Norton circuit, equivalent

Capture and PSpice can be used to easily calculate the Norton and Thevenin equivalents of a circuit. The method we will use is the same as if we were going to find the equivalent circuits in the lab. We will make two measurements, the open circuit voltage and the short circuit current. The Thevenin resistance is then the open circuit voltage divided by the short circuit current. This will require us to create two circuits, one to find the open circuit voltage, and the second to find the short circuit current. In this example, we will find the Norton and Thevenin equivalent circuits for a DC circuit. This same procedure can be used to find the equivalent circuits of an AC circuit (a circuit with capacitors or inductors). However, instead of finding the open circuit voltage and short circuit current using the DC Nodal Analysis, we would need to use the AC analysis. 

For this example, we will find the Thevenin and Norton equivalent circuits for the circuit attached to the diode in EXERCI5E 3-5. The circuit is repeated below ... 

This circuit is difficult because it contains a nonlinear element (the diode) and a complex linear circuit. If we could replace VI, Rl, R2, and R3 by a simpler circuit, the analysis of the nonlinear element would be much easier. To simplify the analysis of the diode, we will find the Thevenin and Norton equivalent circuits of the circuit connected to the diode that is, we will find the Thevenin and Norton equivalents of the circuit below ... 

For determining the diode voltage and current, this circuit is much easier to work with than the original. This example is concerned with finding the numerical values of the equivalent circuit. The analysis of the circuit above was covered in Section 3.C. We will now find the Thevenin and Norton equivalent circuits of the circuit shown below ... 

EXEHCI5E 3-B Find the Thevenin and Norton equivalent circuits for the circuit below ... 

Figure 25.2 Constant-current source using a battery and series resistor, (a) Dummy (resistor) load (b) Norton s equivalent circuit for part a (c) electrolysis cell as the load.

For more complex current sources, it is necessary to employ Norton s theorem0 which states that any linear network of impedances and voltage sources can be substituted by an equivalent circuit containing a current source iN in parallel with an impedance 2 x, where iN is the current which flows when the output terminals of the network are short-circuited and 2EX is the network impedance with all source voltages put equal to zero and replaced by their internal impedances. 

Other methods to simplify the circuit are Thevenin s and Norton s theorems. These two theorems can be used to replace the entire circuit by employing equivalent circuits. For example, Figure 2.34 shows a circuit separated into two parts. Circuit A is linear. Circuit B contains non-linear elements. The essence of Thevenin s and Norton s theorems is that no dependent source in circuit A can be controlled by a voltage or current associated with an element in circuit B, and vice versa. 

Similar to Thevenin s theorem, Norton s theorem states that a section of a linear circuit containing one or more sources and impedances can be replaced with an equivalent circuit model containing only one constant current source and one parallel-connected impedance, as shown in Figure 2.36. 

To determine the Norton equivalent impedance ZG in Figure 2.36, we can kill all the sources in circuit A and then calculate the impedance from n-n terminals by looking back into circuit A. Thus, the Norton impedance ZG is equal to the Thevenin impedance. The Norton current IQ is a constant current that remains the same regardless of the impedance of circuit B. It can be determined by... 

Note that only at the output terminals n-n are the Thevenin and Norton equivalents the same. In other words, at the output terminals n-n the voltage and current of the Thevenin equivalent circuit and the Norton equivalent circuit are identical. 

In a first step, we set the load on the back side of the crystal (left-hand side in Fig. 6a) to zero and short-circuit the respective port. In a second step we apply the Norton transformation (Fig. 13b). The circuit from Fig. 13c is fully equivalent to the circuit shown in Fig. 13a. The equivalence of Figs. 13c and 13d is based on the relation ... 

Fig. 13 Steps in the derivation of the Butterworth-van Dyke circuit, a Same as Fig. 6c with the circuit elements rearranged, b Norton equivalence Zx = aZ/, Zy = aZ, a = Z + Z )jZ. c Norton equivalence applied to b. c Same as c where the relation - 2/ sin(2x) + tan(x) =- cot(x) has been used. The two transformers have been merged...

Calculate the current-flow on a given resistivity on an arbitrary DC circuit. Prove that this task could be solved by the minimization of the sum of dissipation fluxes or by the principle of minimal entropy production, like we did in the case of parallel connection of the Figure 6., when the circuit regarding the contacts of the resistor is replaced with the Norton s current source equivalent circuit [53]. 

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