, but thats whats going on. If you'd like to see the math let me know and I'll write something up.

 

Thanks everyone for such a great explaination.

But I still do not understand about the current flowing from the input to the output in the feedback resistor.

I am sorry I made a mistake in my question. All books that I read shows the curent flowing from the input to the output. The reason was that the opamp draws little current. Makes no sense. If the output potential is positive and the input (inverting) is at zero or lower, the current shall flow from the output to the input than the other way round.

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

A very simple way to understand it is that the input signal and the feedback signal cancel each other out at the inverting input of the opamp.

As there is then effectively no signal at the junction of the two resistors, it's just as if you had connected their junction to ground - so it's a "virtual ground".

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

[quote="Nigel Goodwin"]No no no.... Not virtual ground...

The thing that is troubling me now is the direction of the feedback current. Why does it have to flow from the input to the output? Than the other way round?

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

When doing calculations, we always specify a reference positive direction. If the magnitude of the current comes out to be negative, it just means that the current is flowing in the opposite direction of what was specified.

From what you've read, they only specify the reference positive direction, but not the magnitudes. Besides, opamp calculations mostly aim to find vout/vin. The actual current direction isnt that important. What is important is that you maintain the same reference directions throughout.