if I have a low pass single pole filter feeding a voltage reference, how do I calculate the attenuation of the ripple before the filter, that appears after the filter.
Let's suppose we have 2Vpp of 100Hz ripple from the rectifier/smoothing caps.
Then 10k feeding a 10uF cap into a zener type (shunt) reference.
The RC time constant of the filter is 100mS (-firstname.lastname@example.orgHz).
Do I just add 6db of attenuation for each octave above 1.6Hz until I reach 100Hz i.e. 6octaves =36db of extra attenuation below the -3db starting point? total -39db below the 2Vpp supply.
What happens if two RC filter stages are cascaded? say 5k1 in each half each with 10uF of capacitance. This would seem to give -33db + -33db = -66db below the 2Vpp supply, about 1mVpp .
Does this make any sense?
Yes, your math is OK. But, you also need to consider DC attenuation on the series resistors. In your example with just 5mA drawing into the current reference, the DC attenuation on the resistors would be 50V!
I don't know why you chose 5mA and 50V as an example.
My rough supply will be 64V and the Vref is 10V. Hey presto 5.3mA to reference and leakage thro' the electrolytics and amplifier base input.
Thanks for the confirmation.
I just assumed that the current was 5mA. It follows that serial resistors with the total resistance of 10K would burn 50V. Actually, I wasn't that far off when you think about it.
|All times are GMT. The time now is 01:42 AM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2016 DragonByte Technologies Ltd.
Copyright ©1999-2016 diyAudio