Differentiating Op-Amp basic exercise

[technician]

Put some values in and follow it through.
Suppose the battery connected to the + input is 2V and the input resistor is 1kΩ.
The feedback resistor we will have as 2kΩ and the resistor on the output we will have as 500Ω
The volts at the + and - inputs must be equal (remember) so there is a voltage of 2V across this 1kΩ resistor. This means a current of 2/1000 A flows down the 1kΩresistor. This current must come from the output of the amplifier and flows through the 2kΩ resistor.
This means that the voltage across the 2kΩ resistor = 2x10^-3 x 2000 = 4V.
The voltage at the -input is 2V so the voltage at the output of the amplifier =2 + 4 = +6V
This means there is a current through the output resistor (you could call it the 'load' resistor)
of 6/500 = 0.012A.
What do you think of that?

 

[Femme_physics]

I'll take you on step by step hitting the second reply later.

How come it's an INVERTING amplifier if the only voltage source is connected to the MINUS leg? Doesn't it make it inverting?

 

[I like Serena]

Yes.
It's an inverting amplifier, because it looks most like this picture:

The fact that there is an additional resistor (and battery) after the circuit, and that a battery is used to change the ground potential, does not change the fact that it is an inverting amplifier.

 

[technician]

The -input is also known as the 'inverting' input so if the input is connected to the - input it is automatically an inverting amplifier. It is just a name, don't worry too much about the names...can you follow the voltages and currents...that is the main thing.

 

[I like Serena]

I'll take you on step by step hitting the second reply later.

How come it's an INVERTING amplifier if the only voltage source is connected to the MINUS leg? Doesn't it make it inverting?

When the voltages are worked out, the resulting formula is:
V_{out} = -{R_f \over R_{in}}V_{in}

Note the minus sign?
The minus sign implies it is inverting - it inverts the sign of the voltage.

 

[Femme_physics]

When the voltages are worked out, the resulting formula is:
V_{out} = -{R_f \over R_{in}}V_{in}

Note the minus sign?
The minus sign implies it is inverting - it inverts the sign of the voltage.

Alright, perfectly clear. I guess the "ground" sign confused me. I actually have this circuit in an exercise, I will try to solve it and see if i struggle before I turn to the "sages"...spasiva bolshoy, you two :)

 

[Femme_physics]

this seems about right to me, yes? (ignore the part of the sketch where it says 18)

http://img847.imageshack.us/img847/8926/scan750.jpg

 

[technician]

FP
I see your working and it is correct. Looks neat and well organised

 

[Femme_physics]

FP
I see your working and it is correct. Looks neat and well organised

Really? I thought so too! I don't know why my teacher told me I was mistaken. I showed him the exact page yesterday. Are you 100% sure? I'll be on his case if you are.

 

[technician]

Well... here is the way I approach all of these problems
1) Voltage at the + input = 2V (that battery)
2) Voltage at the - terminal must also be 2V
3) R1 is 2KΩ and has 2V across it. The LH end of R1 is connected to 0volts... ground or earth
4) Current through R1 = 2V/2kΩ = 1mA
5) This current must come from Vout through R2...therefore Voltage across R2 = 1mA x 10kΩ
=10V
6) The LH end of R2 (-input) is at +2V therefore Vout must be 2V + 10V =12V
7) V across LED = 1.4V therefore V across 2kΩ = 12 - 1.4 = 10.6V
8) Current through 2kΩ and the LED = 10.6/2kΩ = 5.3V

 

[Femme_physics]

Right! That's what I thought!

So, I'll be on his case, then! Just you wait. I'm not giving up that easily now. I'll make him write to me his solution.

 

[technician]

I will be interested to see it.
Make certain you post it for me.

 

[technician]

The only possible criticism I can see is that you have the gain of the inverting amplifier as (R1+R2)/R1.
Normally you would expect to see just R2/R1 but with that battery connected to the +input your equation is correct.