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Connecting passive low pass filters in series and with a load, reduces the dc voltage. Maybe it is very simple but would you explain a little bit to me about the low pass filter's effects on dc. Filters don't reduce dc without a load by the way.

Source 12VDC(with little ripple) Load 20 Ohm.

Filter section(RC lowpass passive)

If i take the output of the first filter,

500 Ohm 1mF (cutoff around 0.3Hz)---Out ~7VDC---

If i take the output of the second filter also current draw reduces a lot.

500 Ohm 1mF ---Out ~0.20VDC--- etc.

Thank you.

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    \$\begingroup\$ If you connect a RC lowpass filter to a load, at DC the capacitor is essentially an open circuit so you just have two resistors in series, which is a voltage divider. \$\endgroup\$
    – jramsay42
    Jan 31, 2018 at 22:46
  • \$\begingroup\$ ah.. ok got it now. Thank you. Never thought of it this way. \$\endgroup\$
    – Ali Somay
    Jan 31, 2018 at 22:47

1 Answer 1

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The load voltage will drop by the output_current x filter_resistor. Its just ohms law.

So in your case, when using one filter stage, it will drop by 500mV for every 1mA of load current.

So in your case, when using two filter stages, it will drop by 1V for every 1mA of load current.

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  • \$\begingroup\$ So that makes the RC Filter irrelevant for DC smoothing in my case because my load will be 600mA with almost no resistance(6 Ohms a vacuum tube filament). Am i thinking right? Because i need that amount of current with 6.3 volts potential for a pair of tubes. Any suggestions? \$\endgroup\$
    – Ali Somay
    Jan 31, 2018 at 23:09
  • \$\begingroup\$ @AliSomay Do you really need to make the filter 0.3Hz? If a higher frequency will work you could just lower the resistor value to something a lot less (like 1 ohms). Alternatively you could use an inductor (but it would have to be huge to make 0.3 Hz with 1mF). \$\endgroup\$
    – user4574
    Feb 1, 2018 at 0:08
  • \$\begingroup\$ 0.3 hertz is not a must i just wanted to make it as low as possible with my components. I will try very low ohm value resistors. \$\endgroup\$
    – Ali Somay
    Feb 1, 2018 at 0:20

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