sorry for the late reply, I had made a discussion with my coursemate today, and we seperate the circuit into 2 circuit ... here's the link ,
Circuit Analysis.zip ... we think that the 1st part is high-pass filter, and the 2nd part is low- pass filter ... is that correct ? and how do u think about the seperate circuit ?
You keep doing well, you splitted the circuit into two parts which is a good step on your part. But while I discuss with you the circuit I am afraid I have to refer to the original one and if we will have enough time, we can decide together how to make it, also step by step, a better and complete one.
As we will see, the real problem in your problem is that it is not a completely right circuit in the first place.
So some parts of it need to be discussed more in depth so that its analysis can be helpful.
Let us try clarifying some points:
(1)
The remark of 'LvW' (post #15) about the power supply is right, the first amplifier says it needs a dual supply.
The reason is that R2 is connected to ground, so as we will see, pin 3 (In+) is also at 0 voltage.
Also VR1 is connected to ground, and as we will also see the ac signal coming through C1 will try to let the opamp output be ac too (alternating between positive and negative) while ground is the reference voltage (set by R2). If the opamp is supplied by a single source as +5V and ground, the opamp1 output cannot decrease below ground. But if there is a dual supply (say +5V, 0, -5V), the output can swing into either direction. Though we can alter the circuit to accept single supply, let us assume that we have a dual one instead.
(2)
The pupose of C1 (as a coupling capacitor) is to block the DC voltage (hence DC current) from going to one stage to another in a multi-part amplifier (or alike). Therefore the polarity of C1 shows that, at its left, the input has a negative DC voltage. But if it is positive its polarity should be reversed. And in case, the input DC voltage is also zero (as the reference voltage of our opamp1), C1 should be removed. In fact, when a designer uses a dual supply for his circuit, one of his reasons might be to eliminate many coupling capacitors since almost all ac signals would have the same reference (ground or 0V).
(3)
As you said, another purpose of C1 could be to form a high pass filter. But in this case there might be two design errors. First, since C1 is a unipolar capacitor, the SIGN of its voltage shouldn't change (always positive or negative) for proper operation. Since we have no idea what are the DC and AC magnitudes of the signal at the 'mono input', we can't decide if C1 should be a unipolar or nonpolar type. Second, the input resitance seen by the C1 at VR1 side, varies with VR1 setting, from 100K to zero! (we will see why later). So it is not really a well defined high pass filter since its main RC part is not constant and also it depends on the input resistance (left of C1) which is unknown (so it could be relatively small or large).
(4)
Let us study the function of opamp1. The first thing to know about an opamp is that it has a very high voltage gain Vout/(Vp-Vn) which is too big as 1,000,000 for example (Vp=Vin+ and Vn=Vin-). Since Vout is always limited then Vp-Vn approaches zero which means Vn=Vp when the opamp is in its linear region (that is its output is between its limits, say +5V and -5V in our circuit). The second thing to know is that the input currents at Vp and Vn are relatively very small to the point they can be seen as zero in most applications (as ours). So if ask you what is the voltage at Vp (pin 3), your answer will surely be; ZERO volt, because since the input current is zero the drop on the resistor R2 is also zero. But since Vn = Vp , Vn = 0V too (how nice!). So Vn=Vp=0 is always satisfied even if an ac voltage is applied at the input.
(5)
Later, we will calculate the voltage gain of opamp1 for all settings of VR1. We will denotes by the variable x the ratio of the lower part of VR1 to 100K (its maximum range). That is 0 =< x =< 100K
See you later... if you like :wink:
Kerim