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How to derive these equations???

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chihwt2003

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Hi all,

Please have a look at the attached document.I come across this datasheet from Mitsubishi Electric Volume controller. Does anyone knows how to derive those equations? Please help. Thanks

Regards,

CW
 

echo47

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Some folks can't read Microsoft Word docs. Here's a GIF.
 

v_c

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I think it is lots of algebra. I think you can write some nodal analaysis equations and solve. Assuming this is an ideal opamp, the (-) node of the opamp between R2 and R3 is Vin, so mark that node as Vin and mark the top of R1 as Vx. Then write node equations

sum of current out of the Vin node =0
(Note: current into (-) node of opamp is zero, ideal opamp assumption).

(Vin-Vx)/(R3 + 1/sC1) + (Vin-Vo)/R2 = 0 (Kirchoff's current law)

sum of current out of the Vx node =0
(Vx-Vin)/(R3 + 1/sC1) + Vx/R1 + (Vx-Vout)/(1/sC2) =0

Now eliminate Vx from the equations and you will have the expression of Vo in terms of Vin. How do you eliminate Vx? Take the first equation and solve it for Vx and you will get

Vx = Vin + ( (R3 + 1/sC1)/R2 )*(Vin-Vo)

substitute this in for Vx in the second equation (now you should have only Vin and Vout).

Your resulting equation should be a second order equation.



Does this help?


Best regards,
v_c

Added after 1 hours 13 minutes:

with a little algebra :D it can be shown that
\[ \frac{V_{out}(s)}{V_{in}(s)}= \frac{1 + (C1 R3 + C1 R2 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}{1 + (C1 R3 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}\]

Now this is a standard form of the second order system where the denominator is of form \[1 + (\frac{s}{\omega_0}) \frac{1}{Q} + (\frac{s}{\omega_0})^2\] and \[\omega_0 = 2 \pi f_0\]. So looking at denominator of that equation we see that the term multiplying \[s^2\] must be \[\frac{1}{\omega_0^2}\] so
\[\omega_0^2 = \frac{1}{C2 C1 R1 R3 + C2 C1 R1 R2}=\frac{1}{C2 C1 R1 (R3 + R2)} \] so we get \[\omega_0 = \frac{1}{\sqrt{C2 C1 R1 (R3 + R2)}} \] and \[f_0 = \frac{\omega_0}{2 \pi} = \frac{1}{2 \pi \sqrt{C2 C1 R1 (R3 + R2)}} \]

OK now the term that is multiplying the s term in the denominator must be \[\frac{1}{\omega_0 Q}\] so equating we get
\[\frac{1}{\omega_0 Q}=C1 R3 + C1 R1 + C2 R1\] solving for \[Q\] and noting the value of \[\omega_0\] that we just solved for we get \[Q=\frac{\sqrt{C2 C1 R1 (R3 + R2)}}{C1 R3 + C1 R1 + C2 R1}\]. I think to get the value of \[Q\] you have to assume \[R3\] is small. See if your datasheet does use this approximation. So assuming R3 is much much smaller than other values (limit as R3 goes to zero), you get.
\[Q=\frac{\sqrt{C2 C1 R1 R2}}{C1 R1 + C2 R1}=\frac{1}{C1+C2} \sqrt{ \frac{C2 C1 R2}{R1} }\]

Now the final value is the midband gain. Note that both of the terms multiplying \[s^2\] on the numerator and the denominator is the same they are both \[1/\omega_0^2\], which is the resonant frequency. If the transfer function when evaluated at this frequency at \[s=j \omega_0\], the \[s^2\] terms will cancel with 1, since \[ (j \omega_0/\omega_0)^2 = -1\]. At the resonant frequency the gain is just given by the ratio of the factors that multiply the s term. That is just given by

\[Gv = \frac{C1 R3 + C1 R2 + C1 R1 + C2 R1}{C1 R3 + C1 R1 + C2 R1}\]

Now once again your datasheet must have an approximation, this time they must be assuming C1=C2, because if you do that the capacitance terms cancel from the top and the bottom of the expressions and you get.
\[Gv = \frac{R3 + R2 + R1 + R1}{R3 + R1 + R1} = \frac{R3 + R2 + 2 R1}{R3 + 2 R1} \]. Now if you divide top and bottom by R1 you get
\[Gv = \frac{\frac{R3 + R2}{R1} + 2 }{\frac{R3}{R1} + 2} \]. Now all you have to do is take 20 times the log of this and you have the gain in dB.

I think that is all I have to say ... lots of work for just 3pts :cry:

Best regards,
v_c
 
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    chihwt2003

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chihwt2003

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Hi v_c,

Thanks for the help. Given u 5 points for that :D

CW
 

v_c

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chihwt2003,

I hope it is useful for you. Did you check to see the approximations I made? I derived the full equations and without those approximations you have to use the full (unapproximated) expressions. But the transfer function is correct without approximations.

Best regards,
v_c
 

ee171

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oh wow!! thanks. this is a good pratice for solving circuits. i'll check out this tomorrow to see if i have the same answer as you v_c
 

chihwt2003

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Hi v_c,

I've checked the datasheet and the typical values used are shown in the atachment. For calculating the gain they use C1=C2. I think for calculating the Q factor, the value of R3 cannot be ignored since its value is significant compared to R1 and R2. Maybe the Q factor from the datasheet is just a rough approximation. What do your think?

