Operational amplifier exercise help - Page 1
Forum section: electronics / beginners

author: omglol
author_rank: Regular Contributor
timestamp: August 17, 2014, 07:57:31 pm
content: Hello!I am having a problem calculating output voltage of an operational amplifier and need some help.The circuit is shown on picture #1. I need to calculate output voltage (or voltage across 100k resistor R5). I already know the answer is -7V.How I approached the problem:- I tried to solve the problem using superposition method. In the picture number #2, I "turned off" voltage source U2. Since the voltage across U2 in this case is 0V, I connected resistor R3 to the negative terminal of the OP-AMP, since the voltage there is approx 0V. This way, the resistors R2 and R3 are in parallel and R4 is in series with them. This way, I got equivalent resistance of R2+R3+R4 = 2.5 kOhm. OP-AMP voltage output is -2.5 V, since I used basic inverting amplifier equation (- Requivalent/R1 * U1). I believe my method is wrong, but I don't know why.- In the next step, I "turned off" voltage source U1 - resistor R1 gets connected directly to the ground. I don't know how to approach this case. However, in simulator I get output voltage of -2V. If I combine both voltages (step 1 + step 2) I get -5.5V, which is wrong (should be -7V).Can somebody clarify this for me? Thank you very much in advance!

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 08:39:06 pm
content: I'm not sure which ways you have been taught to solve such a circuit, but the simplest way in my eyes is to walk along the resistor chain starting at U1 and add up voltages and currents as I go along. I can do this because I know that the voltage between R1 and R2 must be zero. From here everything else follows very quickly.

author: omglol
author_rank: Regular Contributor
timestamp: August 17, 2014, 08:42:47 pm
content: Yes, I know that the voltage between R1 and R2 is 0V, but how can I calculate the voltage between R2 and R4, since I don't know output voltage of the OP-AMP?

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 08:43:54 pm
content: Quote from: omglol on August 17, 2014, 07:57:31 pmIf you do this you have changed the topology of the circuit. You are not allowed to do that. If you want to "turn off" U2, you must simply replace it with a short circuit. You must not reconnect it somewhere else.

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 08:45:04 pm
content: Quote from: omglol on August 17, 2014, 08:42:47 pmThe output voltage of the op amp is what you eventually are going to find.The voltage between R2 and R4 must be -1 V. Can you see why?(Hint: how much current is flowing through R1? Where does this current go?)

author: Simon
author_rank: Unknown
timestamp: August 17, 2014, 08:53:24 pm
content: You need to work back from the inverting input of the opamp to work out what output voltage will produce 0V at that point.So for R1 it will have to be at 0V like the input, so 1V/1K = 1mA. Now as no current flows in the inverting input you will have 1mA across R2 as well so that needs to be pulled down to -1V to get the R1/2 point to 0V. With your R2/3/4 junction at -1V that puts 2mA through R3 as +1-(-1) = 2V/1K = 2mA, so now you need to pull 3mA through R4 which at 2K requires 6V or -6V, So I think you were right, -7V. Not that I'm an expert on this sort of thing.

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 08:58:01 pm
content: Quote from: Simon on August 17, 2014, 08:53:24 pmYou are correct, but I was holding back from simply giving away the answer to a homework problem...

author: Simon
author_rank: Unknown
timestamp: August 17, 2014, 08:59:51 pm
content: You mean i got it right ? goodho hopefully the OP has a few more to work on, from the sounds of his post he has already used a simulator as he said he needed -7V but did not understand how.

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: August 17, 2014, 09:10:36 pm
content: Quote from: omglol on August 17, 2014, 08:42:47 pma) The voltage between R1 and R2 must be 0b) The voltage between R2 and R4 must be -1, because a) and R2 = R1 and the current into the opamp must be 0.c) The voltage drop across R4 is I(R4)*R4, where R4 = 2000 and I(R4) = I(R2)+I(R3) which are easy to calculate from a) and b): (-1/1000-2/1000)*2000 = -6

author: omglol
author_rank: Regular Contributor
timestamp: August 17, 2014, 09:33:29 pm
content: Quote from: IanB on August 17, 2014, 08:43:54 pmYes, I know, but if you replace U2 with short circuit, R3 gets connected to the ground. So, to help myself, I connected it between R1 and R2, as there is also (virtual) ground. I believe that virtual thing got me.

author: omglol
author_rank: Regular Contributor
timestamp: August 17, 2014, 09:36:01 pm
content: Quote from: IanB on August 17, 2014, 08:58:01 pmIt is not a homework problem ... it is the middle of the August, all schools are closed.However, I am still having hard time understanding why is there -1V between R2 and R4 ... I will re-think your answers after I get some sleep, thanks for now to everyone!

