# -3 dB point of non-inv op-amp with capacitor in parallel with feedback resistor

Discussion in 'Homework Help' started by Mango Symphony, May 3, 2012.

1. ### Mango Symphony Thread Starter New Member

Mar 1, 2011
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0
Hi

For a non-inverting op-amp, it is easy to show that the gain is given by 1 + RF/R1, where RF is the feedback resistor between the op-amp output and the inverting input, and R1 is the resistor between the inverting input and ground.

With a capacitor added across RF, is easy to see intuitively that as frequency increases, the reactance of the capacitor decreases; therefore the impedance of the cap and RF in parallel becomes dominated by the reactance of the cap. Thus the expression 1 + RF/R1 tends towards 1, because the RF/R1 part tends to 0. So it is clear that a low-pass filter has been implemented - the higher the frequency, the more the gain reduces from 1 + RF/R1 towards 1, where it will stay.

However, using complex arithmetic, I am unsure how to predict where the - 3 dB point will occur, or to be absolutely clear, the frequency where the gain has fallen from the midband gain (which is given by 1 + RF/R1) to 0.707, or 1/√2 of that value.

I assumed initially, that the cap and RF will form the time constant and thus determine where the gain has fallen 3 dB from the midband value. But after deriving the transfer function of the above circuit, in operator j, I get the following:

(1) H(jω) = 1 + (RF/R1 x 1/√(1+jωxRFxC))

I then turned the whole expression on the RHS from rectangular to polar form in order to get an expression for the magnitude of the overall transfer function.

My plan then was to set this equal to 1/√2, and solve for ω.

However, when I did, it seemed a real pickle as I ended up with ω to the power of 4 and all sorts of mess!

It would have made sense to me if the number crunching had returned the following:

(2) H(jω) = (1 + RF/R1) x 1/√(1+jωxRFxC))

NB note the difference between this and (1); in (1) the '1' at the start is outside the brackets.

It would have made sense because... here we have 2 transfer functions, the first is real and constant and is the expression for the midband gain; the second is that of a LPF, whose value will become 0.707 when ωRC = 1, and everything will be nice and simple, and the reciprocal of RC will indeed define the radian - 3 dB point.

Am I overlooking some complex number issue here?

If anyone one can spot where I am going wrong on this I would be most grateful. The derivation does not appear in any of my text books or on the internet!

Thanks

Mango

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It's a good deal of algebraic manipulation to come up with a result but I end up with

$\omega_{(-3dB)}=\frac{A_{DC}}{R_FC_F\sqrt{(A_{DC}^2 -2)}}$

Where

$A_{DC}=1+\frac{R_F}{R_1}$

3. ### Mango Symphony Thread Starter New Member

Mar 1, 2011
7
0
Did you follow a similar procedure to mine, where you convert the a + jb expression to polar to find the magnitude, or some other method?

Here is my working if anyone is interested, with R1 substituted for RF, and R2 for R1 (sorry for the confusion!).

I feel like I am missing a point here on how to go about the exercise...

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4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I checked the first four lines and the gain looks fine. I don't have the patience to look much further but the approach looks much the same as I would take.

What it boils down to is finding the frequency at which the gain has fallen to the DC gain divided by √2.

You will obtain a quadratic equation in ω^2 as the unknown variable - even though it looks like a fourth order equation. You solve for the real root with ω^2 as the unknown then take the square root of the positive real root to give the actual -3dB angular frequency.

My approach was identical with the exception I got rid of a lot of the clutter by deducing that the gain is given by

$A_{v}=\frac{A_{DC}\omega_{F}+j \omega}{ \omega_{F}+j\omega}$

Where for your component tags I have the composite tags

$A_{DC}=1+\frac{R_1}{R_2}$

$\omega_{F}=\frac{1}{R_1C_1}$

I then worked through the process using the composite tags. This made the algebraic burden more bearable & less prone to error for me.

Last edited: May 4, 2012
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But do we really need to do all those calculations to find -3 dB point?
This circuit is a simply first order circuit. So for first order circuit the -3 dB point isn't always equal

$Fc = \frac{1}{2 \pi C R}$

So for the circuit under discussion the transfer function is equal to

$\frac{Vout}{Vin} = 1 + \frac{Rf||Xc1}{R1} = 1 + \frac{\frac{Rf}{1 + j \omega Rf C1} }{R1} = \frac{ R1 + Rf + j\omega Rf C1}{R1(1 + j \omega Rf C1)} = 1+\frac{Rf}{R1} * \frac{1+ j\omega \frac{Rf*R1}{Rf+R1}C1}{1+ j\omega Rf C1}$

So we have one Pole at

$F_p = \frac{1}{2 \pi C1 Rf}$

And one Zero at

$F_Z = \frac{1}{ 2 \pi \frac{Rf*R1}{Rf+R1}C1}$

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6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The -3dB point in this case depends on the DC Gain. At DC gains of less than 5 there is a modest difference in the typical 1/(RC) value for a first order case. At higher gains the difference is negligible.