Regards,

CW
 

v_c

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I don't think there is too much of a problem and I think they have used the values they have used to make an approximation for Q.

The actual value of Q is given by
\[Q=\frac{\sqrt{C2 C1 R1 (R3 + R2)}}{C1 R3 + C1 R1 + C2 R1}\].
The approximated value of Q is
\[Q \approx \frac{1}{C1+C2} \sqrt{ \frac{C2 C1 R2}{R1} }\]

Put in the numbers for the Rs and Cs in the circuit for both equations and see if the two expressions for the Q yield similar results. Maybe saying R3 is small was not the right statement to make. As long as your component values allow the approximation then everything should be OK; othewise you can use the un-approximated full expression for Q.

Best regards,
v_c
 
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    chihwt2003

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david90

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v_c said:
I think it is lots of algebra. I think you can write some nodal analaysis equations and solve. Assuming this is an ideal opamp, the (-) node of the opamp between R2 and R3 is Vin, so mark that node as Vin and mark the top of R1 as Vx. Then write node equations

sum of current out of the Vin node =0
(Note: current into (-) node of opamp is zero, ideal opamp assumption).

(Vin-Vx)/(R3 + 1/sC1) + (Vin-Vo)/R2 = 0 (Kirchoff's current law)

sum of current out of the Vx node =0
(Vx-Vin)/(R3 + 1/sC1) + Vx/R1 + (Vx-Vout)/(1/sC2) =0

Now eliminate Vx from the equations and you will have the expression of Vo in terms of Vin. How do you eliminate Vx? Take the first equation and solve it for Vx and you will get

Vx = Vin + ( (R3 + 1/sC1)/R2 )*(Vin-Vo)

substitute this in for Vx in the second equation (now you should have only Vin and Vout).

Your resulting equation should be a second order equation.



Does this help?


Best regards,
v_c

Added after 1 hours 13 minutes:

with a little algebra :D it can be shown that
\[ \frac{V_{out}(s)}{V_{in}(s)}= \frac{1 + (C1 R3 + C1 R2 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}{1 + (C1 R3 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}\]

Now this is a standard form of the second order system where the denominator is of form \[1 + (\frac{s}{\omega_0}) \frac{1}{Q} + (\frac{s}{\omega_0})^2\] and \[\omega_0 = 2 \pi f_0\]. So looking at denominator of that equation we see that the term multiplying \[s^2\] must be \[\frac{1}{\omega_0^2}\] so
\[\omega_0^2 = \frac{1}{C2 C1 R1 R3 + C2 C1 R1 R2}=\frac{1}{C2 C1 R1 (R3 + R2)} \] so we get \[\omega_0 = \frac{1}{\sqrt{C2 C1 R1 (R3 + R2)}} \] and \[f_0 = \frac{\omega_0}{2 \pi} = \frac{1}{2 \pi \sqrt{C2 C1 R1 (R3 + R2)}} \]

OK now the term that is multiplying the s term in the denominator must be \[\frac{1}{\omega_0 Q}\] so equating we get
\[\frac{1}{\omega_0 Q}=C1 R3 + C1 R1 + C2 R1\] solving for \[Q\] and noting the value of \[\omega_0\] that we just solved for we get \[Q=\frac{\sqrt{C2 C1 R1 (R3 + R2)}}{C1 R3 + C1 R1 + C2 R1}\]. I think to get the value of \[Q\] you have to assume \[R3\] is small. See if your datasheet does use this approximation. So assuming R3 is much much smaller than other values (limit as R3 goes to zero), you get.
\[Q=\frac{\sqrt{C2 C1 R1 R2}}{C1 R1 + C2 R1}=\frac{1}{C1+C2} \sqrt{ \frac{C2 C1 R2}{R1} }\]

Now the final value is the midband gain. Note that both of the terms multiplying \[s^2\] on the numerator and the denominator is the same they are both \[1/\omega_0^2\], which is the resonant frequency. If the transfer function when evaluated at this frequency at \[s=j \omega_0\], the \[s^2\] terms will cancel with 1, since \[ (j \omega_0/\omega_0)^2 = -1\]. At the resonant frequency the gain is just given by the ratio of the factors that multiply the s term. That is just given by

\[Gv = \frac{C1 R3 + C1 R2 + C1 R1 + C2 R1}{C1 R3 + C1 R1 + C2 R1}\]

Now once again your datasheet must have an approximation, this time they must be assuming C1=C2, because if you do that the capacitance terms cancel from the top and the bottom of the expressions and you get.
\[Gv = \frac{R3 + R2 + R1 + R1}{R3 + R1 + R1} = \frac{R3 + R2 + 2 R1}{R3 + 2 R1} \]. Now if you divide top and bottom by R1 you get
\[Gv = \frac{\frac{R3 + R2}{R1} + 2 }{\frac{R3}{R1} + 2} \]. Now all you have to do is take 20 times the log of this and you have the gain in dB.

I think that is all I have to say ... lots of work for just 3pts :cry:

Best regards,
v_c

Lot of time on ur hand? give him 20 point or something
 
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v_c

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david90,
I wrote the nodal equations just for the fun of it! Then I wanted to go one further step and eliminate Vx. It became an avalanche from that point on and I had to see the algebra to the end -- it is nice to exercise the algebraic skills once again. Once you get the transfer function and match it to a standard quadratic, it is pretty easy. Patience is the key.

As they say often "Don't try this at home!"

Best regards,
v_c
 

    chihwt2003

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