Engineering Analyzing an AC Amplifier Circuit with Negative Feedback

[esmeco]

Homework Statement

The problem states that we should draw the Vout curve of an amplifier circuit.

Homework Equations

We have a sinusoidal sign with amplitude=1v and frequency=0.5

v(t)=A*sin(2pi*f*t)

f=1/T
T=2s

The Attempt at a Solution

Since we have negative feedback: Vp=Vn

The currents for this circuit would be:

I1+I2=Iout
I1=(0-1)/0.5=-2
I2=(0-v1)/1=-v1
Iout=(Vout-0)/2

I know that:

V1=1*sin(2pi*0.5*2)=0

So,the final equation would be: Vout=(-1*2)/0.5=-4v

Does this look ok?Also,when drawing the curve for vout,vout is the amplitude of the curve right?Or is the amplitude of the curve related to vout 1volt?

 

[esmeco]

...

After searching a bit I've found that you can't use v(t)=A*sin(2pi*f*t) or the current will be zero, so,in order to calculate the voltage for the alternate current we would use the formula Vrms=V0/squareroot of 2
and Vrms=Irms*R which would be equal to V0=I0*R

So, the final equation for Vout would be: Vout=(-1-2)*2=-6v which would be the amplitude of the curve of the tension on the output of the amplifier...
I think this's right,but if not I would like to be corrected...

 

[esmeco]

Can anyone help me,please?

 

[antonantal]

I didn't do the calculations but since you have a DC component in your input signal you should have one in the output, so V_{out} can't be a simple sinusoid with an amplitude of 4V.

You better use the superposition principle.

 

[berkeman]

I1+I2=Iout
I1=(0-1)/0.5=-2
I2=(0-v1)/1=-v1
Iout=(Vout-0)/2

I know that:

V1=1*sin(2pi*0.5*2)=0

So,the final equation would be: Vout=(-1*2)/0.5=-4v

You need to be more precise in your notation, and remember what voltages are DC and which are functions of time. More like this:

-2 - V1(t) = Vout(t)/2

So Vout(t) = what?

 

[esmeco]

Vout(t)=(-2 -0)/2=-1

IS this right?

 

[berkeman]

Vout(t)=(-2 -0)/2=-1

IS this right?

Nope. V1(t) is a sinusoid and there is a DC 1V input component as well, so the output will be a sinusoid with a DC offset...

 

[esmeco]

I thought that v1(t) would be zero since the frequency is 0.5 and the period is 1 and if we substitute the result of v1(t) will be zero..

 

[antonantal]

I think you should get a basic understanding of AC before you try to solve this problem. Maybe read some of your lecture notes.

 

[berkeman]

I thought that v1(t) would be zero since the frequency is 0.5 and the period is 1 and if we substitute the result of v1(t) will be zero..

Um, no. I'm not following where your confusion is coming from. V1(t) is a sinusoidal source, which is driving the input of the amplifier circuit. You are asked to solve for Vout(t), which is affected by the sinusoidal input V1(t), the DC input offset of 1V, and the gain set by the resistors around the opamp circuit.

You started with the definition of V1(t):

We have a sinusoidal sign with amplitude=1v and frequency=0.5

v(t)=A*sin(2pi*f*t)

f=1/T
T=2s

So to clean up the text a bit, that means V_1(t) = sin(\frac{2\pi t}{2})

Use that value for the V1(t) that is in your equations, and solve for Vout(t). Nothing is going to eliminate that sinusoid from your final answer.

 

[esmeco]

I misunderstood the T in f=1/T with the t in V_1(t) = sin(\frac{2\pi t}{2})
So then vout(t)= -sin(pi*t) - 1 ...And the amplitude is the amplitude given by 1v?

 

[berkeman]

I misunderstood the T in f=1/T with the t in V_1(t) = sin(\frac{2\pi t}{2})
So then vout(t)= -sin(pi*t) - 1 ...And the amplitude is the amplitude given by 1v?

Not quite, but you're getting close. Use that V1(t) equation and plug it back into the equation for Vout(t). Be careful about your factors of two.

-2 - V1(t) = Vout(t)/2

 

[esmeco]

...

So, v(out)=-4 - 2senpi*t
So,what would be the amplitude of this sinusoidal wave?

 

[berkeman]

So, v(out)=-4 - 2senpi*t
So,what would be the amplitude of this sinusoidal wave?

Graph a couple cycles of that function, and you tell me. What is the peak-to-peak AC voltage amplitude, and what is the DC offset voltage?

 

[esmeco]

The peak-to-peak amplitude of the AC voltage is 2 I guess...And the DC offset voltage is 1v I guess...Should we substitute the t in the above expression to determine the amplitude of the output wave?

 

[berkeman]

The peak-to-peak amplitude of the AC voltage is 2 I guess...And the DC offset voltage is 1v I guess...Should we substitute the t in the above expression to determine the amplitude of the output wave?

Almost correct. The convention is to measure the DC offset to the average (middle) part of the sine wave, which in this case is -4V. Just think of what the voltage would be if the amplitude of the sine wave were 0V, and that's the DC offset part of the output voltage waveform.

So now you've answered the original problem. You've drawn the waveform as you've been asked (be sure to label the time ticks on the horizontal axis correctly, and the voltage ticks on the vertical axis), and you know that the AC amplitude is 2Vpp, and the DC offset is -4Vdc. See, that wasn't so hard, right?

 

[KaiZX]

Bumping this thread up (since my question is related)...

1. The problem:

Suppose we're given an AC coupled inverting amplifier (aka. active high pass), and with a DC source at the positive input to provide offset. The circuit diagram is attached. Input Vs is assumed to be some arbitrary sinusoidal input.

How do I prove that the DC source, Vref, is not amplified at the output with such a configuration?

2. My attempt:

I know that you need to use superposition here, with the DC-decoupled AC signal and the DC offset. So I can find that the transfer function for the AC input is just the gain for an inverting amplifier.

But for the DC offset, I assumed this circuit to be a DC amplifier with Vin = 0, then I did this...

(0 - Vref) / R1 = I1
(Vout - Vref) / R2 = I2
I1 = I2

Therefore, Vout = Vref * (1/R2 - 1/R1)

But from lab, I know that Vref shouldn't be have a gain. In other words, the output should just be a sinusoidal waveform with a DC offset = Vref.

Where's my mistake?

 

[berkeman]

For DC, the capacitor is an open circuit, and so R1 goes away as well. What is left?

 

[KaiZX]

For DC, the capacitor is an open circuit, and so R1 goes away as well. What is left?

Ah, so it's almost like a voltage follower left over for DC, then superimposed that with AC gain.

Ok, thank you.