Why does my bandpass filter act as highpass?

I have designed a bandpass filter. Here is the circuit and spectral characteristic:

As you can see from the response curve, it has cutoff with the peak ranging from 260 kHz to 290 kHz.

Here is the actual circuit :

However, the filter works more like a High pass, passing frequencies all above 250 kHz, without any distortion or attenuation.

Here is one of the outputs :

(I observe outputs at all frequencies above 250 kHz. This is just one example)

Where am I going wrong, and what corrections do I need to make?

Update :

I decided to simulate my circuit , and it was clear why my circuit acted as Highpass. (The last freq is graph in 1 MHz). Dunno why the filter design tool showed that (incorrect) graph then.

• It's tough to say if the filter is behaving correctly without seeing the ratio of the output signal to the input signal. Don't expect the higher frequency signals to disappear altogether, they will be subject to attenuation which will depend on how far their frequency is from the passband, and the order of the filter. Commented Oct 10, 2014 at 14:34
• @clabacchio . But it is passing the signals even in the passband with same amplitude ( in fact a little lesser !) Commented Oct 10, 2014 at 14:35
• What DC resistance does each inductor have? Add that resistance in series with each inductor, and re-simulate. Compare with measurement... Parasitic capacitances and inductances may have tiny effects (the top taps are 1nf, which is huge!) but I suspect this is the main culprit. Indeed if you replace the inductors with resistors you have a classic high pass filter...
– user16324
Commented Oct 10, 2014 at 15:09
• @Plutonium smuggler Ah.. I see my problem. The filter is built backwards from the way I would have expected - the output end is closest to the generator rather than being farthest away. That explains it. Sorry for the confusion.
– JRE
Commented Oct 10, 2014 at 15:37
• @BrianDrummond . I used the multimeter to measure DC resistance . I suspect it is too small to show up (Multimeter can measure 1 milli Ohms). What now ? Commented Oct 10, 2014 at 16:11

With the photos, I'd venture that you have stray capacitances all over your circuit because it is on a solderless breadboard. You shouldn't be doing high frequency work like this on one of those. Of course, this doesn't rule out parasitic effects addressed in the comments to your question, but there is a gorilla in the room, and it's the breadboard.

• If not breadboard, then what else can I use ? Commented Oct 14, 2014 at 15:38
• @Plutoniumsmuggler Solder it down to a perforated prototyping breadboard, something like digikey.com/product-detail/en/DMB-4770-CB/377-2076-ND/3829629 Commented Oct 14, 2014 at 15:44
• Are you sure prototyping on breadboard will create problems ? Because I dont really know soldering ( I have planned to solder once my design works on breadboard ; and besides I have only limited components, both in variety and quantity) Commented Oct 14, 2014 at 15:52
• eevblog.com/forum/blog/… -- I'll leave it to you to do the math, but you have 2.5 puff capacitances across every connection at a megahertz. As an aside, you're getting answers about why your passive filter isn't working, but if you really need to get the job done, you should consider an active filter. Commented Oct 14, 2014 at 15:58
• @Plutoniumsmuggler -- as I suggested as a comment to your edit, be careful there. The bode you posted still show QUITE A GOOD DEAL of attenuation at high freqs, like 40 dB. Commented Oct 14, 2014 at 16:25

Filter designs often require a very small component tolerance. So normally, your filter will not work as expected if you just use standard 1% or 5% tolerance components.

Try to simulate the filter again with some of the component values slightly altered -- has this much impact on the frequency response curve?