Help with inverting op-amp

Alustriel

Joined Apr 1, 2009
14
This is is an op-amp circuit from my text book that I'm not able to solve, please bear with my crappy ms paint skills please. Sorry about the image size, dont know how to get it smaller... Anyway,
this is kinda like a basic inverting op-amp circuit, EXCEPT about the R4 in the middle,which makes it kinda confusing for me

Now the question stats this : Find VO when R1 = 80 kΩ; R2 = 50 kΩ; R3 = 70 kΩ; R4 = 30 kΩ and Vin = 5 V

What I did so far is:
( (Vi-Vx)/R1 + (Vo-Vi)/R3 + Vy/R4 + (Vy-Vx)/R2 )= 0

Where Vx = node on the left and Vy = node in the middle.

Now obviously I can't solve this question with my equation, this is where you guys come in. Thanks in advanced. alwayslearning

Joined Feb 27, 2008
22
Node Vx:

(Vx-Vi)/R1 - (Vx-Vy)/R2 = 0
Vx = 0,
Vi =5 V
Vy = 5(R2/R1)

Node Vy:

(Vy-Vx)/R2 + Vy/R4 - (Vy-Vo)/R3 = 0
Vx = 0,
Vy = 5(R2/R1),
Solve for Vo!

Darren Holdstock

Joined Feb 10, 2009
262
Alustriel:
kinda like a basic inverting op-amp circuit, EXCEPT about the R4 in the middle
This is a sneaky way of obtaining high gains without having a high value feedback resistor*. It's quite often cited in textbooks for transimpedance amps with a current input, like a photodiode, but is rarely used in practice as the noise gain is much higher, as is the amplification of the input offset error. It's usually best to cascade 2 more conventional amps if a single TI amp can't give enough gain with a single feeback resistor.

*High value feedback Rs are be prone to noise pickup, and the stray capacitance across the resistor will lower the frequency response.

hgmjr

Joined Jan 28, 2005
9,029
This is is an op-amp circuit from my text book that I'm not able to solve, please bear with my crappy ms paint skills please. Sorry about the image size, dont know how to get it smaller... Anyway,
this is kinda like a basic inverting op-amp circuit, EXCEPT about the R4 in the middle,which makes it kinda confusing for me

Now the question stats this : Find VO when R1 = 80 kΩ; R2 = 50 kΩ; R3 = 70 kΩ; R4 = 30 kΩ and Vin = 5 V

What I did so far is:
( (Vi-Vx)/R1 + (Vo-Vi)/R3 + Vy/R4 + (Vy-Vx)/R2 )= 0

Where Vx = node on the left and Vy = node in the middle.

Now obviously I can't solve this question with my equation, this is where you guys come in. Thanks in advanced. This is good problem to apply Thevenin's Theorem on.

If you convert R3 and R4 to its Thevenin Equivalent Resistance and also compute the Thevenin's Equivalent Voltage based on R3, R4 and Vo then you can turn this opamp problem into a form that you are more familiar with.

hgmjr

PRS

Joined Aug 24, 2008
989
Work it like an inverting op amp where Rf=equiv resistance of R2, R3, and R4. Seems to me that R3 is in parallel with R4 and these together are in series with R2. As

Vo=-Rf/Rs*Vin, you get in this case Vo=-[(R3//R4+R2)/R1]Vi

Thanks for all your help and input guys. Haven't learn Thevenin's Theoram yet tho.. 