R1=400kΩ. R2=500kΩ R3=100kΩ

Transcription

1 Exercise + dd dd5 400kΩ 500kΩ 00kΩ Comute the voltage across resistor.. f you measure such otential difference with a voltmeter featuring an internal resistance 0MΩ. Which is the actual measured voltage? We can redraw the circuit as follows,but get used not to draw exlicitely the bias sources. + dd We can comute the voltage across resistor by alying the voltage divider: resistanceacross which weevaluatethe voltagedro } dd +. 5 totalbranch resistance. The voltmeter can be modeled a san ideal voltmeter (i.e. with infinite inut resistance and, therefore, abosrbing zero current) with the internal resistor 0MΩ in arallel. + dd m, gain we can comute the voltage dro through the voltage divider relationshi: // dd // + +./

2 Exercise + cc C + dd out dd+6 -ss-6 800kΩ 400kΩ 0kΩ CμF - ss Comute the voltage out.. Find the actual outut voltage if we connect a load resistor 0kΩ at the outut. ll the voltage sources are DC sources. Therefore the caacitor acts a san oen circuit: out ss + ( dd + ss ) +. To get a quick answer we can comute the Thevenin uivalent of the circuit and connect the load resistor to it. out out + We are oerating with DC voltages, therefore the caacitor acts as an oen circuit. n order to comute the uivalent resistance,, we switch off all the indeendent source generator and comute the resistance seen at the terminal towards ground. y definition we aly a current (voltage) robe and we measure the voltage at (the current through) the terminals. The resistance is then given by the ratio of that current and that voltage. v // 67kΩ i n order to comute the uivalent voltage source we comute the oen terminals voltage, that is the voltage difference at the two terminals when no current is flowing through them. Such voltage is exactly the one comuted in ss + ( dd + ss ) + Therefore: out 7m!! +./

3 Exercise kω 500Ω 500Ω 4500Ω 8m Comute the Norton uivalent at terminals and.. Find the value of the load resistor that, once inserted at terminals and, dissiates the maximum ower and comute the value of that ower. n order to comute the uivalent resistance,, we switch off all indeendent sources and comute the resistance at the terminal towards ground. y definition, we aly a current (voltage) robe and we measure the voltage at (the current through) the terminals. The resistance is then given by the ratio of that current and that voltage. n a quicker way we can look by insection all the aths that the current can follow to go from to (look at the blue arrows in the next figure). 4 v // 4 5 i ( + // ). Ω n order to comute the Norton uivalent current, we have to comute the short circuit current at the terminals, that is the current flowin through the terminals when they are ket at the same otential. Since it is a linear network, we can aly the suerosition theorem. +) m + // ) // 4. m + // m.m m./

4 . The ower dissiated in resistor is given by: Pdiss ( ) + + et s comute which value of the resistor maximizes the dissiated ower: P { ( + ) ( + ) } diss + P diss 0 ( ) and the dissiated ower is Pdiss. mw.4/

5 Exercise 4 controlled generator + i +0 00Ω.9kΩ 50Ω i0 Comute the current sourced by the voltage source.. Find the Thevenin uivalent circuit at the terminals and. We face a circuit with a current-controlled current-source. To comute the current sourced by the voltage generator (+) we make the current balance at node X. + + i + i controlled generator X i m + i + The current sourced by the voltage source is the sum of the current in the two branches: + m TOT i. To calculate the Thevenin uivalent we have to comute the uivalent resi stance and the oen circuit voltage. To evaluate the uivalent resistance we have to switch off the indeendent sources (WE DO NOT SWTCH OFF THE CONTOED SOUCE!!). v i v i i i + i v 0 + i + 0 v i The oen circuit voltage is v i + 5 where we used the value of the current,, already calculated in The uivalent Thevenin circuit will then be:.5/

6 Exercise kΩ 8kΩ 8kΩ 44.7kΩ 55.6kΩ Find the Norton uivalent at the terminals and.. Find the value of the ower dissiated by a load resi stance 8kΩ, at the outut terminals and.. n order to increase the ower dissiated by the load is it needed to increase or to decrease the value of the resistor? The Norton uivalenti s given by: 5 //( 4 + // ). 8kΩ 89μ 4 // et s rofit of the Norton uivalent already calculated in i lying the current divider: i 544μ + The dissiated ower is given by P i 5 W diss μ. n order to increase the dissiated ower we have to lower the value of resistor. n fact: Pdiss i + + ( ) ( ) ( ).6/

