op-amp basics

So I'm reading up and trying to understand op-amps. I would like to actually have some comprehention to this instead of just bumbling around.

I'm sure it's simple but it's just not sinking in.

I was following the example at this site...
**broken link removed**

I was just wondering if someone could explain the math in the formula for the relationship between gain/Rin/Rf/Vin/Vout?
In example #1 they seem to drop the - sign... why is that? Does it have something to do with it being on the inverting input?

Example:
**broken link removed**

But then the answer to the problem is not negative. it's just 10.

-Rf/Rin = 100k/10k = 10.

Click to expand...

Shouldn't it be -100k/10k = -10 ???

Could someone explain all this?

The negative sign shows that the amplifier inverts.
The input of an opamp draws an extremely small current. Opamps with FET inputs have inputs that draw no current.
The internal gain of an opamp is very high at 200,000 or more so both input voltages are trying to be exactly the same.

If the (+) input of an opamp is 0V then the (-) input will try to be 0V. Then if the input to a 1k input resistor at the (-) input is +1V then its current is 1mA. The output will be -10V if the feedback resistor is 10k because the 1mA in the input resistor will also be in the feedback resistor and 1mA x 10k ohms is 10V.

The formula for gain starts from the basic equations for the circuit.
Using that link's drawing, V2 is the inverting terminal and V1 is the
non inverting terminal, and Vout is the output voltage and RF is the
feedback resistor and Rin is the input resistor and Vin is the input
voltage.

Writing the equation for the voltages of the op amp we get:
Vout=(V1-V2)*AOL
where
AOL is the open loop gain of the op amp (typically 1000 or more).

Now writing the equation for V2 we get:
V2=(Vout-Vin)*Rin/(Rin+RF)+Vin

Now substitute V2 into the equation for Vout and we get:
Vout=(V1-((Vout-Vin)*Rin/(Rin+RF)+Vin))*AOL

Now eliminate the inside parens by distributing the minus sign and we get:
Vout=(V1-(Vout-Vin)*Rin/(Rin+RF)-Vin)*AOL

Now distribute the leftmost minus sign and we get:
Vout=(V1+(Vin-Vout)*Rin/(Rin+RF)-Vin)*AOL

Now distribute the Rin/(Rin+RF) factor and eliminate those parens:
Vout=(V1+Vin*Rin/(Rin+RF)-Vout*Rin/(Rin+RF)-Vin)*AOL

Now distribute the AOL and eliminate those parens:
Vout=AOL*V1+AOL*Vin*Rin/(Rin+RF)-AOL*Vout*Rin/(Rin+RF)-AOL*Vin

Now to explicitly solve for Vout get all terms with Vout to the left side:
Vout+AOL*Vout*Rin/(Rin+RF)=AOL*V1+AOL*Vin*Rin/(Rin+RF)-AOL*Vin

Now isolate Vout:
Vout(1+AOL*Rin/(Rin+RF))=AOL*V1+AOL*Vin*Rin/(Rin+RF)-AOL*Vin

Now divide both sides to get Vout alone on the left side:
Vout=(AOL*V1+AOL*Vin*Rin/(Rin+RF)-AOL*Vin)/(1+AOL*Rin/(Rin+RF))

Now we simplify that equation a little and we get:
Vout=(AOL*RF*V1+Rin*AOL*V1-Vin*AOL*RF)/(RF+Rin*AOL+Rin)

And now we have the equation for Vout, but that is the exact expression
and many times we want to deal with the approximate expression because
it simplifies the math and it still works out ok in practice. To get the
approximate expression we assume that the open loop gain AOL is infinite,
so we take the limit of Vout as AOL approaches plus infinity:
Vout(approximate) = limit [AOL --> +inf]((AOL*RF*V1+Rin*AOL*V1-Vin*AOL*RF)/(RF+Rin*AOL+Rin))

and after taking the limit we get:
Vout(approx)=(RF+Rin)*V1/Rin-Vin*RF/Rin

That's the general approximate equation for Vout, but since for this application
V1 happens to be zero, we can substitute zero for V1:
Vout(approx)=(RF+Rin)*0/Rin-Vin*RF/Rin

and since that eliminates one term that leaves us with the last term only:
Vout(approx)= -Vin*RF/Rin

To get the gain alone we simply divide both sides by Vin and we get:
Vout/Vin= -RF/Rin

WOW!

Thanks for going over that. It has been a while since I did anything with limits. Takes me back to the days of Calculus.

I appreciate you going over the math. I ain't dun ta good wit da numbers.

I'll go over this and see if it helps me better understand these topics.

Thanks again, I really appreciate it.

N

redepoch7 said:

WOW!

Thanks for going over that. It has been a while since I did anything with limits. Takes me back to the days of Calculus.

I appreciate you going over the math. I ain't dun ta good wit da numbers.

I'll go over this and see if it helps me better understand these topics.

Thanks again, I really appreciate it.

N

Click to expand...

You're welcome