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 09:53:41 pm
content: Quote from: omglol on August 17, 2014, 07:57:31 pmIf you don't change the circuit around when removing U2 you can still solve it with superposition.First remove U1 (replace it with a short circuit). That sets the op amp output to -2 V.Next put back U1 and remove U2 (replace that with a short circuit). That sets the op amp output to -5 V.Add both voltages together and you get -7 V.

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2014, 09:58:34 pm
content: Quote from: omglol on August 17, 2014, 09:36:01 pmThe inputs of an ideal op amp have infinite impedance. There is a current flowing through R1. That current has to go somewhere. It cannot go into the op amp. Therefore...

author: omglol
author_rank: Regular Contributor
timestamp: August 17, 2014, 10:32:55 pm
content: Oh man, I feel really stupid. I know about OP-AMP input impedance, but somehow I failed to realize that current through R1 is the same as through R2. Everything clear now.Thanks and stay tuned for next case I am going to need help on!Sent from my iPad using Tapatalk

author: Simon
author_rank: Unknown
timestamp: August 18, 2014, 05:47:25 am
content: There are a few good youtube videos around, that is how I learnt, nothing to do with that shit L3 course I abandoned.......

author: omglol
author_rank: Regular Contributor
timestamp: August 21, 2014, 10:00:51 pm
content: OK, here is another one. Output voltage is 10V, I need to find input voltage of U1 and U2 (ignore they are 0V in the picture). I tried superposition, but got wrong answers (U1 = -10V and U2 = 4.125 V). Can somebody give me a hint how to approach this one? Tnx.Sent from my iPad using Tapatalk

author: Andy Watson
author_rank: Super Contributor
timestamp: August 21, 2014, 11:43:50 pm
content: Are there more constraints to this question?You could pick any value for U2, then make U1 = 2.4 U2 - 10.Edit: Hint, there are several ways to solve this problem. I simplified the inputs to obtain the voltage at the positive input = 0.8 U2, and then combined the two resistors at the negative input to realise the thevenin equivalent in terms of U1. The rest readily drops out by inspection. Also, working the maths as fractions rather than decimals will save reaching for the calculator!

author: IanB
author_rank: Super Contributor
timestamp: August 22, 2014, 02:09:26 am
content: A simple, brute force way to solve this is to consider the unknown voltage at the inputs of the op amp:U+=U-=UYou can write an equation forUin terms ofU1. You can also write an equation forUin terms ofU2. EliminateUbetween these two equations and you have a relationship betweenU1andU2.Since you have one equation in two unknowns there is no unique answer.

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: August 22, 2014, 08:05:58 am
content: Quote from: omglol on August 21, 2014, 10:00:51 pmWhat program is that you're running on the iPad?

author: omglol
author_rank: Regular Contributor
timestamp: August 22, 2014, 08:18:06 am
content: iCircuit it is.Thanks for help!

author: omglol
author_rank: Regular Contributor
timestamp: August 22, 2014, 10:18:10 am
content: I am having trouble with a non-inverting Schmitt trigger. I need to calculate DC voltage of U2 source (ignore 0V in the picture) and ratio between R1\R2 resistors. It has to switch to +15V at U1 = 9V and to -15V at U1 = -6V.I am stuck on this one, really no idea. I know that non-inverting Schmitt trigger switches from negative saturation to positive at the upper trigger point and vice versa.Sent from my iPad using Tapatalk

author: IanB
author_rank: Super Contributor
timestamp: August 22, 2014, 08:37:18 pm
content: This problem is solved in much the same way as the previous ones. Write equations and solve for the unknowns. Maybe you should go back and practice solving the previous problems by yourself until you really understand them, and then come back to this one?We can keep giving you the answers but I don't think it is going to help you until you can solve one without us giving you the answer first.

