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 1. Rs vL + -Vs + -+ -RLRin Rout vin A vin Figure 1. 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 criterion 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 2 we will be able to investigate the details of each part for a deeper understanding. We are particularly interested to determine the relation between the input signal Vs and the output voltage or current. The amplifier part of the circuit consists of the measurement part and the output equivalent circuit. By describing the circuit in terms of the Thevenin equivalent circuits we have provided 6.071/22.071 Spring 2006, Chaniotakis and Cory 1the most compact form possible for this application. As always the i-v characteristic plots of the input and the output sides of the circuit, as shown on Figure 3, provide additional insight. Rs vLVs + -RL iin Rin Rout vin A vin measurement Thevenin equivalent of the output source amplifier Thevenin equivalent of the inputsource load iout Figure 2. Detailed breakdown of the amplifier circuit Vs Vs/iin /vL Rs Slope=1/Rin vin Avin Avin/Rout Slope=1 RL iout i-v characteristic of input The operating point is a the intersection of the two lines. For a given source, the location of the operating point depends on the value of the input resistor Rin. i-v characteristic of output The operating point is a the intersection of the two lines. For a given amplifier, the location of the operating point depends on the value of the load resistor RL. Figure 3. i-v plots of input and output circuits of an amplifier 6.071/22.071 Spring 2006, Chaniotakis and Cory 2Here 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. Change of Resistance1 Let’s look at the amplifier section of the circuit shown on Figure 2 when A=1. This circuit is shown on Figure 4. Since A=1, the circuit does not provide any voltage gain. So what is the purpose of this amplifier? The answer becomes obvious when we look at the power transfer characteristics of the circuit. + -+ -Rin Rout vin vin measurement Thevenin equivalent of the output source iout iin vout Figure 4. Power amplifier circuit The power at the input side of the amplifier is vin 2 Pin =Rin (0.1) And the power at the output is Pout =vin 2 Rout (0.2) And thus the gain in power is 1 This change in resistance is often called change in impedance. Impedance is a generalized resistance that we will see much more of when we turn to reactive elements (capacitors and inductors). For now we will equate it to the characteristic (Thevenin or Norton) resistance of a port. 6.071/22.071 Spring 2006, Chaniotakis and Cory 3Pout Rin = (0.3)Pin Rout It is a common situation, to have the input resistance high (so that we are effectively measuring a voltage), and the output resistance low (so that we are effectively providing a power). Decibels The decibel is used to measure the ratio of two powers or equivalently the gain in power. The gain in decibels is given by Poutgain dB () =10 log (0.4)10 Pin For an amplifier system as shown on Figure 5, the gain may also be written in terms of the input and output voltage or current. i1 i2 + v2 + -Ampifier R2v1 R1 -R2 = 10 log10 ⎜ ⎟−10 log ⎝ v1 ⎠10 R1 Notice that if the input and output resistances are the same, (R1=R2) rewritten in terms of voltage, v2() =20 log gain dB 10 v1 Or the current i2() = 20 log gain dB 10 i1 Figure 5 Amplifier system P2(gain dB ) = 10 log = 10 log 10 P1 10 ⎛ v2 ⎞2 v22 / R2 v12 / R1 (0.5) then the gain can be (0.6) (0.7) 6.071/22.071 Spring 2006, Chaniotakis and Cory 4Use of Dependent Sources Let’s consider the voltage controlled voltage source shown on Figure 6 i1 v1 + -vs = A v1 Figure 6. VCVS In practice these devices are made to have a very large but imprecise gain A. The variability in the value of A from device to device and due to various environmental condition (change in the operating temperature for example) results in undesirable effects. In order for these devices to be useful they must be arranged in a circuit in such a way that the actual value of the gain A does not affect the operation of the circuit. In practice this is achieved with feedback, resulting in a smaller overall, but precise, gain, and a stable operating condition We will explore many applications of this in the near future, here we just give a first exploratory example. Consider the circuit shown on Figure 7. + --Rf R + + -Vs Rin vin A vin iin vout Rout iout vout feedback path vinvin Figure 7. Amplifier with feedback The feedback path is indicated by the dashed line on Figure 7. As we will see next by analyzing the circuit, the feedback is used to set the gain of the amplifier. Notice that the feedback path samples the output and delivers this to the negative terminal of the input. This is an example of negative feedback. 6.071/22.071 Spring 2006, Chaniotakis and Cory 5Let’s make three assumptions related to the quality of the input and output parts of the amplifier. 1. assume that A is very large (typically A ~ 500,000), 2. assume that Rout is small (a few Ω) 3. Rin is very large (~ MΩ) Together the above assumptions mean that the output stage has high gain and can deliver power, and that the input stage has no influence on the measurement of the output voltage. The key to this type of “negative” feedback is to investigate the voltage at the negative