# Wideband Amplifier help

Discussion in 'Homework Help' started by sandeep156, Jan 24, 2014.

1. ### sandeep156 Thread Starter New Member

May 17, 2013
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Hey

We are just trying to figure out what the cut off frequency and gain is for the non-inverting half of the wideband amplifier (image attached).

So for the non-inverting configuration, C4,C8 & C10 are not connected.
We calculated the cutoff freq using 1/(2*pi*R9*C11) and got 160Hz and the gain was found to be (1 + (R8/R9)) = 6.6.

Can someone verify if this is correct please, because we tested our circuit and got very different values? Also, if it isn't,can someone please explain how to obtain the correct gain and cutoff frequency values.

Thank you.

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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If it has cutoff frequency, then it is some kind of filter. The amplifier feature is just an extra.

What kind of filter is it?

3. ### sandeep156 Thread Starter New Member

May 17, 2013
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From what I see it's a high pass filter.

But I'm not sure if the cutoff frequency I have is correct.

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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You have same signal going into inverting amplifier and none-inverting amplifier, then you combine the results, it seems to me that two results would simply cancel each other.

5. ### sandeep156 Thread Starter New Member

May 17, 2013
7
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Hey again,

We don't use both amplifiers at the same time. So, for now, I'm just testing the non-inverting one. To do that, I have disconnected C4,C8 & C10.

Do you have any suggestions how to calculate the cut off frequency?

Thanks

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I am doing it same way you are doing: fc=1/(2*pi*R9*C11), I am getting 159 Hz, so figure 160 Hz.

What did you get from the test circuit? Did you get 1.6 kHz?

7. ### sandeep156 Thread Starter New Member

May 17, 2013
7
0
Hey,

So from our test, we get a constant gain of 2.5 for frequencies above 500Hz - the characteristics of a high pass filter (whereby the gain is constant after a certain frequency). At 500Hz, the gain is 0.707*2.5 (which is the cutoff point).

So both the gain and frequency are off. But there is a pattern.
Any ideas why?

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I am not sure.

May 17, 2013
7
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Thanks

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11. ### vk6zgo Active Member

Jul 21, 2012
677
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Philosophically,yes.but all amplifiers have a cut off frequency.

R8,R9,& C11 control the amount of negative feedback,& hence the gain.
The gain can only be (1 + (R8/R9)) when the reactance of C11 is negligible.
Of course,there are other ramifications---when C11's reactance becomes significant,the feedback is phase shifted away from 180 degrees.

12. ### LvW Active Member

Jun 13, 2013
674
100
sandeep156,

I see no reason why the gain (above the corner frequency) shouldn`t be 6.6.
Regarding the frequency dependence:
I wouldn`t say that it is a "filter" - because you do not intentionally introduce some corner frequencies. Each amplifier has an upper frequency limit (low pass action) and in case of single supply also a high pass charcteristic with a lower cut-off (necessity of input coupling capacitors). Thus, in total it is a very broad bandpass - if you like.

However, as in your circuit the time constants of both the input circuitry (c12 and connected resistors) and the feedback circuitry are of the same order, both wll contribute to the resulting lower cut-off frequency.
As an approximation you can geometrically combine both cut-off frequencies (sqare root of squared values).

13. ### sandeep156 Thread Starter New Member

May 17, 2013
7
0
Hey everyone,

Sorry to get back to you guys this late.
But thank you all for your help!
@LvW - I will add both cut off frequencies as you have mentioned. Thank you.

-Sandeep