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Op Amp Feedback and Design

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R3 is the opamp's load that has nothing to do with its output voltage since most opamps can drive a load as low as 2k ohms perfectly. The 5k load is nothing.
The voltage gain is 1+ (R1/R2).

Even if it was a gargantuan value such as 1MΩ, it wouldn't have any effect on the output voltage?

Thanks,

Austin
 
If the load on an opamp is anywhere from 2k ohms to infinity (nothing) then the output signal is not changed.
 
Interesting, so let's change that resistor to 1k. How would I calculate the output voltage now?

Austin
 
You must look at the datasheet of the opamp to see how high it can make the voltage on the 1k load resistor. It depends on the supply voltage and how much voltage loss occurs when the output transistors saturate.
If the load resistor is reduced to 1k then the output signal level still will not change (unless the output transistors saturate) because of the extremely high internal voltage gain of an opamp and the resulting very high negative feedback.
 
Alright then, let's consider another example in which we can solve for Vout.

Austin
 

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Your example is really a resistor divider problem, not a opamp thing... the R1, R2 node is 1V... what is the Vin (output of opamp) needed to make that true? Answer that question and you just solved for the opamp output voltage.
 
I suppose it isn't all that difficult, but what would be a complicated resistor network type problem that I need to learn how to solve?

How many resistors are typically used when designing with op amps? I assume 2 is probably the most common quantity, but what's the usual quantity?

My goal is to take the schematic form of any op amp circuit and anyalyze it in mathematical (forming the equations as a function of the input and gain) form.

Thanks,

Austin.
 
I got the hint that you wanted me to post here, so here's what I have to say. When I design using op-amps, I usually use at least 2 resistors, however you can ge away with using no resistors at all (make the op-amp a comparator). I can't see any practical point in using more than 7 resistors in an op-amp circuit.

If I can add my two sense in, I think that you may be overthinking this just a little bit. :D Like I said; calculating the cut off frequency using the slew rate is the hardest part of op-amps, but I've never used this because my op-amps only have to tolerate very low frequency opperations (μHz).

Also, why do you need to analise op-amps with math? Are you making an op-amp simulator with a PIC, or is this for the little do-hickie you showed me and Mike?
 
Read 'The Art of electronics' This book has answered all my questions.
 
I suppose it isn't all that difficult, but what would be a complicated resistor network type problem that I need to learn how to solve?

How many resistors are typically used when designing with op amps? I assume 2 is probably the most common quantity, but what's the usual quantity?

My goal is to take the schematic form of any op amp circuit and anyalyze it in mathematical (forming the equations as a function of the input and gain) form.

Thanks,

Austin.


Hi again,

I posted a method for analyzing any op amp circuit in this thread already back a few posts or so, but you seem to have overlooked it completely. You should take another look as that will help you get through just about any circuit out there using a standard op amp...
Vout=(Vinv-Vnoninv)*Ao

The circuit you have posted is a bit overcomplicated for your first op amp circuit. You should look at some of the two resistor op amp circuits and try to understand them first before moving to other circuits that involve more concepts like that. The circuit you posted also requires some other circuit techniques which will detract you from the main theme of understanding the op amp better, so go back to the two resistor circuits first then go from there. You may also find that you have to go back and look at some basic circuit analysis concepts in the future too like for example superposition.
 
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I got the hint that you wanted me to post here, so here's what I have to say. When I design using op-amps, I usually use at least 2 resistors, however you can ge away with using no resistors at all (make the op-amp a comparator). I can't see any practical point in using more than 7 resistors in an op-amp circuit.

If I can add my two sense in, I think that you may be overthinking this just a little bit. :D Like I said; calculating the cut off frequency using the slew rate is the hardest part of op-amps, but I've never used this because my op-amps only have to tolerate very low frequency opperations (μHz).

Also, why do you need to analise op-amps with math? Are you making an op-amp simulator with a PIC, or is this for the little do-hickie you showed me and Mike?

Hi Dragon_Tamer,

I admit I have a knack for making things harder than they are; somehow I expect things to be more difficult. I suppose my demeanor should represent a more "simplistic" attitude.

If I can't analyze op-amps mathematically, why analyze them at all? Using math I can select the parameters I want such as my Vout, gain, etc.

I plan to design a waveform generator using op amps, and also perhaps a PSU.
 
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Hi again,

I posted a method for analyzing any op amp circuit in this thread already back a few posts or so, but you seem to have overlooked it completely. You should take another look as that will help you get through just about any circuit out there using a standard op amp...
Vout=(Vinv-Vnoninv)*Ao

The circuit you have posted is a bit overcomplicated for your first op amp circuit. You should look at some of the two resistor op amp circuits and try to understand them first before moving to other circuits that involve more concepts like that. The circuit you posted also requires some other circuit techniques which will detract you from the main theme of understanding the op amp better, so go back to the two resistor circuits first then go from there. You may also find that you have to go back and look at some basic circuit analysis concepts in the future too like for example superposition.

Hi MrAl,

I apologize for not responding! I just recently discovered that equation you posted in a MIT lecture video: Vout = A(V+ - V-). I see how it can be used to apply mathematically, thanks!

I took your advice and I now know how to apply equations to non-inverting and inverting amplifiers.

Thanks!
 
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