ECE 3144 Lecture 21 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University
Applications of Thevenin and Norton’s Theorems • Note that this systematic transformation allows us to reduce the network to a simpler equivalent form with respect to some other circuit elements. • Although this technique is applicable to networks containing dependent sources, it may not be as useful as other techniques and care must be taken not transform the part of the circuit which contains the control variables. • How to apply these theorems depends on the structure of the circuits. • Case 1: if only independent sources are present, we can calculate the open-circuit voltage and short circuit current and then the Thevenin equivalent resistance. • Case 2: if both independent sources and dependent sources are present, we will calculate the open-circuit and short circuit current first. Then determine the Thevenin equivalent resistance. • Case 3: For circuit only contains dependent sources, since both open-circuit and short-circuit current are zero (no independent sources here to provide the controlling variables), we cannot determine RTH in this case by using voc/isc. Remember during the discussion of Thevenin theorem, we can apply an external voltage source vS(t) (usually a constant) and measure the resulting current iS(t) => RTH = vS(t)/iS(t)
Case 1 example 1 Find the Thevenin equivalent circuit at terminal pair a and b for the circuit shown. + This specific problem can be solved by using different approaches. We solve the problem by using source transformation technique. - Thus we have RTH = 12 and vTH = -8V
Case 1 example 2 I I 2 1 Find Vo in the circuit given in figure (a). (a) Vois the voltage across the 6k resistor. Thus 6k resistor can be considered as the load resistor and the rest network can be replaced by the equivalent Thevenin circuit. The open circuit voltage Voc is found from figure (b). Vo (b) Apply mesh analysis technique: Mesh 1: -6 +4kI1+2k(I1-I2) = 0 Voc => I1 = 5/3 mA Mesh 2: I2 = 2 mA Applying KVL Voc = 4kI1+2kI2 =4*5/3 + 2*2 = 32/3 V (c) RTH can be derived by calculating the short circuit current Isc. Then RTH = Voc/Isc. For the circuits containing only independent sources, RTH can also be derived by zeroing out all sources as shown in (c). RTH RTH = (4k//2k)+2k = 10/3 k
Case 1 example 2: cont’d Thus the original circuit is equivalent to the network shown in figure (d). (d) Thus using voltage divider => Vo (e) The equivalent Norton equivalent circuit is shown in figure (e). Vo
Case 2 example Find Vo in the circuit shown using Thevenin’s Theorem. Find Voc + Vx = 6*2k/(4k+2k) = 2V Vy = 12-4k*Vx/1000 = 12-8= 4V Voc = Vx-Vy = -2V - - Find Isc (6-Vx)/4k=Vx/2k+Isc Isc = Vx/1K+(Vx-12)/4k Vx => Isc = -0.1875mA Rth = Voc/Isc=10.67k Vo = Voc*2k/(2k+10.67k) = -0.32V
Case 3 example Find the Thevenin equivalent of the circuit given a b Since the rightmost terminals are already open-circuit, i=0. Consequently, the dependent source is dead. For this network, since there is no independent source in the network, both voc and isc are zero. Apply a 1-A source iS externally, measure the voltage vSacross the terminal pairs. Then we have RTH = vS/iS = vS. We can see that i=-1A. Apply KVL nodal analysis at node a: a + vS => - vS = 0.6V => RTH = vs/i= 0.6
Problem solving strategy by using Thevenin theorem/Norton theorem • Remove the load and the find the voltage across the open-circuit terminals, Voc. All the circuit analysis techniques presented (KVL/KCL, current/voltage divider, nodal/loop analysis, superposition, source transformation) • Determine the Thevenin equivalent resistance of the network at the open terminals with the load removed. Three different types of circuits may encountered in determining the resistance RTH. • If the circuits contains only independent sources, they are made zero by replacing voltage sources with short circuits and current sources with open circuits. RTH is then found by computing the resistance of the purely resistive network at the open terminals. • If the circuit contains only dependent sources, an independent voltage or current source is applied at the open terminals and the corresponding current or voltage at these terminals is calculated. The voltage/current ratio at the terminals is the Thevenin equivalent resistance. Since there is no energy source, the open circuit voltage is zero here. • If the circuits contains both independent and dependent sources, the open circuit terminals are shorted and the short-circuit current between these terminals are decided. The ratio of the open-circuit voltage to the short-circuit current is the Thevenin resistance RTH. • The load is now connected to the Thevenin equivalent circuit, consisting of Voc in series with RTH, the desired output is obtained. • The problem solving strategy for Norton theorem is essentially the same as that for the Thevenin theorem with the exception that we are dealing with the short-circuit current instead of open-circuit voltage.
Operational Amplifier • An operational amplifier is a linear circuit network. • Based on Thevenin theorem or Norton theorem, any linear circuit can be considered as equivalent to a Thevenin circuit or Norton circuit at a specified terminal pair . The following is the detailed modeling for an opamp. Riis the input Thevenin resistance and Ro is the output Thevenin resistance. i+ -> ->iout i- ->
Ideal Operational Amplifier A • Now we want to derive the ideal conditions for the op-amp. • If an external network A is connected to the op-amp at the input terminals, then op-amp is considered as the load of the external network A. • In order to gain maximum output voltage (v+-v-) for op-amp with any external network A => the input Thevenin resistance for the ideal op-amp is infinite. Ri-> => i+ = i- = 0 • If an external network B is connected to the op-amp at the output terminals, the external network B is considered as the load of the op-amp. • In order for the load to gain the maxim output vofor any external network B=> the output Thevenin resistance for the ideal op-amp is zero, Ro = 0. • Finally, the gain of an ideal operational amplifier is infinite => A-> and v+ = v- = 0. B + vo -
Ideal Operational Amplifier Ideal model for an operational amplifier Modeling parameters i+ = i- = 0 and v+ = v-
Example : Differential operational amplifier Apply KCL in the “-” terminal(inverting terminal) of the opamp: (1) Apply KCL in the “+” terminal (noninverting terminal) of the opamp: (2) v+ = v- (3) Using equations (1), (2) and (3) to solve vout The output of this amplifier clearly is proportional to the difference of the voltages of the driving circuits. For this reason, this amplifier is called a differentialamplifier.
Homework for Lecture 21 • Problems 4.61, 4.67, 4.69.