Dependent Sources

Dependent Sources: Introduction and analysis of circuits containing dependent sources. So far we have explored time-independent (resistive) elements that are also linear. We have seen that two terminal (one port) circuits can be modeled by simple circuits (Thevenin or Norton equivalent circuits) and that they have a straight line i-v characteristic. Here we introduce the idea of a dependent source. We will see that the use of dependent sources permits the use of feedback. Feedback can be used to control amplifiers and to build interesting transducers. Dependent Sources A dependent source is one whose value depends on some other variable in the circuit. An illustrative example of a dependent source is, i black box equivalent of input circuit + v1 - g v1 black box equivalent of output circuit Here we see that there is an “input” circuit that develops a voltage, v1. In a separate part of the circuit there is a linear, voltage-dependent, current source that delivers a current given by i = g v1 (1.1) Where g is a constant with the units of A/V. So the current that flows into the output circuit depends on the measurement of a voltage on the input circuit. Now clearly we could mimic a dependent source by looking at a meter and changing a potentiometer (for example) in relation to the reading. Here we will introduce circuits that carry out this function without any intervention. Notice that the above circuit is still linear since the output current depends linearly on the measured voltage. For now we will concern ourselves with only linear dependent sources. Later, we will see examples of non-linear dependent sources where the analysis will be somewhat more complex. 1 There are four general classes of linear dependent sources. Their names, acronyms and associated symbols are: Voltage Controlled Voltage Source: VCVS i1 + v1 - vs = A v1 Current Controlled Voltage Source: CCVS i1 + v1 - vs = r i1 Voltage Controlled Current Source: VCCS. i1 + v1 - is = g v1 Current Controlled Current Source: CCCS i1 + v1 - is =β i1 The parameters A, r, g, β, are real numbers, and v1 , i1 are voltages/currents in some circuit. 2 Circuit Analysis with Linear Dependent Sources. Linear dependent sources provide no new complications to circuit analysis. Kirchhoff’s laws still apply, and formal circuit analysis goes ahead just as before. The dependent source only introduces a constraint on the solution. The simplest example is where the measurement and dependent source are in two isolated circuits. Let’s consider the current amplifier circuit shown on Figure 1 . The circuit has one independent current source and one dependent current source. The dependent current source is a CCCS. We would like to determine the voltage vc as indicated. ib Is Rs + Rb β ib Rc vc - Figure 1. Current Amplifier Circuit The left hand circuit is a current divider, and ib = Rs Is Rs + Rb (1.2) The right hand circuit is a current source. The output voltage vc is given by vc = β ib Rc (1.3) So now we see that the output voltage vc depends on the measured current ib of the input circuit. Combining Equations (1.2) and (1.3) we obtain RsRc vc = β Is Rb + Rc  (1.4) gain So, the overall circuit behaves as an amplifier with the gain dependent on the resistors and the proportionality constant β. 3 Let’s now consider the slightly more interesting circuit shown on Figure 2. R1 R2 Is Vs= 2 v3 + v3 - R3 Figure 2. Circuit with dependent voltage source Let’s use nodal analysis to solve for the currents and voltages in this circuit. Figure 3 shows the nodes of interest, the variables and the polarities. v1 node1 R1 + - Vs= 2 v3 R2 + i1 v2 node2 - i2 Is + R3 i3 - + v3 - Figure 3. Nodal analysis of circuit with dependent sources KCL at node1 gives i1 + Is − i 2 = 0 Vs − v1 v1 − v 2 + Is − =0 R1 R2 (1.5) KCL at node2 gives 4 i 2 − i3 = 0 v1 − v 2 v 2 − =0 R2 R3 (1.6) In matrix form, Equations (1.5) and (1.6) become 1 ⎛ 1 ⎜ R 2 + R3 ⎜ ⎜ − 1 ⎜ ⎝ R2 1 ⎞ Vs ⎞ ⎛ R 2 ⎟ ⎛ v1 ⎞ ⎜ Is + ⎟ ⎟⎜ ⎟ = R1 ⎟ 1 1 ⎟ ⎝ v2 ⎠ ⎜ + ⎟ ⎝ 0 ⎠ R1 R 2 ⎠ − (1.7) and the solution is given by v1 = ( R 2 + R3)( IsR1 + Vs ) R1 + R 2 + R3 (1.8) R3( IsR1 + Vs ) R1 + R 2 + R3 (1.9) v2 = Now need to include the constraints of the dependent sources. These constraints are And v 2 = v3 (1.10) Vs = 2v3 (1.11) Substituting Equations (1.10) and (1.11) into Equations (1.8) and (1.9) we obtain v1 = IsR1( R 2 + R3) R1 + R 2 + R3 (1.12) IsR1R3 R1 + R 2 + R3 (1.13) v2 = 5 Analysis of Circuits with Dependent Sources Using Superposition When employing the principle of superposition to a circuit that has dependent and independent sources we proceed as follows: • • Leave dependent sources intact. Consider one independent source at the time with all other independent sources set to zero. Let’s explore this with the following example: For the circuit on Figure 4 calculate the voltage v. Av R + v Is Vs - Figure 