Hello,

I am asked to design a bandpass filter with the following spefications:

Q=1
passband gain (Ko) =1
fo=5MHz

I have to use RC, CR, LC, CL circuits to build the filter.

I built an RC and CR circuit, each have fo=5MHz, and cascaded them. so, the bandpass filter now have fo =5MHz, but the bandwidth is too big. How can I reduce it ?

I tried to cascade the circuit with itself (several times) so that the roll off increases, and as a result the bandwidth decreased to 5MHz, but the circuit became too big. Is there an efficient alternative way to reduce the bandwidth ?

About the Ko, I will use an amplifier at the output, so it should not be a problem , I hope.

This circuit is not going to be implemented, it is only going to be simulated using Multisim.

Thanks in advance.

 

Hello,

I am asked to design a bandpass filter with the following spefications:

Q=1
passband gain (Ko) =1
fo=5MHz

I have to use RC, CR, LC, CL circuits to build the filter.

I built an RC and CR circuit, each have fo=5MHz, and cascaded them. so, the bandpass filter now have fo =5MHz, but the bandwidth is too big. How can I reduce it ?

I tried to cascade the circuit with itself (several times) so that the roll off increases, and as a result the bandwidth decreased to 5MHz, but the circuit became too big. Is there an efficient alternative way to reduce the bandwidth ?

About the Ko, I will use an amplifier at the output, so it should not be a problem , I hope.

This circuit is not going to be implemented, it is only going to be simulated using Multisim.

Thanks in advance.

The last time you posted this project you gave the Q as 10. But if it is 1 then we shall have a bandwidth of BW = center freq, fc divided by Q, which means the bandwidth is 5 MHz. Therefore f(low)=2.5MHz and f(high)=7.5MHz.

Since the gain is so small, you can use a 2N2222 or equivalent as an amplifier. Use this amplifier to isolate f(low) from f(high). To do this make the input capacitor and the resistance it sees an RC low pass filter with it f=1/(2*pi*R*C) with f 2.5MHz and solve for C. Then for f(high), put a capacitor in parallel with the load. If you use a common emitter amp, then Rc might be the load. If not RL'=Rc//RL. And to solve for the f(high) use the same RC equation given above with 7.5MHz as f. Solve for C.

In more detail, since you are dealing with frequencies greater than 1MHz, you might have to include the transistor's input and output capacitance values as part of your equation. More on this if you want.

By the way, this is a very simple circuit consisting of a common emitter amp with an input capacitor between the source and the base of the transistor and an output capacitor in parallel with Rc to ground. Don't bypass RE. There is no need and in doing so you inadvertently increase the your output capacitance as the product of Cu and the gain (Miller effect).

 

Thanks, I understand now.