Would you explain why op-amp behave like that, Please?

Thread Starter

Michael George

Joined Feb 8, 2015
53
Hello,
I build a low pass filter on a breadboard using op-amp as a non-inverting amplifier.
This is the schematic of the circuit:
download.png


The op amp was in the saturation mode and it needs two biasing resistors. The waveform was this:

Untitled.jpg
The input is the green waveform. The output is the red waveform.


Then I added a coupling capacitor at the input.

download cap.png


Then, The op-amp is biased without putting a biasing resistors. The waveform was like this:

Untitled.jpg

How is one capacitor able to bias the op-amp? If this cap blocks DC, The op amp would still saturated. How the capacitor biased the op-amp without using two biasing resistors?
 

Papabravo

Joined Feb 24, 2006
12,906
It is not saturation. You can't expect an opamp with a single supply voltage to have the output swing below ground -- or can you?
BTW -- what is the purpose of C2?

Hint: Bias the opamp at 4V = Vcc/2, then apply the AC input and feed back to the negative terminal. Did you dream up this configuration or did you copy it from some source? If so I think you need another source.
 
Last edited:

Thread Starter

Michael George

Joined Feb 8, 2015
53
Hello, @Papabravo
Thank you for your answer,

Of course I don't expect an opamp with a single supply voltage to have the output swing below ground.
What I think is:
The input voltage is 1.5v peak to peak.
Before adding the capacitor, the input voltage swings from -0.75V to +0.75.
The opamp output responds to the positive half cycle only. So, the output wave form is due to the range of input voltage from 0 to +0.75.

I wanted to bias the opamp at Vcc/2. So, I first inserted the capacitor to block any DC signal that might come form the input source. Then, I surprised that the output wave form has changed and the opamp responded to both positive and negative half cycles. I think that the input signal is affected by the capacitor and the input swings from 0 to 1.5 volts instead of from -0.75 to +0.75. But I don't know how could this be? That's why I'm asking here.

I did not dream up this configuration :) I found the circuit here:
http://www.electronics-tutorials.ws/filter/filter_5.html

But in this website the opamp is not biased and it does not work for the negative half cycle. That's why I started adding a capacitor and biasing resistors. But I'm surprised that the op amp works for the entire cycle after adding the capacitor only!!
 

crutschow

Joined Mar 14, 2008
24,343
..................
How is one capacitor able to bias the op-amp? If this cap blocks DC, The op amp would still saturated. How the capacitor biased the op-amp without using two biasing resistors?
In real life it wouldn't.
Either the op amp model doesn't simulate the op amp input bias current, or you didn't run the simulation long enough for the bias current to saturate the op amp.
 

MrAl

Joined Jun 17, 2014
7,139
Hi,

Spice sometimes assumes that you dont do anything that is too far "off the wall", which means too far out of the ordinary. Not using a negative supply or not biasing the input to 1/2 of Vcc would be too far from ordinary here.

To see more realistic effects, try using a clamping diode from the non inverting input to ground. That will ensure that the input is not driven to a non spec'd value which would actually blow out a real chip.

It could be that the cap powers the circuit from the non inverting input or something strange like that. We actually see this happen in real life with some chips but i dont think this chip will allow that i think it would blow out in real life.

So the best bet is to bias the input at 1/2 Vcc with two resistors and go from there. Let us know how you make out if you can.

Another simpler example of this shows up in some diode models, where the reverse voltage breakdown is sometimes set wayyy higher than the actual spec on the data sheet. For example, 500v for a 40v part. In simulation the diode would work up to at least 450v, but in real life it would soon blow out.

So when you see strange phenomena like this always check your assumed limit parameters.
 

RichardO

Joined May 4, 2013
2,271
In real life it wouldn't.
Either the op amp model doesn't simulate the op amp input bias current, or you didn't run the simulation long enough for the bias current to saturate the op amp.
Actually it might -- sort of. The coupling cap might DC restore against an intrinsic / protection diode. My battery is failing. More details later.
 

hp1729

Joined Nov 23, 2015
2,304
Hello,
I build a low pass filter on a breadboard using op-amp as a non-inverting amplifier.
This is the schematic of the circuit:
View attachment 100229


The op amp was in the saturation mode and it needs two biasing resistors. The waveform was this:

View attachment 100233
The input is the green waveform. The output is the red waveform.


Then I added a coupling capacitor at the input.

View attachment 100231


Then, The op-amp is biased without putting a biasing resistors. The waveform was like this:

View attachment 100234

How is one capacitor able to bias the op-amp? If this cap blocks DC, The op amp would still saturated. How the capacitor biased the op-amp without using two biasing resistors?
Is this a better design?
I don't think it is mandatory that R2 = R3. We can adjust the operating point by adjusting this????
 

Attachments

Thread Starter

Michael George

Joined Feb 8, 2015
53
Hello, @MrAl
Actually, it is not spice. I build the circuit on a real breadboard. The oscilloscope is PC based Soundcard Oscilloscope.

Let us know how you make out if you can.
I did biased the input at 1/2 Vcc with two resistors and it works very well. The biased circuit was similar to the circuit that is provided by @hp1729

It is a very very good idea to use a clamping diode. I will try it as soon as possible. Thank you very for the information you gave me.
 

MrAl

Joined Jun 17, 2014
7,139
Hello, @MrAl
Actually, it is not spice. I build the circuit on a real breadboard. The oscilloscope is PC based Soundcard Oscilloscope.


I did biased the input at 1/2 Vcc with two resistors and it works very well. The biased circuit was similar to the circuit that is provided by @hp1729

It is a very very good idea to use a clamping diode. I will try it as soon as possible. Thank you very for the information you gave me.
Hello again,


Ohhhh, so it's a sound card scope :)
Does that have DC coupling or just AC coupling? With just AC coupling it will always look like a zero based sine signal even when there is a large DC bias.
 

Thread Starter

Michael George

Joined Feb 8, 2015
53
@MrAl It is AC coupling. I don't think it has the option of DC coupling. When I apply a pure DC, The wave form is zero voltage.
I think the issue of my circuit is the capacitor was previously charged before inserting it in the circuit. So, when I inserted it in the circuit, it somehow provided some bias (Without using biasing resistors).
Now, I tried the circuit again. I discovered that: when I apply and then remove the biasing resistors, the wave form is still biased and I have to wait for about half a minute so that the capacitor discharges and then the wave form will be changed.
That's why I thought that the capacitor was able to bias the circuit without the two biasing resistors :)
 

atferrari

Joined Jan 6, 2004
3,780
Hola Miguel,

The necessity of bringing the non-inverting input to Vcc/2 has been covered already.

I downloaded and saved this useful image created by Audioguru, a member of this forum. It covers the possible conditions you could encounter. Take your time to understand it. It is worth the effort.

upload_2016-2-8_10-56-36.gif
 

atferrari

Joined Jan 6, 2004
3,780
Is this a better design?
I don't think it is mandatory that R2 = R3. We can adjust the operating point by adjusting this????
I recall using an opamp with the Vref to the non-inverting input quite closer to one of the rails to accommodate a signal. Worked flawless. Sure, it does not seem a common case.
 
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