7 Exercise m kω.kω.kω 45kΩ 5kΩ 68kΩ Find the Norton uivalent at the terminals and.. Find the Thevenin circuit at the terminals and. e. The Norton uivalenti s given by: 6 //( )//( + + ) 908Ω n order to comute the current we can comute the Thevenin uivalent voltage (if it seems easier and since we have to comute it for question.) and find the current rofiting of the relationshi between Norton and Thevenin uivalents: Since the network is linear we can aly the suerosition theorem to comute : +) ( + + )//( ) ) + 4 α 5 6 β [ //( + + )] current divider at nodeα current divider at nodeβ +.5m.9 The Equivalent Thevenin circuit will be given by:.7/

8 Exercise 7 in C out in ΔT 00 μs m kω 6kΩ C0F - - Draw in a time diagram, roviding values for all the relevant oints, the curve of the voltage out(t) when the inut current is the one shown in the figure.. Determine the charge stored in the caacitor C when the steady state is reached following the alication of the signal shown in the figure. Provide justification for your answer.. How the value of the stored charge in the caacitor C (already calculated in.) would change a long time after the alication of the ulse shown in the figure if the ulse duration would be μs. Provide justification for your answer. t The circuit time constant is: τ ( + ) C 98μs so the outut waveform reaches the final value within the time interval ΔT. We can aly the suerosition theorem since the circuit is linear. gives no contribution. lso the voltage source gives no contribution to out since it is a DC voltage, so the caacitor acts a san oen circuit and the voltage out is measued with resect to ground. et s now consider the current in. On the transitino edge the caacitor cannot change the voltage difference at the terminals instantaneously, therefore the contribution of the current in only, to the outut voltage is: + out ( 0 ) in( // ) 6 n order to determine the full behaviour of the outut voltage let s comute its final value. The final value for the voltage due to the current generator in only, can be obtained considering the caacitor an oen circuit: out regime in, max 9 OUT (t) in (t) τ98μ s m t When the steady state is reached following the alication of the signal shown in the figure, the caacitor acts a san oen circuit. Therefore to comute the charge stored on its lates it is enough to comute the voltage difference at the caacitor terminal and exloit the relationshi Q C..8/

9 When the steady state is reached following the alication of the signal shown in the figure, the voltage across the caacitor is due to the voltage generator only, and is, therefore, ual to. The charge stored in the caacitor is, therefore, ual to Q C 0 F 0 C. Since we would like to know the charge stored when the steady state is reached following the alication of the signal shown in the figure, we do not mind the ulse duration. Therefore the charge stored is in any case ual to 0 F..9/

10 Exercise 8 + in C out out in T 0. ms m - / / kω kω C00F Find the average value of the current out when the inut current is the one shown in Fig. b.. Draw in a time diagram, roviding values for all the relevant oints, the curve of the voltage out(t), when the inut current is the one shown in Fig. b. Provide justification for your answer. We want to comute the average value (that is the DC comonent of the signal), the caacitor acts a san oen circuit. The circuit is linear, therefore, we can aly the suerosition theorem and consider for the current generator in only its DC value. + in t/t out,a - out,b The voltage source + gives no contribution since it is in series with a current source. out b + out a in + Considering that T in in, max m, T the average current flowing in is given by: out out + out in m. a b + +. The circuit time constant τ ( // ) C 75ns therefore the outut waveform reaches the final value within every fraction of the eriod. Since the circuit is linear, we can aly the suerosition theorem. Obviously + gives no contribution. et s now consider the current in. t the transition the caacitor cannot change instantaneously the voltage across the caacitor, therefore the outut voltage, due to in only, is zero. To determine the full behaviour of the outut voltage we comute the final value of the outut voltage. The final outut voltage, again considering the current generator in only, can be obtained by considering the caacitor as an oen circuit: // out regime, in in, regime ( ) f we now consider the contribution of the voltage source -, it is a DC source and the caacitor acts as an oen circuit: out +.0/

11 The time diagram of the outut voltage is, therefore, the following: OUT (t) in (t) m -0.5 τ75ns t -./