author: omglol
author_rank: Regular Contributor
timestamp: August 24, 2014, 05:01:50 pm
content: I understand the previous ones. I am having problem with this one ... especially because ratio, no given resistor values etc.For example, I tried to solve exercise 1.C:http://elearning.algonquincollege.com/coursemat/saurioc/ELE-3/ELE-3-ASS/SCHMITT-ASS.pdfAnswers are 2 pages below in that file. I calculated U- voltage, and then from U- = U+ I calculated current trough 30k+1k resistors. Then I got output voltage and input voltage. I don't understand why the input voltage in this case is LTP? Also, why is there X on output voltage on diagram in answers? Moreover, I don't understand how did they get that UTP. Can anyone shed some light on this? Thanks!

author: Andy Watson
author_rank: Super Contributor
timestamp: August 24, 2014, 05:26:57 pm
content: Quote from: omglol on August 24, 2014, 05:01:50 pmDid you notice the that the questions are using an LM311 ? It is a comparator with an open-collector output - so it can only sink current. The X in the diagram is indication that the output is high and therefore no current flows down that path.

author: IanB
author_rank: Super Contributor
timestamp: August 24, 2014, 06:21:30 pm
content: Quote from: omglol on August 24, 2014, 05:01:50 pmYou are given two known states: the state when the output switches high, and the state when the output switches low. For each state you have two unknown variables: the trigger voltage U2 and the resistor ratio R1/R2. This gives you two equations in two unknown variables, so you can solve them simultaneously for both (for example by eliminating one of the variables).Note that if you have a voltage divider with resistor ratio R1:R2, then you can divide both sides of the ratio by R2 and this gives a ratio of R1/R2:1. Let the ratio R1/R2 be called r, then you have a voltage divider in the ratio of r:1. If you do this, your problem then has the two unknown variables U2 and r.

author: omglol
author_rank: Regular Contributor
timestamp: August 24, 2014, 07:07:09 pm
content: Quote from: Andy Watson on August 24, 2014, 05:26:57 pmNo, I did not. Thanks for pointing that out, I do not have a lot of experience with analogue electronic components.Sent from my iPad using Tapatalk

author: omglol
author_rank: Regular Contributor
timestamp: August 24, 2014, 07:13:03 pm
content: Quote from: IanB on August 24, 2014, 06:21:30 pmI understand, but I still don't know how to start writing the equations. I am missing U1, U2 and Uout voltages ... and also R1/R2 ratio. I have no idea where to put that +9 volts or -6 volts in any equation ... Output voltage is -15/+15 V, but the rest is a big question for me. Could you maybe give me a hint how to start? Thanks!Sent from my iPad using Tapatalk

author: IanB
author_rank: Super Contributor
timestamp: August 24, 2014, 08:00:14 pm
content: Quote from: omglol on August 24, 2014, 07:13:03 pmQuote from: omglol on August 22, 2014, 10:18:10 amThere you are. In your own post you have said that U1 = 9 V. So why do you say you are missing U1, and that you don't know where to put the +9 V? The answer is right there in your own words.You know it is supposed to trigger when U1 = 9 V, and you know that when this happens Uout is -15 V (it is just about to switch to +15 V).By the voltage divider formula you therefore can write this equation:(U2 - Uout) / 1 = (U1 - Uout) / (r+1)Putting in values you have:(U2 - (-15)) = (9 - (-15)) / (r + 1)There is one equation. Write the other equation similarly and then find U1 and r.

author: omglol
author_rank: Regular Contributor
timestamp: August 25, 2014, 08:58:47 am
content: Quote from: IanB on August 24, 2014, 08:00:14 pmThank you for clarifying the variables, now I get it. I thought that at +9V input, the output needs to be already +15V.About the equations - I did not understand yours, especially not the right part. So I written my own:(U1 - U2) / R1 = (Uout - U2) / R2... Which got me to:r (Uout - U2) = -U1 + U2 (r is ratio)I then entered the variables for first case and second case (UTP and LTP). I got two equations as you said and got answers:U2 = 1Vr = 1/2I think they are OK ... or so it seems in simulator.Sent from my iPad using Tapatalk