©©©

Branch Unspecified

07 Feb 2009

**Inverting Op-amp questions.**

Hi there, can anyone tell me how to figure out this question? the confusing part for me is after getting the gain value -(56K/8K), the gain circuit suddenly cross into an intersection between 2 resistors ; 8K & 2K.

Im not sure what to do after that. can anyone help me out? thank you.

Im not sure what to do after that. can anyone help me out? thank you.

silverscorpion

Branch Unspecified

07 Feb 2009

well, apparently, the 56K resistor is not the only feedback resistor. There is another parallel combination of 8K and 2K. So, that would form another 4K and so the total feedback is 60 K. so, the total gain is, -60/8 or -7.5.

so, V0 is -1.5V. I think this is the correct answer.

so, V0 is -1.5V. I think this is the correct answer.

©©©

Branch Unspecified

07 Feb 2009

Hey there, thanks alot - that's much help for a simple clarification. Been trying to figure it out for a day after checking out lots of uneccasary references, I can't believe its that simple again. -_- .

reachrkata

Branch Unspecified

12 Feb 2009

I would beg to differ. This is how I understand the circuit -

Considering that the voltage at -ve pin of op-amp is 0 (virtual GND concept) and the input current into opamp is very low (negligible), the effective current at -ve terminal of the opamp would be zero.

That means 0.2/8k = (0 - Voltage @ intersection of 8k and 2k)/56k

=> 0.2/8k = (-Vout*2k/(2k+8k))/56k

From this we get Vout = -7V.

Please check if this answer is right.

@Scorpion - the 8k and 2k are not in parallel.

-Karthik

😁

Considering that the voltage at -ve pin of op-amp is 0 (virtual GND concept) and the input current into opamp is very low (negligible), the effective current at -ve terminal of the opamp would be zero.

That means 0.2/8k = (0 - Voltage @ intersection of 8k and 2k)/56k

=> 0.2/8k = (-Vout*2k/(2k+8k))/56k

From this we get Vout = -7V.

Please check if this answer is right.

@Scorpion - the 8k and 2k are not in parallel.

-Karthik

😁

priyank_180

Branch Unspecified

13 Feb 2009

i think, only 56k+8K resistors are take part in feedback path., 2K resister will be neglated, bcoz of ground connection.

so your ans will be [{-(64/8)}*0.2]=1.6v

so your ans will be [{-(64/8)}*0.2]=1.6v

priyank_180

Branch Unspecified

13 Feb 2009

I think only 56K+8K resisters will take part in feedback path, and 2K will be neglated bcoz of ground connection, so your ans should be [{-64K/8K}*(0.2v)]=-1.6V

josenit1787

Branch Unspecified

14 Feb 2009

Hi,

I think reahrkata has come closest to the answer in his analysis of the circuit.But I think it is slightly in error because he made a mistake calculating the Voltage @ intersection of 8k and 2k by not considering the effect of the 56K in Parallel with the 2k.

Considering this also the Voltage @ intersection of 8k and 2k comes out to be equal to -1.4V.Hence the output voltage will be -1.4-.725*8=

This is the answer to your question.

I think reahrkata has come closest to the answer in his analysis of the circuit.But I think it is slightly in error because he made a mistake calculating the Voltage @ intersection of 8k and 2k by not considering the effect of the 56K in Parallel with the 2k.

Considering this also the Voltage @ intersection of 8k and 2k comes out to be equal to -1.4V.Hence the output voltage will be -1.4-.725*8=

**-7.2V**This is the answer to your question.

reachrkata

Branch Unspecified

14 Feb 2009

Whoopsy !!! Thats right ! I did miss out the 56k in parallel to 2k.

You are right josenit. At least we both are in agreement with regards the concept. He He! 😁

@©©© - Perhaps you can clear the confusion by giving us what the right answer is.

-Karthik

You are right josenit. At least we both are in agreement with regards the concept. He He! 😁

@©©© - Perhaps you can clear the confusion by giving us what the right answer is.

-Karthik

josenit1787

Branch Unspecified

15 Feb 2009

Hi Karthik,

Thanks for your prompt reply.

Regards,

Jose.

Thanks for your prompt reply.

Regards,

Jose.

priyank_180

Branch Unspecified

15 Feb 2009

I think, there is no need to consider 2K resister... bcoz it allready grounded.!!!!

josenit1787

Branch Unspecified

16 Feb 2009

priyank_180I think, there is no need to consider 2K resister... bcoz it allready grounded.!!!!

Dear Priyank,

I think you have got some concepts of circuit theory wrong.I cannot understand what you mean by the idea of completely neglecting a resistance in the analysis of a circuit, simply because it is grounded on one side.

Ok..Consider that in the same circuit I am replacing the 2k resistor with a 0k resistor, i.e.I am shorting that node to the ground..Do you think the answer will still be the same since we can neglect the 0k resistor as it is already grounded?

Regards.

Jose.

raithrovers1

Branch Unspecified

09 Mar 2009

What you have is two important building blocks for any circuits. An op amp with negative feedback and potential divider resistor network.

If you take the op amp part then you could say that the gain of the op amp circuit is 56k/ 8k which is 7.

The problem now is that the feedback is also connected to the potential divider. Therefore a percentage of the voltage will be dropped across the 8k resistor and the remaining across the 2K resistor (As the 56k is connected to "virtual earth" at the -ve input to the op amp this has to be considered as being in parallel with the 2k).

Therefore if the gain is seven without the potential divider it would give a -1.4V output.

But as the 2K resistor in parallel with 56k (use product of sums) only has 0.194 of the output voltage at the mixing point you will have to divide -1.4 by 0.194. This gives an output voltage Vo of approximately -7.2V.

You can then see how it works. If the 8k resistor drops 0.806 of the output voltage that equals -5.8V. This means that at the top 2K resistor you will have -1.4V. The gain is 56K/8K which is 7. Therefore -1.4v/7 = 0.2V for the input.

I hope I haven't confused matters more with my explanation but if you apply the values to the circuit you will see that it is correct.

For all UPS and power needs contact Calibre Power at Calibre Power Electronics, UPS, uninterruptible power supply, home page

If you take the op amp part then you could say that the gain of the op amp circuit is 56k/ 8k which is 7.

The problem now is that the feedback is also connected to the potential divider. Therefore a percentage of the voltage will be dropped across the 8k resistor and the remaining across the 2K resistor (As the 56k is connected to "virtual earth" at the -ve input to the op amp this has to be considered as being in parallel with the 2k).

Therefore if the gain is seven without the potential divider it would give a -1.4V output.

But as the 2K resistor in parallel with 56k (use product of sums) only has 0.194 of the output voltage at the mixing point you will have to divide -1.4 by 0.194. This gives an output voltage Vo of approximately -7.2V.

You can then see how it works. If the 8k resistor drops 0.806 of the output voltage that equals -5.8V. This means that at the top 2K resistor you will have -1.4V. The gain is 56K/8K which is 7. Therefore -1.4v/7 = 0.2V for the input.

I hope I haven't confused matters more with my explanation but if you apply the values to the circuit you will see that it is correct.

For all UPS and power needs contact Calibre Power at Calibre Power Electronics, UPS, uninterruptible power supply, home page

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