Thank you sir! For interest I am A Biomed engineer, and in the middle to late 1990's we had training about this exact type of circuit as used by Schiller from Switzerland. I can send you another analog circuit from my old manuals that you can analyze that is handling the measurement of oxygen concentration in the blood, namely pulse oximetry by using dual wavelength.
The diodes clamp the Voltage at the MID point and R5 acts like a (bad) current source to feed the capacitor. This limits the output signal amplitude. The problem is that the current depends on the voltage already present on C2. So a square wave will not produce a triangle waveform but a sequence of RC (e^-t) charge discharge curves. For low amplitudes around 0v I guess that's just some distortion. But if there's a DC offset then the output is severely distorted and assymetric.
If the time constant is many times (rule of thumb 10times) the period of signal being considered, it works as an approx integrator, thus a trianglewave comes out for square input
@@sambenyaakov the final result looks nice with all the spikes removed. But sometimes we might be overdesigning for the desired behaviour. My question is, did you feed that same corrupted input signal to a circuit with no diodes and no 10megs? My guess, it should give an equally clean signal. It might be good to analyze or check via simulation, the case where really this design visibly outperforms the simpler one without diodes or 10megs. Otherwise, you could use a bomb to kill a mosquito. It works. But do you need it? The verdict is not super clear for me. A followup video addressing "why you really need this" is a good idea because this video only really proves "this will work".
Awesome explanation professor! If it's for medical appliances, only breathing gets so low frequencies. I would have really appreciated if you would have run the simulation from 0.1Hz. When you say it's a good schematic, maybe it should be compared against the classic amplifier-integrator designs.
Nice explanation¡. On 12:00 there could be also some fluctuations of the output saturation voltage of the opamp based on voltage supply specially if there's noise. Don't think that would be much of a problem in this case though
cool circuit. looks like a very good LPF. Are there similar active HPFs that operate in the opposite way ? only take an input signal with a slew rate gtr than some rate?
Great explanation. I was really close to correct answer but I didn't suspect that it's slew rate limiter circuit. Anyway I figured out why there's diode bridge instead of two zener diodes connected in series but with opposite direction (like anode-anode or cathode-cathode). Obviously this solution should be more symmetrical due to relatively high zener voltages differences between of the various particular diodes of the same type. For example: BZX55C6V2 has: zener voltage 5.8 ... 6.6 V at Izt1=5 mA test current. Second issue is that zener diodes has high reverse current below zener voltage. For example BZX55C6V2 has: Ir
@@sambenyaakov The thing is, it only works for cleaning up signals. I wonder what one could do to clean up a power line supply from power line spikes. Is it even possible to modify this circuit to that end?
Unless I am missing something, the diodes can do nothing to increase symmetry, A positive voltage goes D2, D5, D3, a voltage drop of 3 X .7v, or 2.1 total limiting pos volts assuming perfect diodes. A negative voltage goes D1, D5, D4, or .7, , 6.2 , .7, a total 7.6. volts negative. Not withstanding the +/_voltage discrepancy, the very low currents here, is poor design practice given the temperature dependence of the diode leakage currents, any actual clipping level is ill defined. If it relies on "soft" clipping, random diode noise will "overtake" any bias currents rendering the design non repeatable. I suspect a SPICE run without the diodes will give a similarly good result.
R5 together with R3 forms a voltage divider between the output of U2 and capacitor C2. The voltage drop across R6 is negligible. It seems to me that this voltage divider determines the operating point of the diode bridge.
@@GluonToo Ok, I understand that. But what is the purpose of resistor R5 exactly? If it is not there (0 Ohm), then nothing will change fundamentally, right?
NO! 1 It doesn't make sense because there are no such high amplitude pins in the input signal. 2. A good audio amplifier carries a signal with a frequency of even 100 kHz, which corresponds to a period of 10 us. That is, the filtered pins would have to have a period of, let's assume, 10 ns. You won't find such pins in an audio signal, and besides, an audio amplifier won't carry them anyway. They will be filtered out “naturally”. 3. Such additional filtering circuit would introduce a lot of distortion in the audio signal, thus disqualifying the preamplifier. It would also be a source of unnecessary noise.
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Besides the technical explanstion, what i appreciate much is the approach of analysing step by step instead of simply explaining.
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Thank you sir! For interest I am A Biomed engineer, and in the middle to late 1990's we had training about this exact type of circuit as used by Schiller from Switzerland. I can send you another analog circuit from my old manuals that you can analyze that is handling the measurement of oxygen concentration in the blood, namely pulse oximetry by using dual wavelength.
Thanks for sharing
Very nice circuit proffessor. Thanks for introducing & explaining it!
