Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

passive Band pass filter design

Status
Not open for further replies.

manju_gowda

New Member
Hi all,
i have designed bandpass filter and its working fine but for large input signal also its giving small output voltage(0.5V) . can anyone tel me how to alter this design to get large output voltage signal say 3V.
 

Attachments

  • bpf.png
    bpf.png
    135.5 KB · Views: 206
What are the -3dB frequencies of your filter?
How did you decide on (calculate) the L and C values?

You say that there is a small output for a large input, how large is the input?
Have you plotted the response of the filter (attenuation vs frequency)?

JimB
 
OK, what did you specify in the online calculator?
What filter type?
What impedance?

Why does the input signal vary from 3 to 10 volts? If you are trying to measure the response of a filter, you keep the input voltage constant and measure the output.

JimB
 
Simulation shows the band centre frequency is 160kHz. At that frequency the filter gives 6db attenuation when feeding into an infinite impedance. What is the input impedance of the circuit following your filter?
 
The obviously simple solution is to parallel two senders in the tank, if there's room on the module. A fixed resistor paralleled to the sender will likely degrade over time and this situation will re-emerge.
 
My Simulation clearly shows that this passive filter puts out 66mV (at resonance) for an input of 1V (-24db loss).


D183.jpg



Perhaps you would be happier with an active filter.

Is the load at the output of the filter really 78Ω?
 
Last edited:
A likely reason for your apparent drop in voltage is that the input impedance of the filter is 78 ohms when the filter is operated in the passband. Because of that, the voltage at the output drops to one-half the source voltage. The filter is ideally lossless, but the load resistor is pulling the input voltage to the filter down to one-half. You can verify this by making the 78 ohms resistor on the left side (source resistor) zero ohms and running your simulation again.

MikeMi's simulation makes the loss appear horrible because C2 value is off by a factor of 1000.
 
Last edited:
...MikeMi's simulation makes the loss appear horrible because C2 value is off by a factor of 1000.

Only because the OP's figure was so small I couldn't read it. Here it is again with C2 corrected. I plot the output amplitude per 1 volt of input, showing the one-half output at resonance.
D190a.jpg
 
Hello,

Without considering any parasitic effects,
the resonant frequency is: 160571.255Hz
and the amplitude at resonance is 0.500.

R1 is probably the generator internal impedance or the source impedance, and R2 is probably the load impedance. That would be a somewhat standard way of showing those two and including them into the whole circuit.
 
Status
Not open for further replies.

New Articles From Microcontroller Tips

Back
Top