Operational amplifier exam questions

[Samtheguy]

Homework Statement

See attached file for relevant questions. I'm finding the Op-amp questions in section A very difficult. Any help with these questions would be much apprietiated.

Homework Equations

The Attempt at a Solution

 

[uart]

I think you should choose one particular question (part) to ask about. You can always ask another question later.

 

[Samtheguy]

Okay, that seems more sensible.
I have attemped Question 1. Am I correct in saying that you would draw the Inverting Op-amp circuit diagram, remove the op-amp component since the inputs have a very large input impedance (and thus no current flows into them), and label the virtual Earth point between the feedback and input resistors?
Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10

 

[uart]

Also, the gain would be given by -Rf/Rin, therefore, -1M/100K = -10

Yes that is correct, however this question is more about frequency response and gain bandwidth product.

From the question see if you can figure out the following.

- What is the opamp DC gain (without feedback)
- What is the opamp 3dB bandwidth (dominant pole)
- What is the opamp gain bandwidth product.

* BTW. There is a bit of a trick with this question in that the relevant closed loop gain (for calculating gain-BW product) for the inverting opamp is actually 1+|A_cl|.

 

[Samtheguy]

Thanks for your reply. To caluate the DC gain (without feedback) would I use:
A_d = 2x10^5 / (1+jf/7) and set f=0 since it is DC. Therefore, I get A_d = 2x10^5. In dB, this is a gain of 106.02 dB. To find the -3dB point, do 106.02 - 3 = 103.02dB and convert back to magnitude using, 10^(103.02/20) = 141579.
Am I on the right path to the solution?

 

[uart]

Thanks for your reply. To caluate the DC gain (without feedback) would I use:
A_d = 2x10^5 / (1+jf/7) and set f=0 since it is DC. Therefore, I get A_d = 2x10^5. In dB, this is a gain of 106.02 dB. To find the -3dB point, do 106.02 - 3 = 103.02dB and convert back to magnitude using, 10^(103.02/20) = 141579.
Am I on the right path to the solution?

Yep A_0 = 2x10^5 is good. :)

The -3dB point does occur at approx 141400 (gain) but that's a really long winded way to go about finding it. The -3dB points are defined as having a gain of 1/sqrt(2) times the maximum, so you could just divide 2x10^5 by sqrt(2) to get there more easily.

More importantly however you don't actually need to find that -3dB gain (141400) in order to find the -3dB frequency. Just look at the transfer function A_d = 2x10^5 / (1 + j (f/7) ) and tell me what value of "f" you need to make the magnitude of the denominator equal to sqrt(2).

Hint : What is the magnitude of (1+j) ?

 

[Samtheguy]

The magnitude of (1+j) is sqrt(2) or 1.414. So therefore f would be 7Hz at the -3dB point.

 

[uart]

The magnitude of (1+j) is sqrt(2) or 1.414. So therefore f would be 7Hz at the -3dB point.

Yep that's correct. So therefore the gain bandwidth product is 7x2x10^5 = 1.4x10^6 Hz.

 

[Samtheguy]

Okay. So Gain-bandwidth product = Open-loop gain X cut-off frequency. Is this the answer to the small signal bandwidth or is there a lot more computation to do? (I'm guessing there is more to do)

 

[uart]

Okay. So Gain-bandwidth product = Open-loop gain X cut-off frequency. Is this the answer to the small signal bandwidth or is there a lot more computation to do? (I'm guessing there is more to do)

No there's not much more to it, that's the simplicity and beauty of the "gain bandwidth product" method.

The open loop gain times the open loop -3dB freq is approx equal to the closed loop gain times the closed loop frequency. That is, the gain bandwidth product is approximately a constant.

BTW. The gain bandwidth product is most accurate for the non-inverting configuration, so make sure you review my hint in reply #4 above.