Differential amplifier not functioning properly

Thread Starter

Tyler Bradbury

Joined Mar 18, 2015
13
I UNDERSTAND THAT THIS IS JUST AN INVERTING AMP. I AM JUST TRYING TO GET THE DIFFERENTIAL PART WORKING BEFORE I PUT THE OTHER SOURCE ON THE POSITIVE TERMINAL OF THE AMP.

I am attempting to create a differential amplifier using an LM324, and four 100k resistors. The schematic is below.

upload_2015-4-30_17-19-24.png

I am taking the output at Node3, and getting the correct SIMULATION results.

upload_2015-4-30_17-19-50.png

The problem is coming when I put it all together and physically simulate it. I have the hardware set up correctly, but I am not getting what I want out of the circuit.

Node 1 is performing correctly, but I am getting nothing out of the output. I am using an LM324n. The simulated circuit is an ideal opamp. Could this be the problem? I have the hardware hooked up correctly (as inspected by two other engineers, as well). I cannot figure out why it will not give me the output I desire. Is there something obvious that I am doing incorrectly?

The one part of the circuit I do not understand is that when, on my Agilent 33250A Function Generator, I switch from "High Z" output impedance to 50ohm impedance, I am getting effectively double. Why is this?
 

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tindel

Joined Sep 16, 2012
659
Have you put decoupling caps on your breadboard?
Have you replaced the IC to make sure it's not damaged?
Have you connected your function generator ground to your power supply ground?
Be sure to properly terminate unused op-amps on the chip.

These are the most common errors.

As far as your output of your function generator driving twice the voltage - function generators have 50 ohm output impedance ... if you drive a high impedance load (>>50ohms) then you will get twice the voltage because it is not being loaded with 50 ohms.
 

Thread Starter

Tyler Bradbury

Joined Mar 18, 2015
13
Thank you for your response.

- Where would the decoupling caps go? Near Vcc pin?
- I did not try replacing the IC, but I will try when I get back to the lab.
- I will try to connect the function generator ground to my power supply ground. An additional question on this: where do I ground the oscilloscope ground probe? I had it going to the gnd of my +5v power supply. Is this correct?
- I will terminate the additional op amps. Would you mind telling me why this might be necessary?

Thank you for the explanation. You have really cleared things up for me.
 
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Thread Starter

Tyler Bradbury

Joined Mar 18, 2015
13
Also, in looking for an answer, I came across this configuration:

upload_2015-4-30_19-26-29.png

I tried this, and got a different result, but not the one that I desired. I thought this might work because of the explanation about the impedance. Any explanation would be greatly appreciated.
 

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crutschow

Joined Mar 14, 2008
23,149
Explanation of what? It's hard to know what's wrong without a description. :confused:
What is a "different result"?
 

tindel

Joined Sep 16, 2012
659
Thank you for your response.

- Where would the decoupling caps go? Near Vcc pin?
- I did not try replacing the IC, but I will try when I get back to the lab.
- I will try to connect the function generator ground to my power supply ground. An additional question on this: where do I ground the oscilloscope ground probe? I had it going to the gnd of my +5v power supply. Is this correct?
- I will terminate the additional op amps. Would you mind telling me why this might be necessary?

Thank you for the explanation. You have really cleared things up for me.

- yes as close to the Vcc and Gnd pins as possible. 0.1uF ceramic is pretty common for this sort of thing.
- The ground of your scope should be connected to circuit ground, typically.
- Unused amplifiers can mess up the operation of the other op-amps on the chip - especially if there is shared circuitry on the op-amp. It can also damage the part.

A couple other common problems:
-Not operating within the common-mode input range of the op-amp.
-Too much gain saturating the output of the op-amp.
 

Reloadron

Joined Jan 15, 2015
4,465
I switch from "High Z" output impedance to 50ohm impedance, I am getting effectively double. Why is this?
Because when you go to 50 Ohms you have a 50 Ohm output impedance designed to work into a 50 Ohm load. Think a 50 Ohm, 50 Ohm voltage divider. So when the load is high like 1 Meg or 10 Meg Ohm what becomes of your divider? :) Your source goes to 50 Ohms and the load is still very high. That is a quick answer, obviously there is more to it. You could read up a little on Thévenin's theorem.

Ron
 
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