4. Circuit with dependent source. Analysis using superposition We proceed by first considering the effect of the current source acting alone. The circuit of Figure 5 shows the corresponding circuit for which the independent voltage source Vs has been suppressed. A v1 R + Is v1 - Figure 5. Circuit with the voltage source suppressed By applying KVL we obtain: v1 − IsR + Av1 = 0 (1.14) And v1 becomes v1 = IsR 1+ A (1.15) 6 Next we evaluate the contribution to the output with the independent voltage source acting alone. The corresponding circuit is shown on Figure 6. A v2 R + Vs v2 - Figure 6. Circuit with the current source suppressed. Again applying KVL we have Av 2 + v 2 − Vs = 0 (1.16) (Note that the voltage drop across R is zero since there is no current flowing in the circuit.) And v2 becomes Vs (1.17) v2 = 1+ A And so the total voltage is written as the superposition of v1 and v2. v = v1 + v 2 1 = (Vs + IsR ) 1+ A (1.18) Let’s now look at the slightly more complicated circuit shown on Figure 7 with multiple dependent and independent sources. We will determine the voltage vo by using superposition. Rs + v1 + Vs1 - + A v1 + R1 A v2 v2 Vs2 - vo - + Rs Figure 7. Circuit with dependent and independent sources 7 The procedure is the same as before: leave dependent sources intact, calculate the contribution of each independent source acting alone. Figure 8 shows the circuit with Vs2 suppressed. The indicated output vo1 is the contribution of voltage source Vs1. Rs + v1 + Vs1 - + A v1 R1 A v2 vo1 - v2 + Rs Figure 8. Since the voltage source Vs2 has been suppressed, the voltage v2 is zero. Therefore the current provided by the voltage controlled current source Av2 is zero. This is shown schematically on the circuit of Figure 9. Rs + v1 + Vs1 - + A v1 R1 vo1 - v2 + Rs Figure 9 Therefore the voltage vo1 is vo1 = A v1 R1 = A Vs1 R1 (1.19) Next we will suppress the voltage source Vs1 as shown on the circuit of Figure 10. 8 Rs + v1 + A v1 + R1 A v2 - v2 Vs2 - vo2 + Rs Figure 10 Now the voltage v1 = 0 and v2 = −Vs 2 , and the circuit reduces to the one shown on Figure 11 Rs + v1 + R1 + A v2 - v2 Vs2 - vo2 + Rs Figure 11 The voltage vo2 is now vo 2 = − A Vs 2 R1 (1.20) And by combining Equations (1.19) and (1.20) the voltage vo is vo = vo1 + vo2 = A (Vs1 − Vs 2 ) R1 (1.21) 9 Determining the Thevenin/Norton equivalent circuit of circuits containing dependent sources. For a given two-terminal port the equivalent circuit is defined exactly as in the case of independent sources. Measure or calculate the open-circuit voltage and the short circuit resistance. These two points define the i-v characteristics of the port. The characteristic resistance is the ratio of the open circuit voltage to the short circuit current. Given a schematic, the characteristic resistance can also be found by suppressing all independent sources and calculating the effective resistance between the terminals. NOTE, do not suppress the dependent sources. Let’s consider the circuit shown on Figure 12. We would like to calculate the voltage vo. Ra R2 Rc R3 + + Vs1 - Rb v1 Vs2 + Rd v2 A1v1 R1 A2v2 R4 - vo - Figure 12 We will do this by first finding the Thevenin equivalent circuit seen by resistor R4. We will calculate the equivalent circuit across the terminals X-Y as shown on the circuit of Figure 13. Ra Rc R2 R3 X + + Vs1 - Rb v1 Vs2 + Rd v2 A1v1 R1 A2v2 R4 - vo - Y Figure 13 In turn we will find vo by considering the simple voltage divider circuit shown on Figure 14. 10 RTh + VTh R4 vo - Figure 14 First let’s calculate the Thevenin equivalent resistance RTh seen at port X-Y (Figure 15). Ra + Rb v1 Vs2 Vs1 - R2 Rc + Rd v2 A1v1 R3 X RTh A2v2 R1 Y Figure 15 In order to calculate RTh we will suppress all independent sources in our circuit. (Note: leave dependent sources intact). Figure 16 shows the circuit with the independent sources suppressed. Ra Rb v1 R2 Rc Rd v2 A1v1 R1 R3 X RTh A2v2 Y Figure 16 Note that sending both Vs1 and Vs2 to zero causes v1 and v2 to also go to zero. Therefore, the dependent sources also go to zero and the corresponding circuit is now shown on Figure 17 where the dependent sources are now zero. 11 Ra R2 Rc Rd v2 Rb v1 R3 X RTh R1 Y Figure 17 So we now see that the Thevenin equivalent resistance across terminals X-Y is RTh = R 2 + R3 (1.22) Next we will calculate the Thevenin voltage VTh, or the open circuit voltage, across terminals X-Y. R2 Ra Rc R3 X + + Vs2 - Rb v1 Vs2 + Rd v2 A1v1 A2v2 R1 - VTh Y Figure 18 We can calculate VTh by employing any of the circuit analysis methods. We will use superposition for this case. VTh=VTh1+VTh2 as shown on corresponding circuits of Figure 19. Ra R2 Rc R3 X + + Rd v2 A1v1 Rb v1 Vs1 VTh1 R1 - Y Ra Rc R2 R3 X + Rb v1 Vs2 + Rd v2 - R1 A2v2 VTh2 Y Figure 19 12 From Figure 19 we see that, VTh1 = A1v1 = A1Vs1 Rb Ra + Rb (1.23) and VTh2 = A2 v 2 R 2 = A2 Vs 2 Rd R2 Rc + Rd (1.24) And