Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Calculating gain and Vout for this opamp circuit

Status
Not open for further replies.

cjlectronics

New Member
I have been pulling my hair out trying to derive the output voltage equation for this simple circuit. What is throwing me off is the 10.2k resistor in the feedback circuit. (well, I think this is what's throwing me off). I'm wanting to derive the gain and Vout for this circuit.

I've been looking at it so long I'm seeing lines in my vision.

Thanks in advance.
 

Attachments

  • Opamp_a.JPG
    Opamp_a.JPG
    27.6 KB · Views: 121
Hi. Try to redraw your circuit so that R2|R3 act as one resistor with one voltage source with output resistance equal to R2 in parallell with R3.

Then it is much easier to calculate.

[edit]
Like this (see image)
 

Attachments

  • Opamp_bb.png
    Opamp_bb.png
    12.8 KB · Views: 125
Last edited:
C3 is just some "beep cap" to filter noise. Pin 3 will be at whatever resistor divider voltage is set by the 50K pot. Pin 2 will be at virtually the same voltage as Pin 3. The voltage at TP11 with resistive divider R4 & R2||R3 will add to the voltage source mentioned by Grossel to produce the correct voltage at Pin 3.
 
Also, I'm not sure what gain you would be looking for. You have to have an input as well as an output to have a gain. Once you've done the math on the voltage at TP11 = Vout, you could find Vout as a function of the (variable) resistance on the 50K potentiometer....
 
Hi. Try to redraw your circuit so that R2|R3 act as one resistor with one voltage source with output resistance equal to R2 in parallell with R3.

Then it is much easier to calculate.

[edit]
Like this (see image)

Cool, thanks for the reply. Of course, this makes perfect sense. This is the redrawn circuit with the Vout formula. I'm not sure if the second equation should be added to the voltage divider circuit (R1 and VR1) or should it be subtracted? my thought is subtracted because it is on the negative input of the op amp?

Opamp_b.jpg
 
When an opamp is operating as a linear amplifier (its output is not saturated high or low) then the voltages of its inputs are almost exactly the same.
Its negative feedback adjusts its output voltage up or down a little so that its input voltages are almost exactly the same.

Your 5.51k resistor at the positive supply makes the (-) input at a voltage that is much higher then the (+) input so the output of the opamp will be saturated as low as it can go (because it is trying to reduce the voltage at the (-) input so it is almost the same as the voltage at the (+) input).
 
OK, I have updated the drawing and I am getting closer to understanding this bridge amplifier single op amp circuit. The formula and updated drawing is attached. The circuit will be used to monitor the changing resistance of a Thermistor so the bridge amplifier works well. Something I need to do now is calculate the bridge resistance. Do you randomly choose a value for the thermistor to calculate the bridge resistance?

Opamp_c.jpg
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top