Summing op-amp circuit

[@MrMLehman]

After watching just a half dozen op-amp videos, I now have a very good understanding of the fundamentals of op-amps. Thanks.

[@rominadoubnia742]

Same here

[@Plipas]

amazing ! ! ! ! first time in my life I understand so easily the op amps . I appreciate so much how you share your knowledge so generously .

[@anwitmondal6417]

Thank you for making our life a little bit easier. I appreciate it.

[@azureorbit]

Really good explanation, thank you.

[@ElectricalEngineeringApp]

Well explained

[@arneg0]

Is it possible to make a difference amplifier circuit? These videos are amazing to follow .

[@danielsampong6607]

Yep it’s a thing

[@agstechnicalsupport]

Excellent explanation of summing op-amp circuit.

[@cameo64]

What if the summing terminal is the positive terminal instead of the negative terminal?

[@jdga375]

It would be a non-inverting right? The negative terminal means its on the inverting side as far as the wave form goes I believe.

[@xarhspapapadatos]

So what's the point in using an op amp when you just short-circuit the input.

[@qzorn4440]

Gee........... analog op-amp computers must of been a world of it's own? thanks.:)

[@otiebrown9999]

Let me add that an OFFSET function, requires the use of the + terminal.

[@petteral94]

thank you so much!

[@jaikrish50]

May i know which software is this u r teaching

[@hasnainyousaf5084]

very nice way to explain

[@nickace4373]

Hey i have a question (that might be kind of stupid, sorry if that's the case). Since The amplifier has close to zero voltage and current and to my understanding doesn't seem to affect the equations, why use it at all?

[@NicolasSchmidMusic]

what would change if the op-amp was the other way (plus and minus inverted)?

[@TGeo7]

how do you know what supply voltages V+ and V- to use for a summing op-amp? Thanks in advance

[@AhmedKMoustafa2]

hey ,, can you do a video about butterworth filters "second and forth order" please Best Regards One of your loyal fans

[@palesamokgadi7005]

This was very helpful

[@maymyatmyatoo2795]

Thanks a lot.

[@maharaniu4313]

could it be simplified as Vo = -(Va + Vb) in the beginning? since we know Vo has negative i

[@cameo64]

only if Ra and Rb are equal can you say Vo = -(Rf/Rb)(Va + Vb) then if Rf and Rb and Ra are all equal, then it can be -(Va + Vb), because you are summing with a 1:1 ratio of resistance.

[@ralphberg2942]

In which way these calculations would change ifV(a) and V(b) are connected to negative input of amp? Seems like everything would remain the same? Isnt it? Where am I wrong? And why it is called summing if we get a sum with minus (reverted sum)?

[@ralphberg2942]

Sorry. Correction of the question. In which way these calculations would change if V(a) and V(b) are connected to positive input of amp?

[@hasnainyousaf5084]

good job sir

[@limpie6959]

we can just use millmann theorem , this could be more simple than your method , thank u btw

[@akimiya1307]

Should we explain: i1= Va/Ra + Vb/Rb and i2=Vo/Rf. Then, i1+i2=0 => Va/Ra + Vb/Rb + Vo/Rf = 0 => Vo = - ((Rf/Ra)Va + (Rf/Rb)/Vb) ???

[@yezayarlin5917]

It is not i1+i2=0, it is i1=i2=0

[@sanjaydwivedi7167]

Sir , in the previous video you defined A=r2/r1 and said A is a very large number.But if A is a ratio of resistors se cant it be low , if i take r1=r2 then A=1;