😊🙏
I subbed❤ Thanks for these videos
👍🙏
Great video, thanks professor
Thanks
As always, very informative, thanks!
🙏😊
The diodes clamp the Voltage at the MID point and R5 acts like a (bad) current source to feed the capacitor.
This limits the output signal amplitude.
The problem is that the current depends on the voltage already present on C2.
So a square wave will not produce a triangle waveform but a sequence of RC (e^-t) charge discharge curves. For low amplitudes around 0v I guess that's just some distortion.
But if there's a DC offset then the output is severely distorted and assymetric.
If the time constant is many times (rule of thumb 10times) the period of signal being considered, it works as an approx integrator, thus a trianglewave comes out for square input
Indeed
@@sambenyaakov the final result looks nice with all the spikes removed. But sometimes we might be overdesigning for the desired behaviour. My question is, did you feed that same corrupted input signal to a circuit with no diodes and no 10megs? My guess, it should give an equally clean signal. It might be good to analyze or check via simulation, the case where really this design visibly outperforms the simpler one without diodes or 10megs. Otherwise, you could use a bomb to kill a mosquito. It works. But do you need it? The verdict is not super clear for me. A followup video addressing "why you really need this" is a good idea because this video only really proves "this will work".
Awesome explanation professor! If it's for medical appliances, only breathing gets so low frequencies. I would have really appreciated if you would have run the simulation from 0.1Hz.
When you say it's a good schematic, maybe it should be compared against the classic amplifier-integrator designs.
Good points, If I have the time
Nice explanation¡. On 12:00 there could be also some fluctuations of the output saturation voltage of the opamp based on voltage supply specially if there's noise. Don't think that would be much of a problem in this case though
I agree, Thanks for input,
cool circuit. looks like a very good LPF. Are there similar active HPFs that operate in the opposite way ? only take an input signal with a slew rate gtr than some rate?
Not sure
Great explanation. I was really close to correct answer but I didn't suspect that it's slew rate limiter circuit.
Anyway I figured out why there's diode bridge instead of two zener diodes connected in series but with opposite direction (like anode-anode or cathode-cathode). Obviously this solution should be more symmetrical due to relatively high zener voltages differences between of the various particular diodes of the same type. For example: BZX55C6V2 has: zener voltage 5.8 ... 6.6 V at Izt1=5 mA test current. Second issue is that zener diodes has high reverse current below zener voltage. For example BZX55C6V2 has: Ir
Thanks for input
Very interesting circuit. So, essentially it removes spikes. I wonder if clicks removers for record players also used this technique.
Good question! Useful for low frequency perhps can be modified
@@sambenyaakov The thing is, it only works for cleaning up signals. I wonder what one could do to clean up a power line supply from power line spikes.
Is it even possible to modify this circuit to that end?
@@Stelios.Posantzis perhaps
Unless I am missing something, the diodes can do nothing to increase symmetry, A positive voltage goes D2, D5, D3, a voltage drop of 3 X .7v, or 2.1 total limiting pos volts assuming perfect diodes. A negative voltage goes D1, D5, D4, or .7, , 6.2 , .7, a total 7.6. volts negative. Not withstanding the +/_voltage discrepancy, the very low currents here, is poor design practice given the temperature dependence of the diode leakage currents, any actual clipping level is ill defined. If it relies on "soft" clipping, random diode noise will "overtake" any bias currents rendering the design non repeatable. I suspect a SPICE run without the diodes will give a similarly good result.
The zener diode is reverse biased in both cases
D2, D5, D3, a voltage drop of 3 X .7v, or 2.1 Sure about that?
@@sambenyaakov You got me there, nice one , but you are forgiven. I will try harder next time, I promise.
What is the purpose of resistor R5?
R5 together with R3 forms a voltage divider between the output of U2 and capacitor C2. The voltage drop across R6 is negligible. It seems to me that this voltage divider determines the operating point of the diode bridge.
@@GluonToo Ok, I understand that. But what is the purpose of resistor R5 exactly? If it is not there (0 Ohm), then nothing will change fundamentally, right?
Can`t we use this circuit with audio preamplifier.?
NO!
1 It doesn't make sense because there are no such high amplitude pins in the input signal.
2. A good audio amplifier carries a signal with a frequency of even 100 kHz, which corresponds to a period of 10 us. That is, the filtered pins would have to have a period of, let's assume, 10 ns. You won't find such pins in an audio signal, and besides, an audio amplifier won't carry them anyway. They will be filtered out “naturally”.
3. Such additional filtering circuit would introduce a lot of distortion in the audio signal, thus disqualifying the preamplifier. It would also be a source of unnecessary noise.
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