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Daniel Ka

Other Parts Discussed in Thread: OPA827

Hi there,

I'm just going through the tutorials regarding Op Amps. I've a question regarding one video of the course. I hope this is the right place to ask these questions.

In video 5.1 "Bandwidth", regarding the "phase shift response" of the second example, at about16:41 the author states that

"At this point, the phase begins to decrease at minus 45 degrees per decade because of the zero."

Shouldn't that be "pole" instead? Phase begins to change one decade before the pole with a slope of minus 45 degrees per decade.

Thanks?

Best regards.

Dan

over 2 years ago

0 Ron Michallick over 2 years ago

TI__Guru*** 140186 points

Hello Dan,

Yes, at 16:41 that should have been 'pole'. However, there were many times when 16kHz was called a pole. This means you understand poles and zeros well.

0 Daniel Ka over 2 years ago in reply to Ron Michallick

Intellectual 497 points

Hi Ron,

Thanks for answering my question. I'm glad that I understand poles and zeros well, now

By the way, I also completed the short answer exercise and I think there's also a minor typo there. On pages 8 and 9, the gain curve beyond 100kHz should be labeled with "-20db/decade". That's also what the calculation shows.

Now while we are at it, I wonder why the circuit simulated in this course (see page 2) does not provide a perfect symmetrical output:

The course explains why the amplitude is less than expected (closed loop gain follows Aol and that decreases gain) it does not explain why the positive and negative maximum of Vout are not symmetrical to zero.

Could you please explain this or will this be revealed in a different course?

+1 Ron Michallick over 2 years ago in reply to Daniel Ka

TI__Guru*** 140186 points

Daniel,

-20 -20 + 20 = -20, so it should say -20dB/dec

The asymmetry is a DC cause. The op amp input offset error gets 100V/V gain (because AOL at DC is very high). The average output is 7.48mV; divide that by 100 and you get the actual DC input offset error, 74.8uV

0 Daniel Ka over 2 years ago in reply to Ron Michallick

Intellectual 497 points

Hello Ron,

Thanks for the hint. So basically you are saying that the datasheet of OPA827 shows an input offset voltage of typ. 75 uV

This value times the DC gain of 100 V/V will give 7.5 mV.

7.5 mV is almost exactly the offset seen in the diagram above: (84.45 mV - 69.49 mV) / 2 = 7.48 mV

Thanks.

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