Simple op-amp circuit help.

  • Engineering
  • Thread starter Petrucciowns
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  • #1
I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg [Broken]

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.
 
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Answers and Replies

  • #2
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Have you considered that the output might saturate?
 
  • #3
berkeman
Mentor
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I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg [Broken]

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.
Opamps are not used open-loop. You need to add feedback to have an opamp "circuit".
 
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  • #4
1,762
59
I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg [Broken]

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.
You made a slight mistake in your formula. It's a common mistake and I've noticed experienced engineers in this forum making the same mistake. The formula should be (2*V2 - V1) * Aol though opamps are rarely used open loop. This formula holds even for a standard inverting amplifier with feedback. For instance in your circuit, if you put a 1k resistor between V1 and the inverting input and another 1 k resistor between the output and the inverting input the amplifier becomes an amplifier with a gain of -1. If we put 5V at V2 and 4 V at V1, what is the output voltage? Is it 5V - 4V = 1V or is it 2*5V - 4V = 6V? Try it.

In your example, because the gains are so high your outputs were always at the + or - rail except once, when you got -200mV. Lets plug your input voltages into the formula and calculate the result. (2*1uV - 3uV) * 200K = (2uV - 3uV) * 200k = -200mV. Okay?
 
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  • #5
I am still having problems with both formulas. For example lets do the first two. First with my formula and then with yours.

V1=4mv V2=1mv

my way- (1mv-4mv)* 200k= -600 volts
your way - (2x1mv-4mv)* 200k = -400 volts

So why is the answer 9 volts? Does it have something to do with the sources?



V1=2 micro v V2=1 micro volt

my way- (1 micro volt- 2 micro volts)*200k = -200mv
This actually is the answer that is written down why?

Your way (2X 1 micro volt- 2 micro volts)*200k = -400mv
Which is not the answer written down.


Can someone try to explain to me what I am not getting?
 
  • #6
Opamps are not used open-loop. You need to add feedback to have an opamp "circuit".
Yes, this is just academic to see how one works.
 
  • #7
berkeman
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57,718
7,754
my way- (1mv-4mv)* 200k= -600 volts
your way - (2x1mv-4mv)* 200k = -400 volts

So why is the answer 9 volts? Does it have something to do with the sources?
Yep. How would the output of the opamp swing to hundreds of volts, using only +/-10V power supplies?
 
  • #8
The Electrician
Gold Member
1,249
155
So why is the answer 9 volts? Does it have something to do with the sources?
It's probably because the opamp isn't a so-called rail-to-rail output opamp. This means that when the opamp output is saturated, the outputs can't swing all the way to the rail (supply) voltages. They can't make it all the way to 10 volts; only 9 volts.
 
  • #9
Oh, I get it guys. Thanks a lot.
 

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