Recent content by altruan23

So after using superposition and setting the ratio R2/R1 = R4/R3 the same or R2=R4 and R1=R3,i come to the eq. for output voltage Vout= R2/R1 * (V2-V1) or R4/R3(V2-V1). And in the book foundations of analog and digital electronics by agarwal and lang, they are saying that this circuit is a...

so that means i1+i2= i3 and that is what my KCL is saying. i set i3 as leaving the node, because there's a virtual ground so it needs to go to R3.

isnt it that the sum of the currents flowing into the node, is the same as the sum of the currents flowing out of the node. and schouldnt the sum of both currents flow to the R3?

So i used KCL and both currents are flowing into the node, and then leaving together to go to the resistor R3. So my eq can be seen in the picture. I was looking in a book and they had a minus infront of the parantheses. Is the current flowing from R3 into the node??

Hello LvW. I am sorry that you were waiting for my reply, i just forgot so that is my mistake. Yeah the diagram is wrong, because i forgot to give a ground to the + terminal of the op amp. Thank you anyways for your help and that you cared and again sorry that i didnt answer your question.

So this is the circuit. And here i tried to calculate Uout/ Uin , any suggestion how to simplify this term?? I used Uout= Uin * (1+ Z2/Z1)

alright I am going to have a look at this book. thanks!

Hello I have one more question. How can I draw the Bode-diagramm? I simulated in MultiSim it schould probably look like this: Yeah i forgot also to give a logarithmic scale on the upper diagram on Y. On the lower diagram it schould be the phase.

ok thanks for both answers, I am going to try both tommorow:)

So basically we can have a constant current but i don't understand this circuit. for example: i can have the Ia what ever I want with current divider rule: Ia = Iq * R2/R3. So Ia is proportional to the ratio of R2/R3. And if i give a resistor at the collector terminal, if i change it between...

wow, really nice way to solve circuits like this. thanks , i will try to use it in the future.

thanks again for the help!

can someone check, if i calculated the output voltage correct? i used virtual ground because V+-input of the op amp is 0 --> V- input is then also 0.

sorry i wanted to write 12V instead of 10V

So i start with 10V, i get a voltage drop on R1 --> 7.2V Remainder is 4.8V which drops on R2. How do we have in node A 4.8V and then they drop across R3 and R4, if the 4.8V already dropped across R2 in the first loop??