passive bandpass filter loading

Thread Starter

AF_Maxwell

Joined Dec 12, 2018
23
this bandpass filter designed has this response and pretty much behaves like this when breadboarded

there is only one exception - the output doesn't jump to 45dB as spice is saying but that wasn't expected in the first place -> it goes up to about 10dB

BandpassSpice.PNG
there is a problem though

you can't load it with either a bjt based voltage voltage follower (2n3904) or a diode (1N4148) without killing the signal

i don't have other components atm, it's clearly an impedance problem but idk why voltage follower isn't working

for the diode literally have no idea why it won't pass a signal - it has a 1v forward voltage but the signal is 4Vpp so something should pass. raising the DC level of the signal doesn't help either
 

Thread Starter

AF_Maxwell

Joined Dec 12, 2018
23
someone please explain why output is dying and how to deal with it more than stating it's an impedance matching problem
 

Papabravo

Joined Feb 24, 2006
13,910
1 meg on the output is saying that essentially there is no load. Just out of curiosity, what happens if you make the source and load impedance, both 50 ohms.
How did you design the filter, and what are you trying to do?
My guess is that whatever power is going into the filter from the source is being reflected back to the source with minimal loss.
It is a passive filter, so you can't get more power out than you put in.
I don't suppose you have a VNA. Do you?
 
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Thread Starter

AF_Maxwell

Joined Dec 12, 2018
23
I designed the filter in Qucs, the program suggested nH inductors on the tank but I scaled the C down and the L up so it would have the same resonance but with manageable L values.

With 50 ohms on the output it simulates perfectly, no issue, but in practice it kills the signal which is why I added larger loads.

I haven't been able to simulate the input impedance though, the previous stage is a BJT common emitter gain stage following a clapp oscillator. On this gain stage the emitter resistor is 500 ohms and the collector resistor is 2.2k ohms. Sadly I don't know how to calculate the output impedance of this but have been assuming it's 2.2k ohms...
 

danadak

Joined Mar 10, 2018
4,057

Papabravo

Joined Feb 24, 2006
13,910
When you say "it kills the signal", what exactly do you mean, and what do you expect from a passive filter? You can't get more out than you put in. Think about a child's playground with a row of see-saws. If you lift one end up (by creating a voltage gain) something else must go down (like current for example). Remember also, that AC power does not have to be absorbed by a load -- it can just as well be reflected back to the source.

I'm not familiar with the Qucs program. What were the inputs, and what was the initial output?
 

Thread Starter

AF_Maxwell

Joined Dec 12, 2018
23
When you say "it kills the signal", what exactly do you mean, and what do you expect from a passive filter? You can't get more out than you put in. Think about a child's playground with a row of see-saws. If you lift one end up (by creating a voltage gain) something else must go down (like current for example). Remember also, that AC power does not have to be absorbed by a load -- it can just as well be reflected back to the source.

I'm not familiar with the Qucs program. What were the inputs, and what was the initial output?
It kills the signal just means it attenuates everything far too much, from 3Vpp to 200mVpp and a flat frequency response at the frequency change I am able to affect - 1.1Mhz to 1.4Mhz. I don't have a VNA no. I obviously don't expect a gain on the output, I would just like 0dB at the center frequency. I know power in is power out...

My problem as stated isn't that the bandpass filter doesn't work, just ideas on what else it could be apart from the loading problem because it stops working as expected when loaded. Maybe could you explain power being reflected to the source and what might cause that?

Note also in your sim it appears you are not accounting for non ideal L and C
components, which will affect your results.

Just a thought.....

You can sim an emitter follower, add a current source to the node you are trying
to eval for Z, and using sim math capability get the Z at that node.

Or

http://paulorenato.com/index.php/electronics-diy/127-high-frequency-transistor-amplifier-calculator

https://2n3904blog.com/common-emitter-output-impedance/


Regards, Dana.
Thanks, that's pretty useful.
 

Papabravo

Joined Feb 24, 2006
13,910
Reflections are caused by impedance discontinuities. At each discontinuous impedance change a portion of the power will be transmitted and a portion will be reflected. You can see this on a transmission line when the termination is either short (0Ω) or open (∞ Ω). At other terminations you can see various amounts of absorption and reflection.

To check if the filter you designed is matched to 50 Ω, you should rerun your simulation with a 50Ω load resistor and a 50 Ω resistor in series with the voltage source. To check if the actual prototype matches the simulation you need an RF signal generator with a 50 Ω source impedance and you can still use the 50 Ω resistor as a load. Now as to how you integrate the filter with the rest of your hardware, that might require a bit more work.
 
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Thread Starter

AF_Maxwell

Joined Dec 12, 2018
23
i got rid of the loading problem the normal way - using high input impedance on the bjt through biasing with 1Meg+ resistors, so this was just forgetfulness on my part

now i can get my signal to permit loading... i've started designing filters with my input series resistance taken into account but idk tbh the frequency response when breadboarded isn't doing what i want unlike before...
 

danadak

Joined Mar 10, 2018
4,057
In the Mhz and above region C esr and L losses will affect your
overall loses = lower your BW. Look at datasheets for these com-
ponents, from one manufacturer to another you will see significant
variations.

