Just a little confused..

Circuit - http://i46.tinypic.com/23rpvyr.jpg
3Db cutoff points - http://i48.tinypic.com/sdephh.jpg
Voltage Gain - http://i47.tinypic.com/9aq0i8.jpg

This is supposed to be a high pass filter, and according to the question we should set this from 1hz to 30Khz. But when upping the scale to lets say 900Khz it becomes a bandpass filter. Now I dont know if this was meant to be up until 300Khz(typo???).

We need to get the lower cut off freq which is roughly 295Hz. Now am I to take the upper cutoff freq as infinity(as in ideal conditions for high pass filters) or use the higher 300khz scale and use that 3dB value?


Last question is the voltage gain: Practical we say Av(dB)=20log*(Vout/Vin) where I got that above graph.
Can I just use a mid frequency say 10Khz and use that as a value or am I missing something?

Please help , Im very desperate for all marks I can get, even if its just an extra point.

 

It looks very much like a bandpass configuration.

The presence of C1 gives a zero DC gain.

At mid-band frequency C1 reactance will be much lower than C2 reactance and if we further assume C1 reactance to be much lower than 5.36k then the voltage gain magnitude is essentially unity [R2/R1] or 0dB.

As C2 reactance decreases with increasing frequency C2 shunts R2 [with C1 reactance effectively zero] and the gain tends to drop towards zero at ∞ frequency.

The corner frequencies are notionally about 300Hz and 300kHz.

I don't understand how you are showing 105dB gain at 1Hz in your attachment.

 

Theoretically the gain should be: Av= -R2/R1 = 1=0dB
But with practical I used Av=Vout/Vin and then it gives that graph which makes me ponder even more. The differences between the two is miles out.
My input is set at pin 2.

When i change input probe between R1 and C2 I get:
http://i49.tinypic.com/11ceici.jpg

Should the input part be directly at pin2 or maybe even directly at the AC source point?

 

Theoretically the gain should be: Av= -R2/R1 = 1=0dB
But with practical I used Av=Vout/Vin and then it gives that graph which makes me ponder even more. The differences between the two is miles out.
My input is set at pin 2.

When I use the probe between R1 and C1 I get the following output. This looks much better
http://i48.tinypic.com/dvjx34.jpg

You are confused.

The gain is measured from the AC source terminal to the amplifier output terminal.

The result should look like that shown in the attachment.

 

Tnk, sorry I edited my post because I had a error where I had used the output/output by mistake.

 

There we go. Voltage gain = Output after the filter divided by the input at the source.

http://i47.tinypic.com/2q900g3.jpg

 

OK - I see you edited the image but you are still doing it wrong.

The gain is measured from the source [V1 active terminal] to the amplifier output pin6 - not from the R1.C1 junction to the output.

 

There we go. Voltage gain = Output after the filter divided by the input at the source.

http://i47.tinypic.com/2q900g3.jpg

Yep - You now have it correct.

 

Thank you very much TNK. Really appreciate your feedback!

 

And now I kind of want to reopen this thread.

The lecturer made an error on the schematic (wrong value on C2), but i think for now lets keep the values the same and work from there.

1st let me ask if there is any difference between 3dB points and the upper and lower cut-off freq?

The question directly states:
a)Obtain the lower and upper cut-off freq for this filter.
b) Determine the Bandwidth of this filter.
C)Determine the Voltage gain of this filter.

So as posted here: http://i48.tinypic.com/sdephh.jpg

a)lower cut-off freq = 296Hz
Upper cut-off freq = 222Khz
b) Bandwidth = 222.11Khz
c)Voltage Gain = 0dB

And the lecturer marked all my answers wrong.....!! Please can someone help me as I don't understand what I did wrong.

With the change in C2 the upper cut off freq and Bandwidth should change; The lower cut-off and Voltage gain should remain the same according to my understanding. (this also reflects on pspice)

 

Were you allowed to use simulation to get your answers? Did you also try to get them analytically?
The 741 op amp will significantly reduce the upper cutoff frequency, compared to an ideal op amp. Did the original question specify a 741?

 

Were you allowed to use simulation to get your answers? Did you also try to get them analytically?
The 741 op amp will significantly reduce the upper cutoff frequency, compared to an ideal op amp. Did the original question specify a 741?

We were suppose to only use simulation to get our answers. And yes the schematic is directly from our practical guide as is with the 741 op amp.
The only reason I think is a you say the simulation is suppose to differ from the ideal conditions. So why is the simulation giving me those "wrong/ideal" results then?

 

We were suppose to only use simulation to get our answers. And yes the schematic is directly from our practical guide as is with the 741 op amp.
The only reason I think is a you say the simulation is suppose to differ from the ideal conditions. So why is the simulation giving me those "wrong/ideal" results then?

If the op amp were ideal, the low frequency corner would be very close to Fc=1/(2πR1C1), and the high frequency corner would be very close to Fc=1/(2πR2C2).
The 741 has a gain-bandwidth product of 1MHz. This means that, at 250kHz, the open loop gain is only 4, and the output is phase-shifted 90° from the input. When the gain is this low, the op amp is far from ideal, so the high frequency corner is shifted down from the ideal case.
I don't understand why your answers were marked as wrong. Did you find out why this was done?

 

I did try to find out but then I got into an argument with the lecturer and decided before this goes the wrong way I will leave it. The lecturer will probably tell us in a week or two, but trying to speak to him one on one will get me nowhere.

"the open loop gain is only 4": As this is a single stage does that imply the gain of 4 is low?

 

This is a biquad filter with 1 zero and 2 poles.

By the way, my analysis is consistent with t_n_k's plot in post #4. There is a zero at 0 Hz and poles at 297 Hz and 297 kHz. It resembles a bandpass filter more than a highpass filter.

 

I did try to find out but then I got into an argument with the lecturer and decided before this goes the wrong way I will leave it. The lecturer will probably tell us in a week or two, but trying to speak to him one on one will get me nowhere.

"the open loop gain is only 4": As this is a single stage does that imply the gain of 4 is low?

An OPEN LOOP gain of 4 is far from that of an ideal op amp, whose open loop gain is infinite.

 

This is a biquad filter with 1 zero and 2 poles.

By the way, my analysis is consistent with t_n_k's plot in post #4. There is a zero at 297 Hz and poles at 297 Hz and 297 kHz. It resembles a bandpass filter more than a highpass filter.

Isn't the zero at the origin?

 

This is a biquad filter with 1 zero and 2 poles.

By the way, my analysis is consistent with t_n_k's plot in post #4. There is a zero at 297 Hz and poles at 297 Hz and 297 kHz. It resembles a bandpass filter more than a highpass filter.

Yup sorry I had that part all wrong regarding it being a high pass filter. This is definitely a band pass filter.

 

Isn't the zero at the origin?

Yes, thank you. You're correct.