inverter amplifier with characteristics of a low pass filter

Thread Starter

Filipe___

Joined Feb 19, 2021
7
Can someone explain to me why the fase of my output signal converges to -90 degrees instead of 90 degrees.
Really aprecciate it.
1613781443715.png
 

kennybobby

Joined Mar 22, 2019
75
i can't tell where your nodes are located and what the colored scope traces are showing, but

What do you think is the function of C1 in your circuit, and what happens to it as the frequency increases?

What is your reference for the phase of the amplifier output--is this with respect to the input? If so then at low frequency the phase should be -180 for an inverting amplifier. The capacitor will look like a short to ground at high frequency.
 

Papabravo

Joined Feb 24, 2006
16,091
Although your amplifier has a gain of 10 at low frequency the Gain-Bandwidth Product (GBW) of the 741 shows quite a remarkable reduction in gain as the frequency moves past 10 kHz. you are basically at 0 dB or NO GAIN at all. The 741, all by itself is a pretty good low pass filter covering the audio spectrum. It's slew rate is nothing to write home about either. Check the datasheet for the unity gain frequency -- I'm sure you will be impressed.
 
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Thread Starter

Filipe___

Joined Feb 19, 2021
7
Im sorry for missing some information there ( Green is the node of the capacitor) and Pink is the Vout node(output signal). Although my entry singal is just AC 1V amplitude , so the pink signal represents my transfer funtion right?
I did the transfer Function and i got ;
Vx = node on capacitor:
Function 1- Vx/Vi = 0.5/(1+0.00034jW)
and
Funtion 2- Vo/Vi=-5/(1+0.00034jW).
Considering this results after a certain level of Frequency my gain(Vo/Vi) (Function 1) doesnt match to the graphic.
But for the first 6000Hz so-so it matches perfectly!

So my question is why my function doesnt work for some levels of frequency and how can i rematch my calculus to prove what i see in the graphic, I know capacitor will have a -90degree fase shift comparing with Vin, im not understanding why my output signal oscilates between 180º and -90º for high frequencies.
 

kennybobby

Joined Mar 22, 2019
75
R1 and C1 comprise a low-pass filter at ~700Hz, so anything above that is useful for what? The signal at node 2 starts off -6dB due to the voltage divider action of R1 and R2, and from there it just runs off into the deep end of the pool (what does -81dB even mean?).

Your plot shows an output gain of +13dB at the low end, which is not correct, shouldn't it be no more than 10dB?

If you build a breadboard of this circuit to test on the bench you won't be seeing any signals on your scope at the HF. 700Hz LP filter with 2x attentuation on front end and 10x inverting gain on output--makes no sense to use it at the HF.
 

DickCappels

Joined Aug 21, 2008
7,421
You need to use the parallel combination of R1 and R2 as the resistance for the cutoff frequency calculation since the inverting input has a very low impedance. In real life the output impedance of the signal source will also affect the cutoff frequency.
 

Thread Starter

Filipe___

Joined Feb 19, 2021
7
Thank you i understand i just didnt understand the phase of my output signal, i mean tecnically it will be always the phase of my capacitor + 180º right? Why after a certain level of frequency thats not verified anymore?
Can you explain me whats happeninig in the phase?
Im sorry bu its for a report and im not understanding that graphic.
 

kennybobby

Joined Mar 22, 2019
75
GIGO, garbage in garbage out, the simulation is not correct at the low freq, why would it mean anything at the high? Set your nodes at the Vi input and the Vo output, then Vi and Vcap to see what is happening.

It also makes no sense to look at HF above the cutoff of the LP filter. There is no signal there--what is the voltage at -81dB? It's like trying to divide by zero and take the inverse.
 

crutschow

Joined Mar 14, 2008
27,172
If you compare the signal phase at the capacitor with the op amp output phase, you will see that the op amp also has a change in phase-shift with frequency.
 

Thread Starter

Filipe___

Joined Feb 19, 2021
7
After that 10KHz the phase of my output signal converges to -90º and im not understanding why because the phase shift of my capacitor stays in -90º.
Am i suppose to say: " My circuit doesnt work to high frequency?"
 

kennybobby

Joined Mar 22, 2019
75
There is an error; the output gain is showing about 13 dB--how can that be correct when Vi =1V, Vo ~=<5V? All calculations of phase done using these values will be suspect.
 

DickCappels

Joined Aug 21, 2008
7,421
There is an error; the output gain is showing about 13 dB--how can that be correct when Vi =1V, Vo ~=<5V? All calculations of phase done using these values will be suspect.
That is the low frequency gain of the circuit. 68k/(6.8K + 6.8K) = a gain of -5, or -13.97 db.


Well, now that you, Fillipe, mention it, you will probably want a faster opamp if you want to work at those frequencies. See crutschow's comment in post #10.

As you sweep a single pole RC filter you will see the amplitude drop by about 30% (to -3 db) at 1/( 2 Pi R C) at this point you have a 45° phase shift. Continuing the increase the frequency will cause the phase to shift until it reaches 90°. Beyond this frequency the amplitude will continue to decrease but the phase shift stays at 90°

1613834187264.png

Here is some fun reading that you might enjoy: https://www.electronics-tutorials.ws/filter/filter_2.html
 

Thread Starter

Filipe___

Joined Feb 19, 2021
7
Should I conclude that when exceeding certain frequencies the signal is very attenuated and the phase is wrong? Like it should stay at +90º and it keeps decreasing?
And if i want to work at that frequencies i need other type of ampop
 
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