Op-Amp and resistor network problem

[electronZ]

Hello all!

I hope you are all staying safe

I have an understanding problem: V(n007) voltage!
I can calculate V(n002) and the output is simply V(n007)*2 = 2.3 V.
But, how to calculate V(n007)?

See simulation in attached picture.

Thanks a lot.

 

[danadakk]

Hello all!

I hope you are all staying safe

I have an understanding problem: V(n007) voltage!
I can calculate V(n002) and the output is simply V(n007)*2 = 2.3 V.
But, how to calculate V(n007)?

See simulation in attached picture.

Thanks a lot.

Vn007, since no current (ideal OpAmp, or very very small) flows into the + input
of OpAmp we can ignore that. Also since R10 is in parallel with a V source we
can ignore that. So write a Thevinin loop starting with V1 all the way to Vn002 back
to ground and solve.


Regards, Dana.

 

[crutschow]

V(noo7) is determined by the combination of V(noow) and the V1 voltage.

Do you know how to write node loop equations or do superposition calculations?
(I probably would use superposition).

 

[electronZ]

Thank you very much - both of you. I will have to do some reading, it is long time ago... Am getting old

 

[sjb741]

Hi
First, because the stimulus I1 is a DC value, I'll conveniently ignore the effect of C1.

V(n002) is then I1*(R2+[R1|R3])
In fact, it is approximately true we can ignore R1 and R3 then we have
V(n002) ~= I1*R2 = 1.144 V

This is so because of the 'golden rules': The op amp feedback keeps V- at zero, and essentially no current flows into the input pins.

Because R3<<R1, it is not worth finding the exact value of this parallel network, it will be 10 Ohms to within a very close tolerance - but also 10R << 71500 so the feedback at DC is about the same as R2 alone.
-----
Now for V(n007).

First, ignore R10 because it is directly across V1 and V1 does not care.
Then we can find the "Thevenin equivalent" circuit for V1 feeding R5 & R6, call this Veq
This is equivalent to a voltage Veq=(B.V1) with series resistance R5|R6.
It looks like this

GND---[Veq---Req]---v(n007)

Veq = 1.25*(215/(215+8.87))=1.20
Req = 8.52k

B is found from the potential divider equation: R6/(R6+R5) = 0.96
The equivent series resistance is R5|R6 because an ideal voltage source has no resistance. Thus current in or out of the divider R6/(R6+R5) sees a resistance R5|R6.
----
Now we tie n002 to n007 using R7:

GND---v(n007)---R7---x---Req---Veq---GND
==================
n007 = 1.144 + (1.2-1.144)*(R7/(R7+Req))
= 1.144 + 0.056*0.34
= 1.16304