# Help analysing this circuit - how does it act like a low pass filter?

1. Mar 31, 2014

### applestrudle

In my simulation it did act like a low pass filter so it definitely is one!

<a href="http://tinypic.com?ref=657vdh" target="_blank"><img src="http://i59.tinypic.com/657vdh.jpg" border="0" alt="Image and video hosting by TinyPic"></a>

please tell me if the following is correct:

this circuit works as a low pass filter because since Zcap = 1/jwC

when w is low the impedance is high so the current flows through R1. this means the current will go to the inverting terminal of the op amp (after joining with the output from the opamp?)

however if w is high the impedance is low which means current will flow to the capacitor which charges up so less current goes to the op amp?

thank you

2. Mar 31, 2014

### Staff: Mentor

You are basically on the right track, with a few things that should be changed.

If there were no capacitor, what is the gain of the inverting opamp in terms of the two resistor values?

And as you say, the impedance of the capacitor gets low at high frequencies, so what does the impedance of the parallel combination of Rf and C do at higher frequencies? What does that mean for the gain equation that you wrote for the resistors-only case?

3. Mar 31, 2014

### jim hardy

I think i sense a problem of vocabulary that has OP befuddled. I may be completely wrong so take this as my tossing somebody a lifeline, and if he is an olympic swimmer i aoplogize - i simply didn't know.

Physics guy in EE class, you say?
Familiarity breeds contempt it is said, and sometimes teachers in one discipline forget how unaccustomed people from other disciplines are to the all too familiar vocabulary.

You do understand the basic premise of "operational amplifier" ? (aka OPAMP)

It is "operational" because it is used to perform some mathematical 'operation', the name infers nothing more.

It is just an amplifier that is connected so that it can hold its inputs equal.
Observe that yours has its + input connected to circiuit common, or "zero volts".
So it will do its level best to hold its other input (the - input) at zero volts as well, or within a few microvolts of zero. That's your point B.

By agreement among everybody who's used these things for last several decades,
your point "B" is named "summing junction". That's because the sum of currents entering and leaving B is zero...
Vin/R10 flows into B.
So same current must flow out through parallel combination of R1-Zc.
May i call that combination Zfb ? (Zfeedback) ?
Current out = Vout/Zfb (remember B is at zero volts, and that's the key to understanding this thing)

Aha ! sum of currents at B = 0, so:

Vin/R10 + Vout/Zfb = 0

Vout/Zfb = -Vin/R10

Vout/Vin = -Zfb/R10
and that's the mathematical operation your circuit performs.

The negative sign is why it's called "inverting op(erational)amp(lifier)"
and if Zfb is a function of frequency, so is the "operation".

again--- i hope i haven't offended by belaboring what might have been obvious to you.
But this little point trips up a lot of people.

old jim

Last edited: Apr 1, 2014
4. Apr 9, 2014

### applestrudle

Thank you, this was really helpful! I have one more question, if you don't mind answering it?
If R10 and Zfb form an impedance divider, wouldn't the equation be

Vout = Zfb/(R10 + Zfb) x Vin

therefore the Gain:

G = Vout/Vin = Zfb/(Zfb + R10) !!

this is confusing me!

5. Apr 9, 2014

### jim hardy

Hmmm did your post get changed? I just finished typ[ing this, maybe i got confused. Will try to address your question iin a couple minutes.

okay , look at the math operation it performs on whatever is its input:...... Zfb/R10

Can Zfb/R10 be expressed as a f(frequency) ?

Sure .
R10 is not frequency dependent, it's just a resistance. So denominator is a constant .

Zfb is the parallel combination of R1 and Zc,
R1 is just a resistor too
but the feedback capacitor in parallel with R1 IS frequency dependent

so let's express Zfb as f(frequency).

I like to do parallel combinations by changing impedances(ohms) to admittances(mhos) and adding:
1/R1 mhos + 1/Zc mhos = (Zc + R1)/(R1Zc) mhos
which, inverted to get back to ohms = ZcR1/(R1+Zc) ohms

sadly they gave the capacitor as "Zc" because "Z" by convention means impedance, not capacitance, and is in ohms already.
So may i ascribe value Cfb farads to Zc ?

Aha ! Now we can write the familiar Zc = 1/(2pi X C X frequency) ohms, or, Zc=1/2∏fC ohms f being frequency and C farads;

and realizing that impedance is a complex number that includes both real and imaginary terms

Impedance of a resistor is a real number, ie not frequency dependent
Impedance of a capacitor is an imaginary number, ie it's got 90 degree phase shift for sinewaves.

So Zc = 1/j2∏fC

Now plug that into Zfb = ZcR1/(R1+Zc)

Zfb = R1/j2∏fC / (R1 +1/j2∏fC)
multiply numerator and denominator both by j2∏fC
and you get

Zfb = R1/(R1j2∏fC +1)

NOw our transfer function was Zfb/R10
which as f(frequency ) = R1/R10 X 1/(R1j2∏fC +1)
to make it look less awkward, replace j2∏f with s

[EDIT fixed RsC to be R1sC in 3 places jim]

and you see $\frac{R1}{R10}$ X $\frac{1}{R1sC+1}$

and only the right hand term contains frequency. s = 2∏f

at zero frequency the transfer function is clearly R1/R10 because RsC is zero

and as frequency increases, the multiplier $\frac{1}{R1sC+1}$ grows smaller because frequency is in its denominator.

That's the mathematical explanation of why it attenuates high frequency.

If you do a couple angle calculations on R1sC + 1 you'll see that the transfer function also acquires a phase shift with increasing frequency .

One needs to go through that tedious derivation to grasp the physics, or at least i did. I was fortunate to have a great high school electronics teacher who led us boys through it.

If this helps you, pass it on...

old jim

Last edited: Apr 9, 2014
6. Apr 9, 2014

### jim hardy

nope, R10 and Zfb do not form an impedance divider in the sense we usually think of one..
Recall the summing junction is held at zero.
Then go back to the premise of equal current into and out of the summing junction.

Physically, the capacitor Zc lets the opamp pull higher frequency current out of the summing junction with less effort(less voltage) than would be the case without Zc, all the current having to go through R1. When it becomes easier to pull it out than it is to push it in, you have reached the "breakpoint" of the filter.

Now - if you write Kirchoff's voltage law around the loop Vin - Summinig Junction - Vout, you will of course find the voltage divider law is satisfied. But to my way of thinking that satisfaction is the effect not the cause. It's the result of the operational amplifier being allowed to hold its inputs equal, that is "to operate". It'll put Vout wherever it needs to be to hold summing junction at zero.

As you've doubtless discerned i am not very academic.

I used "s" in post above for frequency because s is also the letter used for Laplace operator, which is how EE's attack these problems nowadays. That's what you'll see in textbooks, my j2∏f silliness is from high school when we were doing these on slide rules. Actually, in college too but i had a better slide rule.

I hope my informality is not annoying to you.
I would have loved to major in physics but couldn't do the math.

old jim

Last edited: Apr 9, 2014
7. Apr 9, 2014

### jim hardy

took sixteen minutes. sorry.