I have been playing around with some basic passive filters (first order), really just to try and get a better understanding of the equations and transfer functions.

I was talking to someone about the transfer function for a band-pass filter - I was thinking that to get the full systems transfer function, I needed to multiply the high pass filters transfer function with that of the low pass filter, but they seemed to think that I would require some sort of buffer between the two stages, so I don't "load" the second stage.

I know this is quite general, so may get closed, but I am hoping someone can explain a bit about loading the second stage, and the problems this might bring up and ways to avoid it (i.e. is a buffer the only way to go).

• Related, arguably a duplicate: electronics.stackexchange.com/questions/38597/… – Null Aug 11 '15 at 14:59
• @null Thanks for that!! I didn't see that one. I will delete my post so it isn't getting in the way. Thanks again! – o.fithcheallaigh Aug 11 '15 at 15:04
• No need to delete your post. It'll get closed if the community thinks it's a duplicate and worth closing (I haven't flagged to close it). I was just pointing that out to get you started. – Null Aug 11 '15 at 15:05
• Ah, okay, cool ...still getting used to how things work around here. Cheers! – o.fithcheallaigh Aug 11 '15 at 15:06
• To more directly answer your specific question, if the output impedance of your first stage is low enough and the input impedance of your second stage is high enough, you don't need a buffer. What constitutes "enough" depends on how much loading (deviation from the ideal) you're willing to accept versus the added cost and complexity of adding a buffer. – Null Aug 11 '15 at 15:09

The loading you talk about is effectively just placing any impedance in series or parallel with the filter. This will always move your poles and change filter characteristics. Read Here for some info on the design equations for a simple RC topology.

As you make multiple ordered filters with only passive components the gain will go down, and as a result, you get less SNR. Multiple-ordered filters are much more easy to design if they are buffered or active, as you have only to worry about one specific frequency, and the math gets pretty hairy if you several filters loading each other.

Accuracy is also a limitation, as 1% capacitors can be expensive. If you have seperated circuits by buffering (or just having the filters active) you can have adjustments does not affect the other stages of the filter.

I suggest starting with a RC low pass filter and put a resistor in parallel with the output and see how the low pass filter is affected by different resistances loading it.

• Hi, thanks for the info! I will try the experiment you suggest. Thanks. – o.fithcheallaigh Aug 11 '15 at 17:16

It is not necessary to use a buffer between both parts (high- resp. lowpass). However, there are some specific considerations:

1.) For a bandpass function, the highpass corner frequency must be BELOW the corner of the lowpass - otherwise you get something like a band-stop (notch).

2.) The selectivity (bandwidth) of the bandpass depends on the configuration. However, in any case a passive RC-bandpass has a rather bad selctivity. The maximum Q that can be achieved is Q=Fo/BW=0.5 (midfrequency/bandwidth). This can be realized only using a buffer. Without such a buffer we always will have Q<0.5.

As with any simple RC filter, the filter response depends usually on driving the filter with a zero ohm source (or the source resistance adds to the filter resistance and re-positions the frequency response) AND not loading the filter with any significant load.

Cascading two filters can therefore create undesired effects - the output impedance of the first filter is not zero and will "modify" the characteristics of the 2nd filter. Conversely, the loading of the 2nd filter will give rise to changes in the 1st filter.

As LvW implies in his answer, there are exceptions and small compromises if you don't use a buffer AND the frequency response you require is quite leniant.