cut-off frequency with circuit given

[mkforlife10]

If R= Ri= 10kΩ, C= 10pF, and Rf= 100kΩ.
the cutt of frequency will be 1/2piRC
so the R will be Ri because cutt off frequency depends on the input frequency..so we use input frequency to find the cutt off frequency,
So Fc in this case is 1/2piRiC
Fc=1/(2pi)(10*10^3)(10*10^-12)=1.59MHz

 

[MikeMl]

Unless the task is to determine the effect of the opamp frequency response, and how it might interact with an inductive load while considering the opamp's pin capacitance and stray circuit capacitances, then your simple analysis is correct. Basically, the input network will have a loss of 2 when the shunt impedance of the capacitor is equal to 10K...

In real life, the frequency response of the posted circuit is much more complicated than just the simple RC single-pole network.

 

[mkforlife10]

ok, just to make sure...this circuit non-inverting so in that case the gain will be
So Av(mid) =1+RF/R and R=Ri
so the gain will be 11

 

[audioguru]

Most opamps do not work at a frequency as high as 1.6MHz.

 

[mkforlife10]

Most opamps do not work at a frequency as high as 1.6MHz.

yeah its too high...thus the ckt is high-pass filter....I wonder how the bode plot will luk like
cut off frequency is 1.6MHz... and mid band gain in db will arund 20...
any1 knws how to do bode plot in multisim for this ckt?

 

[MikeMl]

Most opamps do not work at a frequency as high as 1.6MHz.

Especially with a closed-loop gain of 11.

 

[audioguru]

I don't know if Multisim can see the bode plot of the opamp in its datasheet. Why don't you look?
Here is the response of a lousy old 741 opamp. With a gain of 20dB then its cutoff frequency is about 60kHz but its poor slew rate gives it a maximum output level at 60kHz of only 2.5V peak.

 

[mkforlife10]

I don't know if Multisim can see the bode plot of the opamp in its datasheet. Why don't you look?
Here is the response of a lousy old 741 opamp. With a gain of 20dB then its cutoff frequency is about 60kHz but its poor slew rate gives it a maximum output level at 60kHz of only 2.5V peak.

thanks bro ...vey different graph...cuse the cutt off frequency is all the way down to 20 dB...i think mine will luk similar to this bode plot...

 

[mkforlife10]

I will just draw it in hand...or find it online the plot...rather thn doing it multisim...it is not required...but I was just thinking if I can...

 

[mkforlife10]

for the ckt If you exchange the position of capacitor and the resistor what happens?..thn Capcitor will get the input directly and resistor will go to ground
I wanted to see how the signal wud luk different..thts why wanted to draw it on multisim...but failed to do so...

 

[audioguru]

A resistor in series then a capacitor to ground is a lowpass filter. It passes low frequencies and reduces high frequencies.

But a capacitor in series then a resistor to ground is a highpass filter. It passes high frequencies and reduces low frequencies.

Look in Google Images for lowpass filter response and at highpass filter response.

 

[mkforlife10]

A resistor in series then a capacitor to ground is a lowpass filter. It passes low frequencies and reduces high frequencies.

But a capacitor in series then a resistor to ground is a highpass filter. It passes high frequencies and reduces low frequencies.

Look in Google Images for lowpass filter response and at highpass filter response.

got it...so the original ckt is low-pass filter because C goes to the ground...

 

[audioguru]

so the original ckt is low-pass filter because C goes to the ground...

Correct.

 

[Roff]

yeah its too high...thus the ckt is high-pass filter....I wonder how the bode plot will luk like
cut off frequency is 1.6MHz... and mid band gain in db will arund 20...
any1 knws how to do bode plot in multisim for this ckt?

That doesn't make it a high pass. It just reduces the lowpass bandwidth.

LT6200-10 has a GBW of 1600MHz with a gain of 10, meaning this circuit's op amp would have a BW of about 150MHz. There may be others that are higher.
Not common, but doable.

EDIT: I should point out that LT6200-10 has very high input bias and offset currents.

 

[mkforlife10]

ok....thx all for the answers

 

[mkforlife10]

i have another question related to the filter...the question is about switched-capacitor filters.
for the following circuit
the switching frequency fCLK= 50 kHz. For this circuit the cut-off frequency is calculated from: ωc= fCLKC2/Cf , and the mid-band gain is given by Av(mid) = C1/C2 .

I need to find out the values for the capacitors C1, C2 and Cf to obtain a cut-off frequency of ωc = 500 rad/s, and a mid-band gain of 10.

 

[Roff]

i have another question related to the filter...the question is about switched-capacitor filters.
for the following circuit
the switching frequency fCLK= 50 kHz. For this circuit the cut-off frequency is calculated from: ωc= fCLKC2/Cf , and the mid-band gain is given by Av(mid) = C1/C2 .

I need to find out the values for the capacitors C1, C2 and Cf to obtain a cut-off frequency of ωc = 500 rad/s, and a mid-band gain of 10.

Did you read the guidelines for the Homework help section?

 

[audioguru]

I used some switched-capacitor lowpass filter ICs to make audio signal generators with extremely low distortion. I also used them to make a single sideband suppressed carrier voice encryption transmitter and receiver.
The cutoff frequency was simply 1/100th the clock frequency so the calculation was just moving the decimal point.

 

[MrAl]

i have another question related to the filter...the question is about switched-capacitor filters.
for the following circuit
the switching frequency fCLK= 50 kHz. For this circuit the cut-off frequency is calculated from: ωc= fCLKC2/Cf , and the mid-band gain is given by Av(mid) = C1/C2 .

I need to find out the values for the capacitors C1, C2 and Cf to obtain a cut-off frequency of ωc = 500 rad/s, and a mid-band gain of 10.

Hi,

Maybe you should have asked this by starting a new thread?

To start off, with Fclk=50kHz that is 314159 rads/sec. So FCLK=314159.

If you need a mid band gain of 10 then C1 has to be ten times larger than C2 because the design formula states that Av=C1/C2, so we have:
C1=10*C2

If the cut off frequency is w=500 then using the design formula wc=Fclk*C2/Cf (Fclk here in rads/sec) we have:
Cf=FCLK*C2/wc

or with Fclk in Hertz:
Cf=Fclk*2*pi*C2/wc

So pick a value for C2 that seems reasonable and then calculate C1 and Cf and see if they both seem reasonable too. You just have to be a little careful to use the same units for both frequencies when doing the calculation for Cf. If Fclk is given in Hertz, then you either have to convert that to rads/sec or convert wc into Hertz.

 

[mkforlife10]

Did you read the guidelines for the Homework help section?

Yeah...sorry I shud hve create different thread....but i thought since its was abt filters so I aksed on the same thread...