In the equation in post #2 the factor of interest would be

$\frac{A_{DC}}{\sqrt{(A_{DC}^2-2)}}$

At a DC gain of 10 the the factor is 1.01, or close enough to unity.

Similarly

If DC gain = 5 the factor is 1.04

If DC gain = 1.5 the factor is 3.0

So with the feedback R=100k and C=10nF

The nominal frequency 1/(2∏RC)=159Hz

At Adc=10 the actual -3dB frequency is 161Hz.
At Adc=5 the actual -3dB frequency is 165Hz.
At Adc=1.5 the actual -3dB frequency is 477Hz.

Which is probably all fairly academic, albeit with some vague interest.

Last edited: May 4, 2012
7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I notice that t_n_k says in post #2:

"It's a good deal of algebraic manipulation to come up with a result but I end up with"

And again in post #4:

"I then worked through the process using the composite tags. This made the algebraic burden more bearable & less prone to error for me."

Nowadays when I see a problem like this I say to myself, "It's good for a beginning student to go through all the algebra by hand in the beginning, but for myself, let's see in how few steps I can get a modern symbolic algebra program to produce the answer".

Herewith the Mathematica result in an attachment:

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8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Electrician - maybe I should try Mathematica.

On the other hand the doctors & chiropractors keep telling me that an older person such as myself needs lots of mental gymnastics to reduce the likelihood of early dementia.

9. ### Mango Symphony Thread Starter New Member

Mar 1, 2011
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0
Thank you very much for all your help with this.

t_n_k - could you briefly describe or show how you arrive at your expression

$image=http://forum.allaboutcircuits.com/mimetex.cgi?A_{v}=\frac{A_{DC}\omega_{F}+j%20\omega}{%20\omega_{F}+j\omega}&hash=7dee1c832c062750460266d6c0f7ab68$

Thanks again,

Mango

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
$Z_F=\frac{R_F\frac{1}{j\omega C_F}}{R_F+\frac{1}{j\omega C_F}}=\frac{R_F}{1+j\omega C_F R_F}$

$A_v=\frac{Z_F}{R_1}+1=\frac{\frac{R_F}{1+j\omega C_F R_F}}{R_1}+1=\frac{R_F+R_1(1+j\omega C_F R_F)}{R_1(1+j\omega C_F R_F)}$

or

$A_v=\frac{R_F+R_1+j\omega C_FR_1R_F}{R_1C_FR_F(\frac{1}{C_FR_F}+j\omega)} = \frac{R_1C_FR_F$$( \frac{R_1+R_F}{R_1C_FR_F})+j\omega$$}{R_1C_FR_F(\frac{1}{C_FR_F}+j\omega)}=\frac{ ( \frac{R_1+R_F}{R_1C_FR_F})+j\omega }{(\frac{1}{C_FR_F}+j\omega)}$

or

$A_v=\frac{\frac{1}{R_FC_F}$$\frac{R_1+R_F}{R_1}$$+j\omega }{$$\frac{1}{C_FR_F}+j\omega$$}=\frac{\omega_F A_{DC}+j\omega}{\omega_F+j\omega}=\frac{A_{DC} \omega_F +j\omega}{\omega_F+j\omega}$

where

$\omega_F=\frac{1}{R_FC_F}$

and

$A_{DC}=\frac{R_F}{R_1}+1=\frac{R_1+R_F}{R_1}$

11. ### Mango Symphony Thread Starter New Member

Mar 1, 2011
7
0
Thanks t_n_k

I have to say that is some very nifty algebraic manipulation...

Is there a term/description to this type of parameter juggling, where you are expressing the complex transfer function in terms of the time constant formed by Rf and Cf, and the mid-band/DC gain, in order to get a final expression that lends itself to actually finding out something useful? I couldn't see this wood for my trees when I ended up with the quadratic...

I.e. is this a known standard procedure to try out when dealing with transfer functions or were you just being nifty and clever?

Mango

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Nothing particularly "nifty" in my methods. I try to see patterns in formulae which might lead to simplifications of this type. Algebra can be a drudgery sometimes. In the end it probably comes down to years of 'playing' with these ideas.