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

Discussion in 'General Electronics Chat' started by Michael George, Feb 7, 2016.

1. Michael George Thread Starter Member

Feb 8, 2015
52
2
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:

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

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

Then I added a coupling capacitor at the input.

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

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?

2. Papabravo Expert

Feb 24, 2006
10,021
1,759
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: Feb 7, 2016
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3. Michael George Thread Starter Member

Feb 8, 2015
52
2
Hello, @Papabravo

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!!

4. crutschow Expert

Mar 14, 2008
12,573
3,082
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.

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5. MrAl Well-Known Member

Jun 17, 2014
2,225
438
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.

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6. RichardO Well-Known Member

May 4, 2013
1,161
360
Actually it might -- sort of. The coupling cap might DC restore against an intrinsic / protection diode. My battery is failing. More details later.

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7. hp1729 Well-Known Member

Nov 23, 2015
1,853
207
Is this a better design?
I don't think it is mandatory that R2 = R3. We can adjust the operating point by adjusting this????

• Design 613 NI AC amplifier.pdf
File size:
42.1 KB
Views:
3
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8. Michael George Thread Starter Member

Feb 8, 2015
52
2
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.

9. Michael George Thread Starter Member

Feb 8, 2015
52
2
Yes, The design works very well. Thank you very much for your demonstration by sending me the schematic.

10. MrAl Well-Known Member

Jun 17, 2014
2,225
438
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.

11. Michael George Thread Starter Member

Feb 8, 2015
52
2
@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

12. atferrari AAC Fanatic!

Jan 6, 2004
2,612
738
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.

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13. atferrari AAC Fanatic!

Jan 6, 2004
2,612
738
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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14. Michael George Thread Starter Member

Feb 8, 2015
52
2
Thank you very much