# Amplifier design

Discussion in 'Homework Help' started by Yarrow, Nov 1, 2007.

1. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
Hi, I need some help understanding how to design an amplifier to meet given demands.

Ok, the demands are:
Gain = 40dB (100)
BW = 60kHz

I am supposed to use LM324, so using one opamp is out of the question here. I have to use more. Now I know how to get the gain right, only by putting for example two opamp in series, each with gain=10.

The thing I cant understand is how to make the BW right on the output. Can anyone please explain to me how I can achieve BW=60kHz?

I was thinking about using the two opamps with gain = 10 each and a low-pass filter on the output to set the cutoff frequency. I did not get it to work, anyone know why?

I also had a chat with my professor today, and he confused me abit. I asked him about how one should look at the BW when putting opamps in series. And he said one should look at fc as -6dB if there are two opamps in series. Is that right? I thought that no matter what one should look at fc as -3dB.

2. ### Distort10n Active Member

Dec 25, 2006
429
1
You have to use 2 LM324's because its gain bandwidth product is too low given the gain and bandwidth requirements.

Putting two in series will do the trick, so what is your input? Are you sure you are not slew rate limiting the amplifier? What do you mean that it did not work?

3. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
Ok, I will try to illustrate what I have done with images. To meet the given requirements, I have put two LM324's in series and a LP filter in the end. I have simulated the amplifer (se both schematic and AC Response images)

Now when I look at the AC Responce, is it correct to look at the -6dB point (Which gives 60kHz in this case), or do I still have to look at the -3dB point(which does not give 60kHz)?

4. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
Forgot to add images, here they are.

The red line represents -6dB.

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5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Think about what would happen if you paralleled Rf1 and/or Rf3 with a capacitor.

6. ### Distort10n Active Member

Dec 25, 2006
429
1

I never liked the fact that it is called lead compensation when it is a low pass filter.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
I don't think I had ever heard it called that, so I Googled and found this. While the feedback cap does introduce a pole (lag) in the closed loop function, it introduces a zero (lead) in the feedback term, compensating for a pole due to (stray) capacitance which is present at the summing node. At least, this is my understanding. Correct me if I'm wrong.

8. ### Distort10n Active Member

Dec 25, 2006
429
1
Yep. It is one of those things that I would like to write a short article on. This configuration using an op-amp is a low pass filter, a practical integrator, and also lead compensation.
I am guessing it depends on the nature of the input signal, and what one is trying to accomplish which would lead to particular values of Cf.
Op-Amps for Everyone by Ron Mancini has this circuit under 'lead compensation.' It bothers me because in power electronics you hear about lead/lag or phase boost in controller schemes.

9. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
I did put a capacitor in parallel with Rf3, expecting the same result, but actually i got lower BW.

Anyway noone answered my first question. Does the schematic on the image meet the given requirements (40dB, 60kHz)? When I have two opamps in series, do I look at th -6dB OR -3dB point when analysing AC Response?

10. ### Distort10n Active Member

Dec 25, 2006
429
1
The purpose of the capacitor across the feedback resistor is not to extend the bandwidth of the circuit. Remember, it is a low pass filter. In terms of lead compensation, it provides additional phase margin or phase boost if you happen to be a power electronics engineer.

The BW of the circuit is reference to the -3dB point as far as I am concerned.

11. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
Yes, I know, I was not trying to extend the BW, I was trying to limit BW to a certain(60khz) frequency fc by using a LP filter.

I have tried several configurations (active/passive filters), but I cant hit 60kHz at -3dB. I have even tried to use 3 opamps, but still I keep missing. There must be something I am missing... Any ideas?

12. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
You can't get 60kHz @ -3dB because the GBW product of the LM324 is only 1MHz. A single amplifier with 20dB of inverting gain will only have a bandwidth of 90.9kHz, because the feedback ratio is actually Rs/(Rf+Rs)=1k/11k=0.0909... Therefore, the closed loop bandwidth is 90.9kHz. If you cascade 2 of these, the -3dB point is around 58.4kHz. You need more op amps. If you use 3, you should get around 88kHz bandwidth without adding a filter. You can add filtering by putting a cap across any one of the feedback caps.
Do you have to justify your capacitor choice mathematically, or can you do it empirically, or at least semi-empirically? Either can be done, but the empirical method is a heck of a lot easier!

13. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
There are no requirements for how things should be done.

I am not really familiar with empirical method (i think)

Cant I just calculate a value for the capacitence accros the feedback by
fc=1/(2pi*Rf*C)
to satisfy my cutoff frequency? Or is it more complicated?

14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The problem is, your op amps don't have infinite bandwidth, as you seem to know. Since they are already rolled off (more or less, depending on how many stages you have) at 60kHz, the capacitance you need to add to bring the response down to -3dB at 60kHz would be less than that calculated by the formula. An empirical solution would be one where you iteratively change the value of C in your simulation until you get the desired response. That may not be acceptable to your instructor.

15. ### Yarrow Thread Starter Member

Sep 18, 2006
23
0
Well at the moment I got nothing better going for me then an emperical solution, but if I get time I will try to find out more on how I can calculate the exact value for C.

tnx for all the help!