op amp node voltage analysis

[SeanHatch]

Hello,

There's something I don't think I understand about node voltage analysis with ideal op amps. In the circuit shown below, At node D, I wrote the equation,

((1/1000) + (1/1000))Vd - (1/50000)Vc - (1/1000)Ve = 0

However my prof's equation for node D is:

((1/1000) + (1/1000))Vd - (1/1000)Ve = 0

I do not completely understand this. Unless me made an error. The only thing I could think of is something pertaining to the op amp's constraint Va = Vc, but if that justified getting rid of Vc in the above equation, it would seem like it'd justify getting rid of Va in the other ones, which he didn't do.

help?

Thanks alot;
Sean

EDIT: all resistors besides the one listed are taken to be 1000 ohms

 

[Optikon]

Hello,

There's something I don't think I understand about node voltage analysis with ideal op amps. In the circuit shown below, At node D, I wrote the equation,

((1/1000) + (1/1000))Vd - (1-50000)Vc - (1/1000)Ve = 0

However my prof's equation for node D is:

((1/1000) + (1/1000))Vd - (1/1000)Ve = 0

I do not completely understand this. Unless me made an error. The only thing I could think of is something pertaining to the op amp's constraint Va = Vc, but if that justified getting rid of Vc in the above equation, it would seem like it'd justify getting rid of Va in the other ones, which he didn't do.

help?

Thanks alot;
Sean

You ought to write all the resistor values in... no one reading your post can decide if a 1/1000 term is correct when there is only a 50k value shown.

Here is the other piece of advice, try to be technically correct. In otherwords a statement such as (1/1000) + (1/1000)Vd is invalid. The reason is the units do not match. 1/1000 is unitless and (1/1000)Vd presumably has units of voltage. The two cannot add.

Here a strategy for solving the whole circuit but I wont do it for you.
Start with the current source, assume none of that current flows into the OA, it all flows into the network on A & B (current divider). using ohms law & equivalent resistances, find the voltage at node A. Due to opamp virtual ground, VA wil also equal VC. VC drives a current through that R to ground. That current also flows through the 50k & resistor between D & E. Write an equation for the external source, F , Vout, D & E. SOlve it for all remaining nodes.

 

[SeanHatch]

We were told to solve this problem using strictly node voltage analysis, so while I'd like to use the methods you presented, I should not.

((1/1000) + (1/1000))Vx

is not unitless is it? A voltage times a conductance is a current, yes? Unless you are just hinting that I should have included ohms.

Are these node voltage equations the norm? obviously I've only been taught this by one person, so maybe there are different ways to write them.

 

[Optikon]

We were told to solve this problem using strictly node voltage analysis, so while I'd like to use the methods you presented, I should not.

((1/1000) + (1/1000))Vx

is not unitless is it? A voltage times a conductance is a current, yes? Unless you are just hinting that I should have included ohms.

Are these node voltage equations the norm? obviously I've only been taught this by one person, so maybe there are different ways to write them.

My bad.. whoops you had it right.. I didnt notice the second ")" it's ok.
:roll:

You can solve this with all node voltages. But if you are being taught opamp theory, ideally the opamp draws no input currents and has zero output impedance so you'll have to make use of these facts when writing node equations. One node equation will be that V- = V+ on the opamp inputs due to the "virtual" ground.

Are you still having a problem if you do that?

 

[DigiTan]

**broken link removed**

First, use what Optikon suggested and take the fact that no current enters the + input of the Amp. This tells you that all of the current source's current flows to ground through resistors R1, R2, and R3 (see attachment). The equavalent resistance for all three is: (R3+R2)||R1. Or...

Req = [ (R3+R2)(R1)/(R3+R2+R1)]

...then use V = I*Req to the voltage at node A. Once you know node A, use R2 and R3 as a voltage-divider and find the voltage on node B with...

Vb = Va*[(R2)/(R3+R2)]
---------------------
Now use the "virtual short" property to get Va = Vc. Since no current flows into the Amp, you know that the current through R4 is the same as the current through R5 and R6. Since you know the voltage of ground and Vc, the current through R4 is...

I4 = (Vc - ground)/R4 = (Vc/R4)

Doing node-voltage on R5, you know Vd is Vc, plus the voltage drop on resistor R5 caused by that feedback current I4....

Vd = Vc + (I4/R5)

Doing the same on R6, you know Ve is Vd, plus the voltage drop on resistor R6 caused by our feedback current...

Ve = Vd + (I4/R6)
-----------------------
The voltage Vf is equal to whatever the ideal voltage source on the left is. Finally, to get the current through R7, do one last node-volt for old times sake with Ve, Vf, and R7 to get...

I7 = (Ve - Vf)/R7
-----
Substituting in all the V's and R's, I'm getting...
Va = I * [ (R3+R2)(R1)/(R3+R2+R1)]
Vb = Va*[(R2)/(R3+R2)]
Vc = Vb
Vd = Vc + [(Vc/R4)/R5]
Ve = Vd + [(Vc/R4)/R6]
Vf = Vsupply

 

[SeanHatch]

Guys, I talked to my prof today. Turns out I was on the right track and he made an error on the solution set. Thanks for all of your help though, and sorry for wasting your time

 

[Styx]

Guys, I talked to my prof today. Turns out I was on the right track and he made an error on the solution set. Thanks for all of your help though, and sorry for wasting your time

never a waste of time.
It is good that you stuck by what you think is correct, it helps with yr confidence (whcih 1/2 what engineering is abt trusting yr gut feeling)