gain of inverting op amp?

Thread Starter

no12thward

Joined Jan 10, 2011
30
I'm looking for two resistor values from the table that will give me a gain of -18. Ive tried converting to kilo, milli, micro, nano ohms, etc, then using -(Rf/Ri). Cant seem to find any combo of resistor values on the table that will give me the gain of -18.
 

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JoeJester

Joined Apr 26, 2005
4,074
There are two number combinations from your table that will work ... how do you think you should go about and solve this problem?
 

Thread Starter

no12thward

Joined Jan 10, 2011
30
i know I'd have to change one of the resistor values from Ωs to a larger or smaller ohm conversion, but I've tried that for R1 as well as R2 resistors. Maybe im missing something?
 

Georacer

Joined Nov 25, 2009
5,154
Just look at your values more carefully. You have available all of the powers of 10 of these values. For example, you can use 1Ω, 10Ω, 100Ω etc.

Does that make it any easier?

These papers you post lately aren't the most strict and best-written exercises, granted.
 

Thread Starter

no12thward

Joined Jan 10, 2011
30
Just look at your values more carefully. You have available all of the powers of 10 of these values. For example, you can use 1Ω, 10Ω, 100Ω etc.

Does that make it any easier?

These papers you post lately aren't the most strict and best-written exercises, granted.
Sorry, it doesnt. im not seeing any power of ohm value i can use to get a gain as low as 18. I've used all the conversions
 

Thread Starter

no12thward

Joined Jan 10, 2011
30
how can i derive anything above 9.1ohm (10ohm or 100ohm) from the values on the table?
 

JoeJester

Joined Apr 26, 2005
4,074
9.1, 910, 9100, 91000, 910000, 9100000

Keep multiplying by 10.

There are two groups that satisfy your requirements.
 

Georacer

Joined Nov 25, 2009
5,154
Another hit:

If I wanted a gain of -11, I could use a resistor of 11kΩ and a resistor of 1kΩ.
 

Audioguru

Joined Dec 20, 2007
11,251
Your list is missing the standard resistance value of 1.8, 18, 180, 1.8k 18k and 180k.
A 1k input resistor and 18k feedback resistor or a 10k input resistor and a 180k feedback resistor results in a gain of 18 times.
 

Georacer

Joined Nov 25, 2009
5,154
Since this is homework, let's assume that the teacher has set the restrictions that he feels they will best test the student's knowledge.

It's rare that homework that eludes the restrictions but applies to everyday rules gets rewarded.
 

Thread Starter

no12thward

Joined Jan 10, 2011
30
I'm thinking I could use Rf=2 and Rin=1.1. But then how would I convert the ohm values to get 18?
 

Georacer

Joined Nov 25, 2009
5,154
The closest possible gain you could achieve with those coefficient would be
\[A=-\frac{2*10^1}{1.1}=-18.18\]

It is close, but you can do better.

Remember that if \[|A|=\frac{R_f}{R_i}\] then \[R_f=|A| \cdot R_i\]
You can input various values for \[R_i\] and see if you have the \[R_f\] available.
 

Thread Starter

no12thward

Joined Jan 10, 2011
30
I think were allowed a 1% error. My only question would be, with only being able to use these standard resistor values...if I use the 2ohm, the next size up would be a milliohm right? How could I use 2 X 10^1. Wouldn't that exceed my resistor values from the table?
 

Georacer

Joined Nov 25, 2009
5,154
Your don't need to insert error if you can help it.

A mili is equal to \[10^{-3}\], so in that aspect you are wrong.

Pay a bit more attention. It has been stated numerous times that the exercise wants you to use any power of 10 you want to multiply your values with. It cannot be done any other way.

I 'll give you another hint: You can set \[R_i=1.5K\Omega\]. How much will \[R_f\] be, according to the formula I gave you previously?
 

Georacer

Joined Nov 25, 2009
5,154
Please formulate complete answers.

2.7what? If I assume that you mean 2.7Ω then the gain will be \[A=-\frac{2.7 \Omega}{1.5 \cdot 10^3 \Omega}=-0.0018\] which is clearly wrong.

Try once more.
 

Georacer

Joined Nov 25, 2009
5,154
I gave you \[R_i=1.5K\Omega\], not \[R_i=1.5\Omega\].

With \[R_f=27\Omega\] you will have a gain of \[A=-\frac{27}{1500}=-0.018\], once again not acceptable.
 
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