# Wien Bridge

#### The Bridge

In the figure is presented a Wien Bridge. When its two arms are at equilibrium ($R1*R3=R2*R4$) the voltage on the Wien bridge diagonal cancels. You apply to this bridge a voltage $V(t)=100*\cos(100 t)$ V and to you are using an oscilloscope with $R(in)=1M\Omega$.

a) You determine the voltage between a) and b) directly.

b) You determine the voltage between a) and GND, respectively b) and GND, then subtract these two voltages.

What is the Voltage between a) and b) of the Wien bridge in each case, with and without considering $R(in)$?

#### Differential voltage

To solve correctly one needs to write a system of 5 equations (there are 5 currents I1, I2, I3, I4 and Iab) and solve it (all currents are downwards and Iab is from left to right):

If we consider the resistance of oscilloscope R(ab) = 1 Mohm the one in the figure we have:

$\left\{\begin{matrix}

1*I1 +1*I(ab)-2*I4 =0\\

1*I3+1*I(ab)-5*I2 =0\\

I1 =I(ab)+I2\\

I4+Iab =I3\\

1*I1+5*I2 =100 (=U/1 Mega)\end{matrix}\right.$

This system can be solved easy using the Cramer rule. The matrix of the equation is ($I(ab)$ is the 5th unknown):

$Det=\begin{vmatrix}

1 & 0 &0 &-2 &1 \\

0 &-5 &1 &0 &1 \\

1 &-1 &0 &0 &-1 \\

0 &0 &-1 &1 &1 \\

1 &5 &0 &0 &0

\end{vmatrix}=-45$

The extended matrix for I(ab) is (replace the 5th column with column of free terms from right side)

$Det2= \begin{vmatrix}

1& 0& 0& -2& 0\\

0& -5& 1& 0& 0\\

1& -1& 0& 0& 0\\

0& 0& -1& 1& 0\\

1& 5& 0& 0& 100\end{vmatrix}=-900$

To compute the determinants I used this. The unknown is

$I(ab) = Det2/Det =900/45 =20 \mu A$

$V(ab) = I(ab)*R(ab)=20 \mu*1 Mega = 20 V (=20*cos(100*t) V)$

##### With R(in)

If we consider the resistance of the oscilloscope in parallel with 1 Mohm already in the figure the total resistance $R(ab)= 0.5 Mohm$

The equations are

$\left\{\begin{matrix}

1*I1 +0.5*Iab -2*I4 =0\\

1*I3+0.5*Iab -5*I2 =0\\

I1=0.5*Iab +I2\\

I4 +0.5*Iab =I3\\

1*I1+5*I2 =100

\end{matrix}\right.$

The matrix of this system is

$Det1=\begin{vmatrix}

1& 0& 0& -2& 0.5\\

0& -5& 1& 0& 0.5\\

1& -1& 0& 0& -0.5\\

0& 0& -1& 1& 0.5\\

1& 5& 0& 0& 0\end{vmatrix}=-22.5$

and the extended matrix for determining I(ab) is

$Det2= \begin{vmatrix}

1& 0& 0& -2& 0\\

0& -5& 1& 0& 0\\

1& -1& 0& 0& 0\\

0& 0& -1& 1& 0\\

1& 5& 0& 0& 100\end{vmatrix}=-900$

$I(ab) =det2/det1 =40 \mu A$

$V(ab) =I(ab)*R(ab) = 40 \mu*0.5 Mega =20 V$

#### Absolute Voltage

b)

Now the impedance of the oscilloscope is in parallel with the 5 Mohm resistors

R(ab) = 1 Mohm is missing now and there are 2 1 Mohm resistances in parallel with 5 Mohm.

$Va = U*(5 || 1)/[5 || 1) + 1] =U*0.833/1.833 =0.4545*U$

$Vb =U*(5 || 1)/[(5 || 1) + 2) =U*0.833/2.833 =0.294*U$

The measured voltage is thus

$Va-Vb =(0.4545-0.294)*U =16.05*cos(100*t) V$

Obs.

If there is a 1 Mohm resistance on the bridge diagonal one needs again a system of 5 equations solved.

##### Without R(in)

Without the impedance of the oscilloscope we have

$Va =U*5/(5+1) =5/6*U =0.833*U$

$Vb =U*5/(5+2) =5/7*U =0.714*U$

$Va-Vb =(0.833-0.714)*U =0.1187*U =11.87*cos(100*t) V$