Op Amp resistor question (hwk) - Page 1
Forum section: electronics / beginners

author: mg84
author_rank: Newbie
timestamp: January 23, 2015, 10:10:03 pm
content: So I am in the middle of my first homework assignment for a microelectronics class and even though I have the correct answer for the problem, I have a question:In the attached picture we were to find the input resistance, and gain (20k, -10). In the previous question the op-amp had the same answer, even though the 20k by the output was not there. My question is, in practice would there be any reason for having a resistor pull to ground here? is it some sort of bleed-off resistor? Sorry if it is really naive, but I am an inquisitive person and my textbook (microelectronics:circuit analysis and design 4e, Neaman) is very cut-dry to the point analysis with no explanations for the "why" questions.On that note, any recommendations for a good source that is a hybrid of dave's excellent op-amp video "how and why" with some KCL anyalsis at the same time?

author: w2aew
author_rank: Super Contributor
timestamp: January 23, 2015, 10:20:48 pm
content: Quote from: mg84 on January 23, 2015, 10:10:03 pmProbably not covered in the text at this point - but some op amp output stages include a push-pull arrangement. Have you heard of something called cross-over distortion? How to you think this 20k load/pull-down resistor might affect that?As for basics - did you see my video on the basics of op amp circuits? Note quite as in depth as Dave's but covers the basics...

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2015, 10:41:23 pm
content: I also suppose it's possible, depending on the style of the text, that that was included to reinforce the idea that the feedback loop compensates for output loading

author: mg84
author_rank: Newbie
timestamp: January 24, 2015, 03:32:42 am
content: Quote from: w2aew on January 23, 2015, 10:20:48 pmOk, thanks for the nudge. I was not familiar with either term, but it did provide me with a good rabbit hole to tumble down. This material is from chapter 9, after which we go from chapters 1-8 and cover BJT and mosfets and such, so maybe that knowledge is assumed at this point.I don't fully grasp the concepts yet, but after a little reading I think the basic idea of push-pull is because of the delay with transistors and biasing which causes the distortion at 0v. I assumed this was something separate than the load which would be at Vo, instead of part of the output, hence the confusion. So using your hint and following that line of reasoning, would this resistor serve as a (negative?)bias before the transistors to eliminate or reduce crossover distortion? By providing a point to sink some current to ground, the voltages are never simultaneously zero? I'm not fully aware of how this would affect the wave form, but would it basically "center" it somewhere below zero to mitigate the distortion?Hopefully I went somewhere in the right direction, I know we are attaching a lot of assumptions to such a simple design but it helps me to try to learn the "why" as I go, or find out if something was just a red herring to distract the student. Also, thanks for the video link! It looks like you've got some other good videos I need to check out as well.Thanks for the insights!Edit:After viewinghttp://www.aikenamps.com/index.php/what-is-crossover-distortionI see that negative biasing actually worsens crossover distortion, instead of improving it.

author: w2aew
author_rank: Super Contributor
timestamp: January 24, 2015, 03:52:24 am
content: Quote from: mg84 on January 24, 2015, 03:32:42 amYou basically went in the right direction. You're thinking along the right lines. A couple of mis-statements, but your getting there. Since you stated this is homework, it's best that you discover and learn this on your own (with nudging in the right direction).The negative bias mentioned in the webpage you linked isn't really the same as the pull down resistor in your original circuit. And, the cause and effect of the crossover distortion in the circuit in the article are specific to the amplifier they're discussing, and not generally applicable to all circuits (it all depends on the particular circuit in question).In the op amp case, a pull down or pull up resistor may help reduce cross over distortion by preventing the 'switch' between 'push' and 'pull'... More food for thought.I've got a little over 190 videos on my channel. Many of them are circuit related, either "back to basics", or tutorials, or fun little projects or schematic reviews. Enjoy...

author: Howardlong
author_rank: Super Contributor
timestamp: January 24, 2015, 04:29:21 am
content: Quote from: c4757p on January 23, 2015, 10:41:23 pmMaybe, but I suspect R1=20k is either a deliberate bum steer or a mistake. I guess it depends what level this it at. Perhaps if you were offered a datasheet for the device on question possibly, but not with 20k loading.If I were a betting man R1 is a bum steer.Edit: i am assuming R1 is the 20k resistor from the opamp output to ground, the text presented is not clear.

author: KM4FER
author_rank: Regular Contributor
timestamp: January 24, 2015, 02:51:16 pm
content: Sometimes those who teach can be a bit devious in getting a point across.Working problem 1,withoutthe resistor from output to ground you get an answer, and it is the correct answer.Working problem 2,withthe 20k resistor from output to ground you get the same answer, and it is also correct.What answers do you get if you change the output-to-ground resistor to 10k? 1k? 100k?What conclusion can you draw about the effect of output-to-ground ( load ) resistance on gain?earl...

author: mg84
author_rank: Newbie
timestamp: January 25, 2015, 12:57:58 am
content: Quote from: KM4FER on January 24, 2015, 02:51:16 pmThanks, I think this is ultimately what the question was trying to illustrate when considering all 4 examples (attached below) had the same input resistance and gain. I think my misconception was that Vo was what simulated a load, and the resistor in question would serve some other purpose since the load was already simulated by Vo, which was not the correct line of reasoning seeing as there was no resistance there. I am still intrigued by the whole crossover distortion line of thought, and what effect that resistor to ground would serve, but I guess I'm not quite there yet and it's way past the scope of the homework. Being that this was one question out of 25, probably best not to dwell on it...thanks for all the input

author: dannyf
author_rank: Super Contributor
timestamp: January 25, 2015, 01:02:46 am
content: QuoteFor this exercise, all you need to remember is that an ideal opamp has output impedance of zero. So whatever you hang on the output pin does not matter.In reality, it gets far more complicated than that.