Introduction to Electronic Circuits. DC Circuit Analysis: Transient Response of RC Circuits

Transcription

1 Introduction to Electronic ircuits D ircuit Anlysis: Trnsient esponse of ircuits

2 Up until this point, we hve een looking t the Stedy Stte response of D circuits. StedyStte implies tht nothing hs chnged in the circuit in long time no power hs een turned on or off, no switches hve een flipped, no lightening olts hve hit the circuit, nd so on. In D Stedy Stte conditions, we cn tret inductors s short circuits nd cpcitors s open circuits. However, when switch hs just een flipped from on to off or off to on, we cn no longer mke these ssumptions out inductors nd cpcitors. The timedependent or D Trnsient response of circuits must then e clculted using known ehviors of inductors ( = L di/dt) nd cpcitors (i = dv/dt) nd the pproprite mthemtics/ clculus Ω 50Ω 20mH 1mF c

3 Our discussion of trnsient response is divided into severl sections: 2. Those circuits tht contin only resistors nd inductors (L circuits) 3. Those circuits tht contin resistors, cpcitors, nd inductors (L circuits) 20 75Ω 50Ω 20mH 1mF c nd 1. Those circuits responding to power removed from the circuit (Nturl esponse) 2. Those circuits responding to power delivered/pplied to the circuit (Step esponse) The circuit ove is n L circuit nd Nturl esponse

4 1 This circuit is n circuit which goes into its Nturl esponse t t > 0

5 1 ight efore () (t= 0) = 0 c (t= 0)=

6 1 ight fter () (t= 0) = (ecuse is in prllel with c ) c (t= 0)= (ecuse cpcitor voltge cnnot chnge instntneously)

7 1 For t > 0 in generl: KL t node gives: I I I I = 0 d /dt / = 0 Solving: d /dt = / (1/ ) d = 1/ dt Tking the integrl of oth sides gives: ln = t/ Evluting from c () to c Evluting from t =0 to t = t gives: ln ( ) ln (c ()) = t/ ln ( /(c ()) = t/ e t/ = c / c = e t/

8 1 I I The voltge c is: c = e t/ Some things to note: The time constnt τ of this circuit is, mening tht the voltge reches 36.8% of its initil vlue fter τ seconds, or the voltge decys y 63.2% fter τ seconds. The voltge egins t the initil vlue of the cpcitor voltge nd then decys exponentilly to zero (in the Nturl esponse)

9 I I 1 This circuit is n circuit which goes into its Step esponse t t > 0

10 1 ight efore () (t= 0) = 0 c (t= 0)= 0

11 1 ight fter () (t= 0) = (ecuse is 0) c (t= 0)= 0 (ecuse cpcitor voltge cnnot chnge instntneously)

12 For t > 0 in generl: KL t node gives: I I 1 I = I d /dt = ( )/ Solving: d /dt = ( )/ (1/( )) d = 1/ dt Tking the integrl of oth sides gives: ln ( ) = (t/) Evluting from c () to c Evluting from t =0 to t = t gives: ln ( ) ln ()= t/ e t/ = ( ) / e t/ = ( ) c = e t/

13 1 I I The voltge c is: c = e t/ Some things to note: The time constnt τ of this circuit is, mening tht the voltge reches 63.2% of its finl vlue fter τ seconds. The cpcitor voltge egins t 0 nd increses exponentilly to its finl vlue of. The finl vlue of cn e found y modeling the cpcitor s n open circuit (Stedy Stte D conditions) nd finding the resulting voltge cross it.

14 I 1 The voltge c is: c = e t/ Additionl prmeters cn e clculted from c : The current i s function of time cn e found s: i (t) = i (t) = d c /dt And the voltge cross the resistor is: c (t)

15 I 1 The voltge c is: c = e t/ A generl solution for ll circuits (Nturl, Step) is: Unknown vrile s function of time = The finl vrile s function of time (Initil Finl vlue of vrile) exp [ (t t switching )/time constnt τ)

16 400kΩ 20Ω 90 c 0.5µF 60Ω 40 Looking for the voltge cross the cpcitor: The initil vlue of the cpcitor voltge is 40 * 60/(2060) = 30 The finl vlue of the cpcitor voltge is 90 The time constnt is = (400,000)*0.5µF = 0.2 seconds A generl solution for ll circuits (Nturl, Step) is: Unknown vrile s function of time = The finl vrile s function of time (Initil Finl vlue of vrile) exp [ (t t switching )/time constnt τ)

17 400kΩ 20Ω µF 60Ω 40 The cpcitor voltge is found y pplying the generl solution: 90 (3090) e t/ e t/0.2 A generl solution for ll circuits (Nturl, Step) is: Unknown vrile s function of time = The finl vrile s function of time (Initil Finl vlue of vrile) exp [ (t t switching )/time constnt τ)

18 Our discussion of trnsient response hs een divided into severl sections: 2. Those circuits responding to power removed from the circuit (Nturl esponse) 3. Those circuits responding to power delivered/pplied to the circuit (Step esponse) 1 We hve found generl solution to finding the cpcitor voltge in circuits contining resistors nd cpcitors, whether the cpcitor is experiencing nturl or step response. Once the cpcitor voltge is known, it cn e used to find/solve for ny other current or voltge in the circuit. The sme pproch cn e used in L circuits to find the inductor current.