RLC Bandpass Filter - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: March 20, 2014, 08:20:30 pm
content: I setup this filter but the results I'm getting are way off from the predicted values.VIN_pp= 5VVOUT_ppat f0= 1.064VR = 9.8R (measured)L = 100uH +- 10% (marked, but not measured)C = 1nF (marked, but not measured)f0= 1/(2*pi*sqrt(LC)) = 503kHz (Calculated)f0= 447.5kHz (Measured)Q = XL,0/ RT= (2*pi*f0*L) / RT= ~32 (Calculated)Q = 5.52 (Measured)Bandwidth = f0/ Q = 503kHz / 32 = 15.7kHz (Calculated)Bandwidth = 81kHz (Measured)I measured the resistor to get 9.8R. The cap and inductor I can't measure directly with my DMM. The inductor is marked 100uH +- 10% from digikey. The cap is marked as 1nF. I tried swapping out both the cap and inductor for components from the same package in case one of them was way out or the wrong value but the results were consistent.My voltage source is an FG. My scope was hooked up at TP and grounded to the negative lead of the FG.I found f0by sweeping the frequency until I found the maximum VOUT_pp. I found the bandwidth by finding the frequencies below and above f0that produce VOUT_PP= 0.707*VIN_ppwhich was at 407kHz and 488kHz, respectively.I'm not sure why my results are so far off. Also is my VOUT_pponly ~1V at resonance because of the 50R output impedance of the FG (ie. voltage divider between 10R resistor and 50R FG at resonance)?Lastly I don't see how this circuit is different from a notch filter when there is no load present? With a load they can be distinguished by whether the resistor or the LC is parallel with the load. But with no load present and an AC source I'm not sure how the two differ.

author: Legion
author_rank: Frequent Contributor
timestamp: March 20, 2014, 08:31:40 pm
content: If I assume a worst case scenario of L being over by 10% (110uH) and C being over by 20% (1.2nF) then I get an f0= ~438kHz. This is a lot closer to the measured values.But the Q and bandwidth are still way off with Q =~ 31 and BW =~ 14kHz.

author: jimmc
author_rank: Frequent Contributor
timestamp: March 20, 2014, 08:50:04 pm
content: The function generator is not a pure voltage source, you need to add its 50 ohm output resistance in series with the voltage generator in your equivalent circuit.Things should then start to make more sense e.g. the circuit Q becomes XL0/(50+10) ie approximately 5.2, much closer to your measured result.Jim

author: Legion
author_rank: Frequent Contributor
timestamp: March 20, 2014, 09:17:26 pm
content: Quote from: jimmc on March 20, 2014, 08:50:04 pmThat 50R output impedance gets me every time! Thanks!

author: KJDS
author_rank: Super Contributor
timestamp: March 20, 2014, 09:20:27 pm
content: What's the series resistance and capacitance of your inductor.You can calculate these from the SRF and Q given on the inductor data sheet. Don't use the DC resistance it won't take into account some of the loss.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 20, 2014, 09:22:59 pm
content: What they said. Also,5V is kind of a lot for a "small signal" test. Depending on the inductor and capacitor types and ratings, you may be saturating one or both, with the result that the observed frequency is higher than predicted (and distorted, and the resonance peak is lopsided). If this isn't apparent, then it's fine.When there is no load, there is no load, and thus no filtering. All power transmitted by the source is reflected back to it, wide band. Filters only filter when there is a source and a load, and the impedance matching is appropriate. Then, in the passband, the load draws power from the source, and in the stopband, power is reflected back to the source.Note that a reflection condition occurs when the source is more-or-less short circuited (e.g., a parallel capacitor shorts it out at high frequency) or open circuited (a series choke opens the circuit at high frequency). A filter works by using reactive components to select between these conditions (blocking and passing) over frequency.Shorts and opens are perfectly normal behavior for impedance matched loads. Shorts are not normal behavior in the realm of perfect voltage sources (you'd never short one out to reduce load power!), because a perfect voltage source is an unrealistic model -- the general Thevenin/Norton source (whichever is most appropriate for the analysis) is much more realistic.Tim