Why there is more losses in a band pass than in LPF, HPF, notch ... - Page 1
Forum section: electronics / rf-microwave

author: jujun
author_rank: Regular Contributor
timestamp: August 04, 2018, 02:05:12 pm
content: Hello,I don't understand why when using the same technology, a band pass filter have always more losses than a low pass filter or a high pass filter, or a notch filter.Do you know why ?Thank youJulien

author: KJDS
author_rank: Super Contributor
timestamp: August 04, 2018, 02:55:56 pm
content: Quote from: jujun on August 04, 2018, 02:05:12 pmYes I do, the maths is covers by Dishal's equations.

author: ChristofferB
author_rank: Frequent Contributor
timestamp: August 04, 2018, 03:02:21 pm
content: Well, depending on topology, a bpf could be thought of as an lpf followed by a hpf or vice versa. It'd then sound logical that the filter would have the loss of both. I'm sure there is an actual math-y explanation though.

author: jujun
author_rank: Regular Contributor
timestamp: August 04, 2018, 03:10:42 pm
content: Quote from: KJDS on August 04, 2018, 02:55:56 pmDo you have a link about it ?I quickly googled and found nothing :/

author: jujun
author_rank: Regular Contributor
timestamp: August 04, 2018, 03:13:27 pm
content: Quote from: ChristofferB on August 04, 2018, 03:02:21 pmBut for example (in UHF) it would be lower losses to build two large notch filter than building one band pass filter.(it's only from simulation)But maybe the reality is different ... I have to try

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 04, 2018, 03:20:18 pm
content: https://ieeexplore.ieee.org/document/4050569/is probably relevant but I see little for "Dishal's equations", and without access to the paper it's impossible to say if this is it.Offhandedly, I would say a bandpass is effectively a lowpass and highpass combined, so you should expect double the transition band insertion loss. (Narrowband filters are essentially all transition band, so the whole passband will get that loss.)Tim

author: KJDS
author_rank: Super Contributor
timestamp: August 04, 2018, 05:09:17 pm
content: Quote from: jujun on August 04, 2018, 03:10:42 pmoops, I think the equation is from S.B. Cohn, not Dishal, it's given in the link below.https://www.rfshop.co.uk/Martin.pdf

author: Wolfgang
author_rank: Super Contributor
timestamp: August 04, 2018, 10:16:19 pm
content: A simple but heuristic answer is:I you have a lowpass with N elements, then a bandpass or bandstop filter of the same order will have 2N elements. If half of the elements are coils with losses (there are no others) it becomes plausible that more losses occur because we have two times the number of coils in a bandpass or bandstop than in a lowpass.For a highpass this formal argument does not hold because the number of elements stays the same. Here a practical effect plays a role: If you have an ideal highpass, the coils must retain ideal properties (ESR as low as possible, no resonances up to very high frequencies, ...). Such coils dont exist. Therefore the losses of a highpass can be optimized at the and reasonably above the corner frequency, but hardly to frequencies that are 10 to 100 times above that.

author: G0HZU
author_rank: Super Contributor
timestamp: August 05, 2018, 06:27:19 pm
content: QuoteI guess it depends on what you class as low insertion loss. It's possible to design a simple lumped BPF up at (say) 1000MHz with about 100MHz bandwidth and get less than 1dB insertion loss. This would be using SMD parts for the capacitors and copper wire for the inductors and it could be built with simple ugly construction on a sheet of copper PCB. To get low insertion loss for a BPF like this with just 10% bandwidth you need to select parts with a high unloaded Q. If optimised with decent caps from ATC and with inductors wound for high unloaded Q the insertion loss could be as low as about 0.6 to 0.7dB at 1000MHz for a low order BPF with about 100MHz bandwidth (10% bandwidth).

author: the_janitor
author_rank: Newbie
timestamp: August 05, 2018, 07:18:46 pm
content: Higher filter order translates to higher insertion loss (while considering real models, i.e. with ESR).

author: Antonio84e
author_rank: Contributor
timestamp: August 11, 2018, 08:16:32 pm
content: I haven´t gone through the answers, but just replying to original question I would say you can a lot of the times think of a band pass as a low pass and a high pass - hence the higher loss typically.

author: KJDS
author_rank: Super Contributor
timestamp: August 12, 2018, 06:05:53 am
content: Lets work through the mathsthe loss is 4.34 * (sum of g values) * fractional bandwidth / (Q of resonators)The sum of the g values can either be calculated accurately, or a close approximation is that it is the order of the filter, so a seventh order filter will have the sum of the g values equal to 7, whereas a third order filter will equal 3. Hence a higher order filter will have proportionally more loss than a lower order filter.The fractional bandwidth is one of the keys to the original question. Most bandpass filters are fairly narrowband. If the fractional bandwidth is set to one, then the loss of the bandpass is the same as the lowpass*.The resonator Q also affects loss, so the higher the Q the lower the filter loss.*If each element in a lowpass had the same Q as each resonator in a bandpass then this would hold true, however inductors are usually more lossy than capacitors and half the sections in the lowpass are capacitors with much better Q.