op-amp stray-capacitance

At low freq we have classic G = 1 + R2 / R 1, R2 = 300k.....

So calc that G in db, take away 3 db, and translate back to linear G.
Or just use your rule of thumb 3 db down = .707.....

That then tells you what the z of the combo of R1 and Cstray must
be, in turn you can then find out what freq thats at to create that Z.


Regards, Dana.

What is the impedance of the capacitor at a frequency of 100 kHz?

danadak said:

At low freq we have classic G = 1 + R2 / R 1, R2 = 300k.....

So calc that G in db, take away 3 db, and translate back to linear G.
Or just use your rule of thumb 3 db down = .707.....

That then tells you what the z of the combo of R1 and Cstray must
be, in turn you can then find out what freq thats at to create that Z.


Regards, Dana.

Click to expand...

I'm not sure if i understand, the goal is to determine R1 and R2 so that the gain is not affected by no more than 3 dB at 100kHz. Can you explain in more mathematics?

We actually need to calculate a new values of R2 and R1 for that frequency.

RS_Vimperk said:

We actually need to calculate a new values of R2 and R1 for that frequency.

Click to expand...

The ratio of R1 to R2 must remain fixed, as the DC gain must stay the same.

The question comes down to how low a value must the (presently) 120K resistor be, so adding 43K in parallel changes the ratio by no more that root 2, then setting a proportional value in place of the 300K to maintain the ratio.

eg. A simple and useable solution would be to divide both resistors by ten, fifteen or more, so adding the 43K in parallel has a minimal influence on the ratio - but I suspect the correct solution is to find values that just meet the question criteria.

RS_Vimperk said:

I'm not sure if i understand, the goal is to determine R1 and R2 so that the gain is not affected by no more than 3 dB at 100kHz. Can you explain in more mathematics?

We actually need to calculate a new values of R2 and R1 for that frequency.

Click to expand...

You start here -



So compute Vout / Vin = G for - 3 db.

Then move on the computing the R's for the G, in this case the Z's since its at 100 Khz and a
reactive component is in the fdbk loop.


Regards, Dana.

Here is a sim just for reference before you do the calculations. Note the problem
assumes an ideal OpAmp, eg. infinite GBW, but real OpAmps have limits in many
performance aspects. Notice sever peaking that occurs, you can actually do AC
analysis and predict this.




Regards, Dana.

I tried to calculate R2 and R1, taking into consideration that R2=2.5*R1

\( A_{v}(dB)=20*log (\frac{v_{out}}{v_{in}})= 20*log(3.5)=10.88 dB \)

i took away 3dB and it becomes 7.88 dB

then

\[ 7.88=20* log(\frac{v_{out}}{v_{in}}) \]

when i substituted Z as R1 in parallel with Xc and as Xc=43k

but actually i got a negative value of R1.

There should be something wrong with my calculations.

The G you find in your is the ratio of R2 / |r1||C|,
eg. compute the magnitude of the Z.



Regards, Dana.

danadak said:

The G you find in your View attachment 138776 is the ratio of R2 / |r1||C|,
eg. compute the magnitude of the Z.

View attachment 138777

Regards, Dana.

Click to expand...

But this is non-inverting amplifier so G should be 1+R2/Z.

and z is parallel combination of R1 and XC so Z=(R1*Xc)/(R1+Xc)

There is problem in the mathematics in general, not all is clear.

RS_Vimperk said:

There is problem in the mathematics in general, not all is clear.

Click to expand...

What is the disconnect for you ?


Regards, Dana.

\[ 7.88=20*log(\frac{v_{out}}{v_{in}})\\ \frac{v_{out}}{v_{in}} = 10^{\frac{7.88}{20}}=2.477=1+\frac{R_2}{Z}\\ \frac{R_2}{Z}=1.477\\ \frac{R_2}{\frac{R1*X_c}{R_1+X_c}}=1.477\\ R_2=2.5*R_1 => 2.5*\frac{R_1+X_c}{X_c}=1.477\\ 2.5*R_1+2.5*X_c=1.477*X_c => 2.5*R_1=-1.023*X_c \]

The answers should be R1=60.2k, R2=150.6k

According to the textbook.

So there is something wrong in this mathematics, i think we lack knowledge of theory that leads us to wrong results, resistor can never be negative and moreover even the absolute value is wrong.

300K / 120K = gain of 3.5 with the original circuit.
150.6K / 60.2K = gain of 3.501, near enough.

