[SOLVED] to design the circuit to improve circuit to noise ratio

Without feedback
Vo/Vs = G2

Click to expand...

shoudnt it be Vo/Vs=g1*g2
because we have cascaded two amplifer and ignoring the noize signal taking the assumption Vn<<<VS

You are totally right if we have only the circuit we work on.

But if you read on the pages you sent me... (please refer to them), A2 (noisless) and B (feedback) were added to A1 (noisy) in order to increase the amplifier S/N.
In other words, the original gain is A1 (hence G2 in our problem).
The modified (updated) amplifier should be, in my opinion, equivalent to the previous one when A1 was alone (hence its gain should be also G2).

Hope you got the point.

Kerim

okay i got the point that actually we increase the signal by using A2 but our actuall amplifer is A1
and from that we infer beta is 1/50 that is less than 1
then the Vo would be 50v/v and overall gain with feedback would be 50v/v

And the S/N is increased 56.23 times...

and the s/n is increased 56.23 times...

Click to expand...

yes you are right
but now how are we gona design the feedback circuit with gain of 1/50??

Sorry for the delay... our internet connection had a fault.

You see... your question is very good.

There is the easy way and the hard way

The easy way is to let R8=2K but the lecturer may say that it is not good to run an inverting amplifier using a feedback (R8/R10) that gives a gain less than one (see the pic I attached). The remedy is to add a dummy resistor R11 between IN- and ground and we let R11=2K for example. Now the equivalent resistance seen by IN- with the input sources is R10//R11.The input voltage sources in this case are Vo (via R10) and 0V (via R11) . Now the opamp has an overall gain R8/(R10//R11) close to 1 but only 1/50 of Vo reaches its output :wink: (since R11 is connected to 0V)

The hard way is to rethink to build another topology by eliminating the feedback opamp U4 which is bothering you :grin:

how about potential divider connected in series shunt confriguration??

This is possible but the feedback sign becomes positive... This means the gain of one of the other two blocks should be made positive too.
I think it is better to make the gain of G2 positive and keep G1 negative.

Try to complete the other pieces of your homework... I will prepare for you a dual schematic; one circuit without feedback and another one with feedback so that one can see the difference at their output in presence of the signal Vin and noise Vn. For example, we will assume Vin as 50mV 500 Hz and the Vn as 20mV 4000 Hz.

i am able to understand the circuit but im unclear about the feedback??
how did you decide abt the resistor with 500k??
the beta we calculated was 1/50 ??

You are right... And I already knew you will ask me about it since it took me a couple of minutes before knowing how to calculate it :wink:

The gain of G1 for Vin is R12/R11
The gain of G1 for Vclean is R12/R13
So the output of G1:
Vg1 = -Vin*(R12/R11) - Vclean*(R12/R13)
Vg1 = -R12(Vin/R11 + Vclean/R13)
Vg1 = - (R12/R11) * [Vin + Vclean*(R11/R13)]
Vg1 = - (562.3/10) * [Vin + Vclean*(10/500)]
Vg1 = - 56.23 * (Vin + Vclean/50)

Kerim

you have help me a lot and im very greatful to you
but here is a question what is the function of resistor R24 R21 and R14???
i think R14 is to give the opamp in put resistance to reduce the offset
am i right?
but im unsure about the other two.

---------- Post added at 02:04 ---------- Previous post was at 01:47 ----------

and the feedback added into input signal is a very small noise signal how it is obtained??
is it that when vclean is divided by beta the ratio of signal and noise is reduce and the noise signal become significant ??View attachment untitled.bmp

You are right about R14.

About R21, R24... and R22, R23, R25... they determine the gain of the non-inverting amplifier G2.
It is the same for G0 (at the left).
Try to analyze the amplifier G0 and find its gain (Vnoisy/Vin) after you let Vn=0 (Vn connected to ground so that R05 and R04 become parallel).
I hope you learnt how to find the gain of a non-inverting amplifier using opamp... right?
You will find out that if Vin is gounded, the gain (Vnoisy/Vn of G0) will be the same as the previous one.
Vnoisy = Vin*50 + Vn*50
Vnoisy = 50*(Vin + Vn)

G2 and G0 are similar, but Vin is now the output of G1.

Concerning what we see at IN- of G1:
In theory, the voltage at IN- is equal to the voltage of IN+ which is connected to ground. So it should be 0V but actually we notice that it is not. Why?
Vclean sends a current Vclean/R13 to the node IN-. A part of it goes to ground (through the source Vin) via R11 and the rest of it continues through R12 which creates a rather large voltage at G1 output. If we divide the noise voltage at G1 output by the noise voltage at IN- we get the actual G1 gain alone (called open loop gain of the opamp) but for the frequency of Vn (which is 4 KHz here). This gain is not infinity or too large as we assume in our approximate calculations. Typically, when the frequency increases, the gain decreases. You can test this by changing the 4 KHz to 8 KHz.
Note: To ease the reading on the traces, you can let Vin=0 (or replace its label at R11 with the ground symbol) since the test now is about Vn only.


Kerim

V1(output from G1)
Vo(output from G2)=V1[(R24/R24+R23)(1+R22/R21)] +Vn[(R24/R24+R25)(1+R22/R21)]
so calculating that V1(99.01)+Vn(99.01)
i think R21 should be 10k instead of 5k as for difference amplifier we have condition R4/R3=R2/R1
using R21 5k does not give 50 gain of the amplifier but 100??
what do u think?

Vo(output from G2)=V1[(R24//R25)/(R24//R25+R23)(1+R22/R21)] +Vn[(R24//R23)/(R24//R23+R25)(1+R22/R21)]
I guess you remember the theorem of superposition
for Vo/V1 (Vn = 0 , Vn at ground)
for Vo/Vn (V1 = 0 , V1 at ground in theory)

using R21 5k does not give 50 gain of the amplifier but 100??

Click to expand...

Now it is 50 when R21=5K :wink:

Did you notice what you have missed in your gain formula? I guess you did

i didnt get it??
Vo/V1=99.01
Vo/Vn=99.01
Vo1+Vo2=99.01.+99.01=200????
althuogh we are using theorem of superposition but

---------- Post added at 12:09 ---------- Previous post was at 11:42 ----------

Did you notice what you have missed in your gain formula? I guess you did

Click to expand...

what exactly did i miss??

When you calculated Vo/V1, you removed Vn but you let the node of R25 in the air!
By doing this you considered Vn as a current source not voltage source.
The theorem of superposition says:
Keep one source and remove the effect of any other independent supplies. How? by replacing the voltage source with a short and by opening the current source.

I noticed you added the two gains Vo/V1 and Vo/Vn. This is totally wrong as you add oranges with apples.

i get it


THANKS A LOT!!!!!!!
I HOPE SUMDAY, I MIGHT BE ABLE TO HELP SUMONE , YOU ARE DOING A GREAT JOB!!!!

Thanks... and keep insisting to understand everything... logically

yes i will try my best:grin: