## 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.

Using, Kirchhoff’s Rules, we can find the differential equation that governs this situation.

This equation has the general solution .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?

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

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

We also know that

Using the junction rule we get

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

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

We can write this system in matrix form

**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 is an eigenvalue, then

In our case,

It’s easy to see that and are eigenvectors for .

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

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

From this, we can convert back into linear equations

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

Combining this with the other solutions,

**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 . All of the supply voltage drops across , and that current flows into the capacitor. So our initial conditions are

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

**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.

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

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

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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