Not sure what circuit you were measuring but it is not the one you showed at the beginning. That circuit have a frequency response which is nothing like what your measuring is showing. Either you are measuring a completely different circuit or Siglent have some serious flawed measurement. The circuit in the beginning is actually just a low pass filter. A bandstop filter needs a different topology. The circuit in the beginning have a transfer function: (C1*R1*s + 1)/(1 + C1*C2*R1*R2*s^2 + ((C1 + C2)*R1 + C2*R2)*s), or with the values shown in the sketch, (2.200*10^(-5)*s + 1.000)/(4.400*10^(-8)*s^2 + 0.004*s + 1.000). This gives around -20dB/decade as a low-pass filter and has a cut of freq of roughly 40Hz (3dB limit)
Thanks for your thoughts. 👍 Actually you are right! I misdrew it. The 2nd resistor goes in series from the output to though C2 to ground. Or to put it another way, the output is between the two resistors. Good catch!
@@uni-byte If you are not convinced by the mathematics you can just do a simple thought experiment. According to your schematic (if I understand it correct) the output from the filter is taken as the voltage drop over the last capacitor. Right? For high:ish frequencies we can do an engineering approximation and replace the capacitor with a short-circuit. The output is then measuring the voltage drop over a short circuit, which is 0V. So you can never have any oputput for high frequencies. If I do the math the filter function of the schematic drawn (in Hertz) it is: H(f)=5.757*10^5*sqrt(1.000 + 1.911*10^(-8)*f^2)/sqrt((f^2 + 2.105*10^8)*(f^2 + 1.574*10^3)) and as can be seen you have only f-terms in the denominator which shows that this is a low-pass filter and nothing else. You can do a simple simulation (e.g. using LTSpice which is both free and accurate and supports the Spice ".ac" directive) of the schematics to convince yourselves that the math is correct (easier to play with component values in LTSpice than on the breadboard :-) ) The mistake in the schematics is that you need a LP and HP stage in parallel as the output, your output is just the LP stage. I'm not trying to discourage you from making these type of videos, all spread of electronics knowledge is good! I just want it to be correct for all newbies that might get confused by small mistakes that we all make from time to time. Even I (who do electronics semi-professionally) do mess up (more often than I like to admit!)
Great video! A quick question which you may have figured out yourself already. In your example you only need about 5dB of dynamic range, but is this dynamic range limited by the noisefloor of the scope? In 10 bit mode the DR should be about 60dB, but because of internal amplification and attenuation it can be much higher of course. Does the scope 'switch' the input amplification automatically to make full use of the available resources, or once it sets the output 'vertical controls' it doesn't change them during the sweep? If the latter it would severely limit the usability of the bode plot!
Not sure what circuit you were measuring but it is not the one you showed at the beginning. That circuit have a frequency response which is nothing like what your measuring is showing. Either you are measuring a completely different circuit or Siglent have some serious flawed measurement.
The circuit in the beginning is actually just a low pass filter. A bandstop filter needs a different topology.
The circuit in the beginning have a transfer function: (C1*R1*s + 1)/(1 + C1*C2*R1*R2*s^2 + ((C1 + C2)*R1 + C2*R2)*s), or with the values shown in the sketch, (2.200*10^(-5)*s + 1.000)/(4.400*10^(-8)*s^2 + 0.004*s + 1.000). This gives around -20dB/decade as a low-pass filter and has a cut of freq of roughly 40Hz (3dB limit)
Thanks for your thoughts. 👍 Actually you are right! I misdrew it. The 2nd resistor goes in series from the output to though C2 to ground. Or to put it another way, the output is between the two resistors. Good catch!
@@uni-byte If you are not convinced by the mathematics you can just do a simple thought experiment. According to your schematic (if I understand it correct) the output from the filter is taken as the voltage drop over the last capacitor. Right? For high:ish frequencies we can do an engineering approximation and replace the capacitor with a short-circuit. The output is then measuring the voltage drop over a short circuit, which is 0V. So you can never have any oputput for high frequencies.
If I do the math the filter function of the schematic drawn (in Hertz) it is: H(f)=5.757*10^5*sqrt(1.000 + 1.911*10^(-8)*f^2)/sqrt((f^2 + 2.105*10^8)*(f^2 + 1.574*10^3)) and as can be seen you have only f-terms in the denominator which shows that this is a low-pass filter and nothing else.
You can do a simple simulation (e.g. using LTSpice which is both free and accurate and supports the Spice ".ac" directive) of the schematics to convince yourselves that the math is correct (easier to play with component values in LTSpice than on the breadboard :-) )
The mistake in the schematics is that you need a LP and HP stage in parallel as the output, your output is just the LP stage.
I'm not trying to discourage you from making these type of videos, all spread of electronics knowledge is good! I just want it to be correct for all newbies that might get confused by small mistakes that we all make from time to time. Even I (who do electronics semi-professionally) do mess up (more often than I like to admit!)
@@johanp162 Did you read my updated reply?
Great video!
A quick question which you may have figured out yourself already. In your example you only need about 5dB of dynamic range, but is this dynamic range limited by the noisefloor of the scope?
In 10 bit mode the DR should be about 60dB, but because of internal amplification and attenuation it can be much higher of course. Does the scope 'switch' the input amplification automatically to make full use of the available resources, or once it sets the output 'vertical controls' it doesn't change them during the sweep? If the latter it would severely limit the usability of the bode plot!
Thanks.
You can configure the Bode plot whatever way you want. In this case I let it set most things automatically.