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 Vs vin Rout Rin A vin vL RL 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 1 the 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 amplifier Rs iin iout vin Rin Vs Rout A vin vL RL Thevenin equivalent of Thevenin equivalent of measurement the inputsource the output source load Figure 2 Detailed breakdown of the amplifier circuit iin Slope 1 Rin i v characteristic of input Vs Rs Vs vin 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 iout Avin Rout Slope 1 RL i v characteristic of output Avin The operating point is a the intersection of the two lines For a given amplifier the location of the operating point depends on vL 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 2 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 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 Rout iin iout vin Rin vin vout measurement Thevenin equivalent of the output source 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 3 Pout Rin Pin Rout 0 3 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 gain dB 10 log10 Pout Pin 0 4 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 v1 i2 R1 v2 Ampifier R2 Figure 5 Amplifier system gain dB 10 log10 P2 v22 R2 10 log10 2 P1 v1 R1 2 R2 v2 10 log10 10 log10 R1 v1 0 5 Notice that if the input and output resistances are the same R1 R2 then the gain can be rewritten in terms of voltage gain dB 20 log10 v2 v1 0 6 gain dB 20 log10 i2 i1 0 7 Or the current 6 071 22 071 Spring 2006 Chaniotakis and Cory 4 Use 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 vin iin Vs vin Rin Rout iout A vin vout vout Rf vin R feedback path 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 5 Let 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 3 Rin is very large M 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 terminal of the input vin vout R Rf R 0 8 since vin Vs 0 9 Avin A vin vin vout 0 10 And vout becomes R …