# Homework Help: Op-Amp/Band pass help

1. Mar 12, 2014

### cps.13

1. The problem statement, all variables and given/known data

I have an electronics project to design and build a band pass amp for my HNC. The only thing I have been given is it must use a uA741 or TL071 with an input imp greater than 1k and bandwidth between 200hz to 10khz. And gain must be 10db.

It is to be used in battery powered equipment.

2. Relevant equations

Part of the report write up is that I need to justify my choice of chip and use of either inverting or non-inverting.

3. The attempt at a solution

I have decided to use an inverting op amp setup as I can simply put a 1k resistor on the input and that is my input impedance sorted. Also makes calculating the gain easy!

I will use the TL071 as the output will not drop off at higher frequencies. The uA741 shows it will drop off at around 10kHz.

My problem is we have been taught nothing about a band pass amplifier/filter. So I am unsure if it makes any difference using a inverting/non-inverting setup?

I also intend to put a couple of variable resistors in before R1 and R2 to help get the gain as accurate as possible.

Any help is appreciated, if I am missing anything obvious could you point me in the right direction?

Thanks.

2. Mar 12, 2014

### Staff: Mentor

What order is your filter going to be? What polynomial are you going to use and why? What reading have you done about active opamp filters?

3. Mar 12, 2014

### psparky

Just off the top of my head without any research, I would use two inverting amps. Perhaps differentiator and a integretor in series. Set the RC constant at 200hz for the differentiator and set the RC constant for the low pass filter at 10Khz. Keep the first amp passive, make the second amp active.

Is your 10db gain a voltage gain or a power gain?

If trying to get a 10db gain in power, you will need your gain to be 10.
10^(db/10)= Power Gain.
10^(10/10)= 10
Or going backwards, 10 log 10= 10db

If trying to get a 10db gain in voltage, you will need your gain to be set at 3.16.
10^(db/20)= Voltage Gain.
10^(10/20)=3.16
Or going backwards, 20 log 3.16 = 10db

I'm not an expert on this, but perhaps a few of my comments will stear you in the right direction. You may be able to use just one amp to accomplish the same thing.

I don't think it makes a difference if you use a inverting or non-inverting as long as your transfer function meets your requirements. We generally used inverting amps for filters in school.

Last edited: Mar 12, 2014
4. Mar 12, 2014

### psparky

Just some food for thought....

Some times it pays to go back to the definition. Decibels are defined as:

dB=10Log(Po/Pi) Po=power out; Pi=power in.

When power is expressed as V2/R then: (where V2 means V squared)

Po=Vo2/RL and Pi= Vi2/Ri; when RL=Ri

dB=10log((Vo2/RL)/(Vi2/Ri)) since RL=RI the R's cancel and

dB=10log(Vo2/Vi2)=10log(Vo/Vi)2 since this a log function the square comes forward, and

dB=20log(Vo/Vi)

The are several variations of dB's, dBm in particular you should be familiar with.

To quote a well know author, "You should keep the original definition of the decibel firmly in
mind because it is of fundamental importance in many engineering applications."

5. Mar 12, 2014

### rude man

You can design a simple bandpass filter with 1 op amp, 2 resistors and 2 capacitors. Total.

The transfer function would be T1s/(T1s+1)(T2s+1) with T1 corresponding to 200 Hz and T2 to 10 KHz.

If you don't know laplace transforms, just substitute s = jw in the above transfer function.

I would use a higher input resistor than 1K ohm. Like 10K.

Since you want the gain to be 10 dB = 3.16, think Rf/Ri = 3.16.