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Muting Circuit: Potential Divider Charging

September 4th, 2016


A capacitor charging circuit can be pretty simple. You have a current limiting resistor, a power supply, and a capacitor.
capcharge
Using, Kirchhoff’s Rules, we can find the differential equation that governs this situation.

V_{supply}=R\frac{dQ}{dt}+\frac{Q}{c}

This equation has the general solution Q=CV(1-e^{-\frac{t}{RC}}) .But what do we do if we don’t want to charge the capacitor to the full supply voltage? Let me give an example…

I. Real Example

For an amplifier I am building, I need a mute circuit to protect the headphones from on-rush current. The muting circuit is built using a 12V relay, and the amplifier has a B+ supply of 150V. Ideally, I would have between 13-15V across the relay. We could use a potential divider to get the correct voltage. A 100k on top and a 10k on the bottom gives us around 13.6V from the divider. That’s pretty good. So now we have to calculate how long it takes to charge the capacitor in the muting circuit?

capcharge2

In the first case, we solved the differential equation (well, we didn’t, but someone did, and it’s pretty easy to solve if we wanted to). So let’s try to do the same in this case.

II. Differential Equation(s)

Let’s start by using the loop rule

V=V_{R_1}+V_C=I_1R_1+Q/C

Since we need differential equations, we will differentiate both sides to get

0=\dot{I_1}R_1+I_c/C\hspace{15mm}(1)

We also know that V_{R_2}=V_C

I_2R_2=Q/C

\dot{I_2}R_2=I_c/C\hspace{15mm}(2)

Using the junction rule we get

I_1=I_2+I_c

\dot{I_1}=\dot{I_2}+\dot{I_c}

\dot{I_1}-\dot{I_2}=\dot{I_c}\hspace{15mm}(3)

But wait, we have 3 differential equations, how can we solve them all?

Well, it’s sort of like a system of linear equations. We have 3 equations and 3 variables, so maybe we can  solve it… (hint, we can in this case!)

III. Rearrange Equations

First, let’s get all of our equations into the form

\dot{I_j}=c_1I_1+c_2I_2+c_3I_c

\dot{I_1}=-\frac{I_c}{R_1C}\hspace{15mm}(1)

\dot{I_2}=\frac{I_c}{R_2C}\hspace{15mm}(2)

For the third equation, we are going to substitute equations 1 and 2 into our original equation 3

\dot{I_1}-\dot{I_2}=\dot{I_c}\hspace{15mm}

\dot{I_c}=-\frac{I_c}{R_1C}-\frac{I_c}{R_2C}\hspace{15mm}(3)

We can write this system in matrix form

\begin{bmatrix}\dot{I_1}\\ \dot{I_2}\\ \dot{I_3}\end{bmatrix}=\begin{bmatrix}0&0&-\frac{1}{R_1C}\\ 0&0&\frac{1}{R_2C}\\ 0&0&-\frac{R_1+R_2}{CR_1R_2}\end{bmatrix}\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}

IV. Eigenvalues

To solve this system, we will compute the eigenvalues and eigenvectors for our coefficient matrix. Our matrix is a triangular matrix, so the determinant is just the product of the entries along the main diagonal. So if \lambda is an eigenvalue, then

\text{det}(A-\lambda\cdot I_n)=0

In our case,

\lambda\cdot\lambda (-\frac{R_1+R_2}{CR_1R_2}-\lambda)=0

\lambda=0\hspace{10mm}\lambda=-\frac{R_1+R_2}{CR_1R_2}

It’s easy to see that   \begin{bmatrix}1\\ 0\\ 0\end{bmatrix}   and   \begin{bmatrix}0\\ 1\\ 0\end{bmatrix}   are eigenvectors for \lambda=0.

For the other eigenvalue, we will have to compute the eigenvector, though it isn’t too difficult

\lambda=-\frac{R_1+R_2}{CR_1R_2}

A\vec{I}=\lambda\vec{I}\hspace{7mm}(A-\lambda\cdot I_n)\vec{I}=\vec{0}

A is our matrix of coefficients and lambda is our eigenvector, so let’s substitute in

\begin{bmatrix}\frac{R_1+R_2}{CR_1R_2}&0&-\frac{1}{R_1C}\\ 0&\frac{R_1+R_2}{CR_1R_2}&\frac{1}{R_2C}\\ 0&0&0\end{bmatrix}\cdot\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}

From this, we can convert back into linear equations

 I_1=\frac{1}{R_1C}\cdot\frac{CR_1R_2}{R_1+R_2}=\frac{R_2}{R_1+R_2}\cdot I_3

 I_2=-\frac{1}{R_2C}\cdot\frac{CR_1R_2}{R_1+R_2}=-\frac{R_1}{R_1+R_2}\cdot I_3

I_3=I_3

We then have to remember to multiply by our eigenvalue term. Since I_3 is a free variable, we will let it equal 1. so for the nonzero eigenvalue we get

\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}=e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}}\cdot\begin{bmatrix}\frac{R_2}{R_1+R_2}\\ -\frac{R_1}{R_1+R_2}\\ 1\end{bmatrix}

Combining this with the other solutions,

\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}=c_1e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}}\cdot\begin{bmatrix}\frac{R_2}{R_1+R_2}\\ -\frac{R_1}{R_1+R_2}\\ 1\end{bmatrix}+c_2\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}+c_3\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}

V. Initial Conditions

We now have the general solution, so we just have to solve for the initial conditions. When the power is turned on, the capacitor is completely discharged, so I_2=0. All of the supply voltage drops across R_1, and that current flows into the capacitor. So our initial conditions are

\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}_o=\begin{bmatrix}V/R_1\\ 0\\ V/R_1\end{bmatrix}

Using row reduction, we can determine the particular solution of our system of differential equations

\begin{bmatrix}I_1\\ I_2\\ I_3\end{bmatrix}=\frac{V}{R_1}\cdot e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}}\cdot\begin{bmatrix}\frac{R_2}{R_1+R_2}\\ -\frac{R_1}{R_1+R_2}\\ 1\end{bmatrix}+\frac{V}{R_1+R_2}\cdot\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}+\frac{V}{R_1+R_2}\cdot\begin{bmatrix}0\\ 1\\ 0\end{bmatrix}

VI. Muting Circuit Revisited

Wow, that’s a lot of matrices. The good thing is we only need to look at the current through the capacitor, since the capacitor is what triggers the relay.

I_3=\frac{V}{R_1}\cdot e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}}

Since we are concerned with the charge on the capacitor, we will integrate the current

Q=c_1-\frac{VCR_2}{R_1+R_2}\cdot e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}}

We know the initial charge is 0, so we can solve for the constant of integration, yielding

Q=\frac{VCR_2}{R_1+R_2}(1-e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}})

V_{cap}=\frac{VR_2}{R_1+R_2}(1-e^{-t\cdot\frac{R_1+R_2}{CR_1R_2}})

The voltage across the capacitor approaches the voltage from the potential divider, which is exactly what we want.

we can now input various resistances, capacitances, and times, and decide how to set up our muting circuit.

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Software


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