If you go to Digikey and use parts locator you can see Q values for
the L's. They vary a lot one L to another, for same value inductance.

You might consider using a IF transformer, used in radio receivers, as
they have very hi Q windings. You could modifiy a 455 Khz one to get
to 1 Mhz.

Regards, Dana.
 
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danadak

Joined Mar 10, 2018
4,057
Here I get a result showing two distinct peaks, this I think due to LC product input
series LC not same as output series LC. Not sure if thats what you want.

I used a V source driving network, and a 10K load on output.

1588328125583.png

Regards, Dana.
 
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MrAl

Joined Jun 17, 2014
7,656
someone please explain why output is dying and how to deal with it more than stating it's an impedance matching problem
Hello,

With ideal components the only loss in the circuit is the load. with non ideal components there are other losses that affect the output and that usually affects the max output the most because the Q is very sensitive to pure resistances in the circuit.

Here is what you can do...
Analyze the circuit exactly as shown in your drawing using math not a simulator. That means that every element is ideal and only one resistance which is the output load. Note the output peak amplitude with a frequency sweep.
Next add a small series resistance to one of the series inductors. Analyze that and note the difference it makes on the output from before.

Keep in mind that when you add a series resistance that resistance forms a voltage divider with the output load and that means the output amplitude with be reduced and the Q of the entire circuit comes down. If the series resistance is small it will have less effect of course than if it is large. In real life there will always be small resistances that come with real life components not limited to just the inductors. Capacitors are a little better but still have equivalent series and parallel resistances.

You could use Nodal analysis for this for example.
Doing a complete analysis is the way we find out these things when we need to know what is happening.
 

danadak

Joined Mar 10, 2018
4,057
If you do this via nodel analysis you have a 6'th order circuit, which is
algebraically challenging. Consider using signal flow analysis since
the T(s) falls out quite easily when applying the method. Then you
will need to factor the T(s), there are tools out there that can take
an algebraic equation and generate the factored form. Off hand I do
not remember who/where the tools are available. MATLAB maybe ....?

Totally agree with MrAI, the Ls need to have their finite Q accounted for.

Have you stated what your actual goals are, eg. G, Zin, Zout, BW for the filter ?
Do you care about its phase response ? Do you want a passive only solution ?
Or would active be acceptable ? Latter gets rid of L's, and easily can be tailored
for G. Does design need to handle wide Temperature variation ? Noise of filter
of concern ?


Regards, Dana.
 

MrAl

Joined Jun 17, 2014
7,656
Here is the amplitude equation, it looks correct when tested...

Vout/Vin=
(w^3*C2^2*L3*R1)/(sqrt((w^5*C2^2*C7*L1*L3*R1-w^3*C2*C7*L3*
R1-w^3*C2^2*L3*R1-w^3*C2^2*L1*R1+w*C2*R1)^(2)+(-w^6*C2^2*C7*L1^2*L3+2*
w^4*C2*C7*L1*L3+2*w^4*C2^2*L1*L3-w^2*C7*L3-2*w^2*C2*L3+w^4*C2^2*L1^2-2*
w^2*C2*L1+1)^(2)))

That is with all ideal elements, R1 is the load as shown in the schematic previously in this thread.
Next we could look at the amplitude expression with non ideal elements.
 

Tesla23

Joined May 10, 2009
403
I haven't been able to simulate the input impedance though, the previous stage is a BJT common emitter gain stage following a clapp oscillator. On this gain stage the emitter resistor is 500 ohms and the collector resistor is 2.2k ohms. Sadly I don't know how to calculate the output impedance of this but have been assuming it's 2.2k ohms...
AF, what problem are you trying to solve with this filter? It's a bit unusual to put a filter like this after an oscillator, so if you describe what you are trying to do we may be in a better place to advise, or suggest an alternative approach.
 

MrAl

Joined Jun 17, 2014
7,656
Hello again,

Here are two frequency sweeps derived from pure theory not simulation.
One sweep has Rin=0 and the other Rin=50.
The output loading R1 ranges from 1Meg down to 50 ohms and the respective plots shown.
This is done with no parasitics so this is what we would get with all ideal elements and no secondary frequency effects.

It is interesting to see the degradation of the output even with all ideal elements even with Rin=0.
The frequency was swept from 1MHz to 2MHz.
Note also the development of two humps with decreasing R1 the output load resistance.
Also note that we lose sharpness right away with some non zero input resistance Rin.

The conclusion of course is that it is not the parasitics alone that cause the filter response degradation it is the nature of the filter design itself which would have to be improved in order to reduce these effects. This is the beauty of pure theory it provides answers that are sometimes hard to imagine without it.
 

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