The Algorithm

To calculate the small signal noise of a circuit, the AC noise analysis has to be applied [6]. This technique uses the principle of the AC analysis described in chapter 4 on page . In addition to the MNA matrix one needs the noise current correlation matrix of the circuit, that contains the equivalent noise current sources for every node on its main diagonal and their correlation on the other positions.

The basic concept of the AC noise analysis is as follows: The noise voltage at node should be calculated, so the voltage arising due to the noise source at node is calculated first. This has to be done for every nodes and after that adding all the noise voltages (by paying attention to their correlation) leads to the overall voltage. But that would mean to solve the MNA equation times. Fortunately there is a more easy way. One can perform the above-mentioned steps in one single step, if the reciprocal MNA matrix is used. This matrix equals the MNA matrix itself, if the network is reciprocal. A network that only contains resistors, capacitors, inductors, gyrators and transformers is reciprocal.

The question that needs to be answered now is: How to get the reciprocal MNA matrix for an arbitrary network? This is equivalent to the question: How to get the MNA matrix of the adjoint network. The answer is quite simple: Just transpose the MNA matrix!

For any network, calculating the noise voltage at node is done by the following three steps:

(5.3) | ||||

(5.4) | ||||

(5.5) |

If the correlation between several noise voltages is also wanted, the procedure is straight forward: Perform step 1 for every desired node, put the results into a matrix and replace the vector in step 3 by this matrix. This results in the complete correlation matrix. Indeed, the above-mentioned algorithm is only a specialisation of transforming the noise correlation matrices (see section 2.4.2).

If the normal AC analysis has already be done with LU decomposition, then the most time consuming work of step 1 has already be done.

(5.6) |

I.e. becomes the new matrix and becomes the new matrix, and the matrix equation do not need to be solved again, because only the right-hand side was changed. So altogether this is a quickly done task. (Note that in step 3, only the subvector of vector is used. See section 3.1.3 for details on this.)

If the noise voltage at another node needs to be known, only the right-hand side of step 1 changes. That is, a new LU decomposition is not needed.

Reusing the LU decomposed MNA matrix of the usual AC analysis is possible if there has been no pivoting necessary during the decomposition.

When using either Crout's or Doolittle's definition of the LU decomposition during the AC analysis the decomposition representation changes during the AC noise analysis as the matrix gets transposed. This means:

with and | (5.7) |

becomes

with and | (5.8) |

Thus the forward substitution (as described in section 15.2.4) and the backward substitution (as described in section 15.2.4) must be slightly modified.

(5.9) |

(5.10) |

Now the diagonal elements can be neglected in the forward substitution but the elements must be considered in the backward substitution.

The network that is depicted in figure 5.1 is given. The MNA equation is (see chapter 3.1):

(5.11) |

Because of the controlled current source, the circuit is not reciprocal. The noise voltage at node 2 is the one to search for. Yes, this is very easy to calculate, because it is a simple example, but the algorithm described above should be used. This can be achived by solving the equations

(5.12) |

and

(5.13) |

So, the MNA matrix must be solved two times: First to get the transimpedance from node 1 to node 2 (i.e. ) and second to get the transimpedance from node 2 to node 2 (i.e. ). But why solving it two times, if only one voltage should be calculated? With every step transimpedances are calculated that are not need. Is there no more effective way?

Fortunately, there is Tellegen's Theorem: A network and its adjoint network are reciprocal to each other. That is, transposing the MNA matrix leads to the one of the reciprocal network. To check it out:

(5.14) |

Compare the transposed matrix with the reciprocal network in figure 5.2. It is true! But now it is:

(5.15) |

Because of the original network equals of the reciprocal network, the one step delivers exactly what is needed. So the next step is:

(5.16) |

Now, as the transimpedances are known, the noise voltage
at node 2 can be computed. As there is no correlation, it writes
as follows:

(5.17) | |||

(5.18) | |||

(5.19) | |||

(5.20) |

That's it. Yes, this could have be computed more easily, but now the universal algorithm is also clear.

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