60.2K in parallel with 43K = 25.08K

150.6K / 25.08K = gain of 7.
A 3db gain change from 3.5 would be to about a ratio of 4.95, not 6.


The book answer is apparently wrong; it's not all that rare.

Also:

RS_Vimperk said:

I tried to calculate R2 and R1, taking into consideration that R2=2.5*R1

Av(dB)=20∗log(voutvin)=20∗log(3.5)=10.88dBAv(dB)=20∗log(voutvin)=20∗log(3.5)=10.88dB A_{v}(dB)=20*log (\frac{v_{out}}{v_{in}})= 20*log(3.5)=10.88 dB

i took away 3dB and it becomes 7.88 dB

Click to expand...

Remember that effectively reducing the value of R1 in proportion to R2 with increase the gain, not decrease it!

You should have added 3db in there.

Can you explain in more mathematical way? I just feel lost between the words.

As for the book's answer, it can be wrong, however when i solved other problems i didn't find any single mistake (up till now). I mean there could be errors but not that essential, for example only in the fraction after the whole number.

Even if those numbers are incorrect, i would like to perform a more comprehensible approach to understand all the process, should that be possible, of course.

Thank you all so much for your cooperation.

RS_Vimperk said:

The answers should be R1=60.2k, R2=150.6k

According to the textbook.

So there is something wrong in this mathematics, i think we lack knowledge of theory that leads us to wrong results, resistor can never be negative and moreover even the absolute value is wrong.

Click to expand...

Actually in circuit analysis not uncommon to see a -R element. But this circuit no.

I dont see you doing the magnitude of the Z, which eliminates - signs as every term
is squared and then rooted to get magnitude of the Z. Post # 10


Regards, Dana.

RS_Vimperk said:

Can you explain in more mathematical way? I just feel lost between the words.

Click to expand...

You need to understand the circuit operation before you can do any maths, otherwise the math is irrelevant or incorrect.

eg. Shunting the 120K in the original circuit can only over lower the effective value and increase the circuit gain.
However you calculated for a decrease in gain - which means making the 120K (or whatever replaces it) higher in value, in proportion to the 300K position.

That can only be done in practice by adding a series component - or in maths by adding a negative resistance in parallel, which is why you got crazy results.

That would not have happened if you had properly thought about how the circuit works, before jumping in to (irrelevant, due to a misconception) over-complex calculations.

You need to look at whatever circuit you are working on and visualise its operation, and be able to rough out results in your head, before you start doing precision calcs - if they are ever actually needed, which they are not in a lot of real-world design work; simple mental calculations are often more than sufficient.

eg. My "Divide both by ten" solution in post #6 - it fits the requirements of the question and would be a simple real-world solution to the problem. If the resistor value is small in comparison to the capacitor impedance, the gain change is also going to be small. My ballpark mental calculation put the gain increase at somewhere around 30% or so, safely below the 41% increase limit (100% -> 141%; 1 / 0.7071, which is root 2).

Yes, a calculator is needed for such as capacitive and inductive reactance, resonant frequency and RF calculations - but very rarely for general projects.

It's real-world reality - Even if you choose to use E96 series 1% resistors, that means no resistor is guaranteed to be closer that 1% to it's nominal value and the steps between values are about 1.5%

More likely you would choose from E24 values, which have a step between values of roughly 10%.

And most capacitors are 5% or 10% tolerance, unless you pay crazy prices.

You have to design things to work with lack of precision in normal components, so there is frequently no point calculating at multi-digit accuracy, as long as you understand the principles and can do that if it's ever needed.

If a circuit needs extreme precision, then you use adjustable components in appropriate key locations - trimmer (preset) resistors and capacitors, often in conjunction with a fixed component to limit the adjustment range and increase adjustment accuracy.
Those can be used calibrate things and compensate for inaccuracies in other components, and allow for component aging and drift, so the device or instrument etc. can be recalibrated at times to maintain its precision.

danadak said:

Actually in circuit analysis not uncommon to see a -R element. But this circuit no.

I dont see you doing the magnitude of the Z, which eliminates - signs as every term
is squared and then rooted to get magnitude of the Z. Post # 10


Regards, Dana.

Click to expand...

You are right. I re-wrote the formula.

\[ \frac{2.5*|R_1-j X_c|}{|-jX_c|}=1.477\\ 2.5*\sqrt (R_1^{2}-X_c^{2})=1.477 *|-j X_c| \\ 6.25*(R_1^2-X_c^{2})=2.18*X_c^2 \\ R_1= 50k\\ R_2=2.5*R_1=125k \]

Does that sound reasonable?