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High Pass Filter. how it works.

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mengghee

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hiye,

I have built a sawtooth signal out of a 8038 chip and i have obtained a vp-p of around 5v. but it started from 5v to 10v. not from 0v to 5v. someone told me, i can bring it down to 0v to 5v by using a high pass filter. i am just wondering how it works and how can i choose the value of the resistor. thank you.

regards,
Jeffrey
 

rkodaira

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You do not need a high pass filter but you need to eliminate the dc component.
High pass filters are used to attenuate high frequency components of a signal, not for dc level.

Maybe using a fast Amp Op as a subtractor configuration (one input for your signal and another input with 5V - the dc offset) obtaining a 0 to 5V sawtooth signal output.

In audio amplifiers, we use a large capacitor to block the dc component going to speakers, but your signal would be -2.5V to 2.5V variation.
 

    mengghee

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mengghee

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thank you very much. but instead of using an op amp, can't i just used the high pass filter to eliminate the dc component ? isn't that a cheaper option ? i have done some googling and found a circuit which uses the same exact component to generate the sawtooth and it also uses a high pass filter to get rid of the offset. thank you

regards,
Jeffrey
 

Kral

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mengghee,
The high pass filter will also work. You need to ensure that the corner frequency of the high pass filter is much higher than ( e.g., >10X) the fundamental frequency of the sawtooth in order to avoid waveform distortion.
~
Pick a value for the capacitor C.
Pick a corner frequency Fc > 10 X the fundamental sawtooth frequency.
Then R = 1 / (2Pi Fc C).
Regards,
Kral

Added after 7 minutes:

Please change the ">" sign to "<". Also change "higher than" to "lower than than". Sorry about the mistake.
regards,
Kral
 

    mengghee

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mengghee

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kral,

what is the fundamental frequency ? my frequency of the sawtooth is 50khz. is that the fundamental ? thank you.

regards,
Jeffrey
 

rkodaira

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Sorry Mengghee, I wrote wrong. A high pass filter is used to block the low frequencies components. So when I wrote about the coupling capacitor used in audio amplifier, it is a kind of high pass filter, because you can consider dc component as a 0Hz component.

But I think your problem will not be solved with only the filter (because the offset can be different of the desired 2.5V). I think the subtractor amp op more effective.

Nevertheless, to help you, the fundamental freq of your signal is the 50kHz itself. A sawtooth signal is composed of a fundamental frequency and infinite even and odd order harmonics (multiples of fundamenta freq).

A high pass filter can be designed with capacitors and inductors and resistances.
The simplest is a capacitor followed by a resistor (a kind of voltage divider), with the output the node between capacitor and resistor:

input signal ---- capacitor -----output------resistor------ground
this is a 1st order filter

or

input signal -----capacitor -----output -----inductor -------ground
this is a 2nd order filter (can attenuate better)

Some complex calculations are necessary because you are going to use impedances, so if you know the ac circuits calculus, you can do this. I suggest you to study some filter theory texts.
 

    mengghee

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Kral

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mengghee,
Yes. The sawtooth consists of a fundamental and harmonics. In your case, the fundamental is 50 KHz. The amplitudes of the harmonics are inversely proportional to the harmonic number. They are positive for odd harmonics and negative for even harmonics. So the relative amplitudes are:
Fundamental multiplier = 1/1
2nd harmonic multiplier = -1/2
3rd harmonic multiplier = 1/3
4th harmonic multiplier = -1/4
etc.
~
I just thought of another problem. The high pass will not give you a sawtooth that goes from 0 to 5 V. It will give you a sawtooth that goes from -2.5V to +2.5V, since the high pass eliminates the DC component. So perhaps you need to use the op-amp circuit suggested by rkodaira.
~
Regards,
Kral
 

    mengghee

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mengghee

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hiye,

the technician in the lab has actually tried the circuit with a high pass filter. and it actually works near perfect as the Vp-p goes from -0.2V ( approximate ) to 4.8V which is near perfect. i have tried doing so in the pspice without further analysis on calculations. and it appears to me that if the capacitor is large enough. it will go from 0 to 5V vp-p. but it my practical application, when i have the high pass filter and feed the signal to the comparator, the signal is distorted (by just measuring it without the comparator, it is perfect ) and i assume that i need a buffer ( i have also found an example using the same chip, waveform generator, comparator, high pass filter, buffer in google ). personally, i don't mind using an op amp, i too think that it will be simpler for me not to use filter, but i am afraid my lecturer didn't think so..... do you guys think it will work with the high pass filter ( as i have tried it with the pspice ). i also have pasted a link at the bottom of the circuit i found.

http://www.ece.utexas.edu/~grady/EE362L_PWM_Inverter_Control_Circuit.pdf

page 7, 11 and 12

thank you,

Jeffrey
 

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