superposition gives VTh = VTh1 + VTh 2 Rb Rd = A1Vs1 + A2Vs 2 R2 Ra + Rb Rc + Rd (1.25) From the equivalent circuit shown on Figure 14 the desired voltage is Rb Rd R4 ⎛ ⎞ vo = ⎜ A1Vs1 R2 ⎟ + A2Vs 2 Ra + Rb Rc + Rd ⎝ ⎠ R 4 + R 2 + R3 (1.26) 13 Next, let’s analyze a circuit that contains only dependent sources. For the circuit shown on Figure 20 we will determine the Thevenin and Norton equivalent circuits across terminals a-b. a R1 + v Av R2 b Figure 20 The equivalent circuit for this has to be a resistance since with no input there can not be an output. Normally with a two terminal device we would measure the open-circuit voltage and the short circuit current and from these two measurements determine the i-v characteristic. For this circuit, the short-circuit current is zero since shorting the terminals forces v to zero. The open circuit voltage is not so obvious, one might think that the output voltage is undefined in the open circuit arrangement and thus any voltage would be possible. But let’s look more carefully. In the open circuit configuration the current through R1 is zero, so there is no voltage drop across R1. By KVL this requires that v be equal to the voltage drop across R2. However, KVL requires that the voltage drop across R2 equal Av, or that v = Av . This can be true in only two cases: (1) in the case that A=1, or (2) in the general case that v = 0 . Let’s consider the more general case. The circuit is still a linear, resistive circuit, and so we still need only 2 points to define the i-v characteristic. Since both the open circuit and short circuit measurements probe the same point (i=0 and v=0) we need to define another point. In order to find the second point we may fix either the current or the voltage at the output and determine the other. If we set the voltage across the terminals a-b to Vt (it does not matter what value we use), then we have the circuit shown on Figure 21. The current flow indicated by the arrow results from the application of the test voltage Vt. 14 a i R1 Vt + - A Vt R2 b Figure 21 By applying KVL around the loop formed by Vt, R1 and R2 we have iR1 + AVt − Vt = 0 i = Vt (1.27) 1− A R1 (1.28) The i-v characteristic curve is now shown on Figure 22 i 1-A slope= R1 Vt 1-A R1 Vt v Figure 22 And the Thevenin equivalent resistance is R1 1− A And so the Thevenin equivalent circuit is as shown on Figure 23 RTh = RTh (1.29) a i + v b Figure 23 15 Similarly we could apply a test current source It and determine the resulting voltage v. This is shown on Figure 24. a R1 + v Av R2 It b Figure 24 Again applying KVL around the loop we have ItR1 + Av − v = 0 (1.30) And v = It R1 1− A (1.31) Again indicating that the equivalent resistance (Thevenin/Norton) is RTh = RN = R1 1− A (1.32) 16 Applications of dependent sources Dependent sources provide a convenient means of: 1. converting between voltage and current 2. changing resistance Since dependent sources often appear in the part of the circuit that we are using to make a measurement, they also enable the input and output characteristics of a device to be optimized separately. Look at the amplifier circuit shown on Figure 25. Rs Vs + - + vin - Rout Rin A vin + vL - RL Figure 25. Amplifier circuit The circuit at the far left is a Thevenin equivalent of a voltage source. So this can stand in place of any voltage source regardless of the actual complexity of the physical source. The resistor Rin is used to measure the voltage vin that is provided by this source. Since Rin is the basis of a voltage measurement we desire that Rin>>Rs. This is a general design criteria that we have seen before. The circuit at the right is a Thevenin equivalent voltage source driving a load. Here the good design characteristics require that Rout< < RL. By breaking the circuit into four components as shown on Figure 26 we will be able to investigate the details of each part for a deeper understanding. 17 amplifier Rs iin Vs + Rout iout vin Rin A vin vL RL - Thevenin equivalent of Thevenin equivalent of measurement the inputsource the output source load Figure 26. Detailed breakdown of the amplifier circuit By describing the circuit in terms of the Thevenin equivalent circuits we have provided the most compact form possible for this application. Since it is a linear circuit, the i-v characteristic will provide additional insight. iin Slope=1/Rin Vs/Rs i-v characteristic of input Vs vin iout vin/Rout Slope=1/RL i-v characteristic of output Avin vL 18 Here then we see that depending on the choice of the resistors and the parameter A we can build an amplifier that detects a voltage and delivers power. Notice that Vs would often be a time varying voltage (a signal) and so the operating points would slide back and forth, but the slopes would not change. (We will continue with more details on dependent sources including the concept of feedback next class) 19