ECE53A Introduction to Analog and Digital Circuits Lecture Notes Second-Order Analog Circuit and System Fall 2007 ECE 53A Order of an analog circuit or system depends on the number of energy storing elements Ls and Cs. Simplest second-order circuits are ones containing two Cs plus resistors and their dual networks containing two Ls; e.g. R1 R2 L1 + + C1 vg (t ) C2 - RL R1 vg (t ) R2 RL - Figure 1(a) L2 Figure 1(b) Simplest RLC-circuit is the series RLC-circuit shown in Figure 2(a) and its dual, the parallel (or shunt) RLC network.

V(t) R L + + v g (t ) - i(t) Figure 2(a) Single loop RLC-circuit C R ig (t ) L Figure 2(b) Single node-pair RLC-circuit C ECE 53A Series-RLC Circuit R iL(t) Fall 2007 L + v g (t ) + i(t) C Vc(t) - - Figure 1 Single-loop series RLC circuit First of all, even before we write the network equations to solve for the single unknown which is the loop current i(t), we need to satisfy two initial condition s (I.C) Vc(0) and iL(0). KVL (loop) equation: di

vc (t ) vg (t ) dt 1 t vc (t ) ic ( )d C Ri L Differentiate Eq.(1) using (2), or di (t ) d 2i (t ) i (t ) R L vg (t ) dt dt 2 C d 2i (t ) R di (t ) 1 1 i ( t ) v g (t ) dt 2 L dt LC L (3) (1) [Note iL(t)=i(t)] (2) [Note ic(t)=i(t)] Fall 2007 ECE 53A Eq.(3) is in the form of a generic second-order (n=2) ORDINARY differential equation w ith constant coefficients, d 2 x(t ) dx(t )

a a0 x(t ) f (t ) 1 dt 2 dt Where x(t ) can be a current or voltage and f (t ) is the input or forcing function. The circ uit we are considering is linear and lumped containing two(2) energy storage circuit ele ments (In our case, the circuit contains a inductor and a capacitor). For the series RLC circuit, to solve for i (t ), t 0 , we need two(2) I.C, vc(0) and iL(0). L et i (t ) ke,stsubstituting this into the homogeneous equation d 2i (t ) R di (t ) 1 i (t ) 0 dt 2 L dt LC (5) We obtain the quadratic equation s2 (4) R 1 s 0 L LC (6) This is known as the characteristic equation. For the second-order circuits, the generic characteristic equation can be written as s 2 2n s n 0 (7) Fall 2007 ECE 53A

Where Damping factor n Undamped natural or undamped resonant frequency in rad/sec Comparing Eq.(6) and (7), we have R R LC C R C 2n or 2 R or L L L 2 L (8) 1 1 or n (9) LC LC (No damping) and the characteristic eq. becomes n 2 If R 0, 0 Let be s 1 s 2 n 2 s 2 , then for LC s j ,

R 0( 0) 1 1 2 or in rad/secnatural or the un-damped resonant frequency of This justifies the nnaming of nas the un-damped LC LC the circuit. n characteristic equation are The roots of the generic quadratic s 1,2 2n 4 2n 2 4n 2 2 n 2 1 (10) Fall 2007 ECE 53A If can be shown that does not affect the behavior of the response of the circuit by using frequency, thus, the behavior of the circuit response depends solely on the damping factor . Without lose of generality, let us assume n 1rad/sec . Thus, Eq.(10) reduces to s 1,2 2 1 (11) Case 1: If 1, s1 , s2 ,(roots are real and equal. Double order) Critical damping 2 Case 2: If 1, s1 , s2 1, (both roots are real and unequal) Un-damped

2 Case 3: If 1, s1 , s2 j 1 , (roots are complex and complex conjugal) Overdamped j j Case 1 j Case 2 Case 3 Fall 2007 ECE 53A Standard method is used to solve for the complete solutions of the ordinary second-order differential equation as x(t ) xh (t ) x p (t ) (12) where xh (t ) solution of the homogeneous equation x p (t ) partial solution due to the forcing function xh (t ) k1es1t k2 es2t (13) ECE 53A Parallel-RLC Circuit Fall 2007 + ig (t ) R L

iL(t) C Vc(t) - Figure 2. Single node-pair parallel RLC circuit KCL (Nodal) Equation: Since v(t ) vc (t ) , we have v(t ) dv (t ) iL (t ) C ig (t ) R dt t 1 iL (t ) v( )d L or d 2v(t ) 1 dv(t ) 1 1 dig (t ) v ( t ) dt 2 RC dt LC C dt Note this equation is identical to (3) if In fact the parallel circuit in Fig. 2 is the dual network for the series circuit in Fig. 1. v(t ) i (t ) 1 R G R CL vg ig (14) ECE 53A Parallel-RLC Circuit i1 (t ) ig (t )

R1 Fall 2007 i2 (t ) 0 R2 v v1 (t ) C1 - v vc1 + + C2 v2 (t ) v2 vc2 - Figure 3. Second-order passive RC-circuits Write KCL (Nodal) Equation by inspection, we have v(t ) vg (t ) R1 C1 dvc v(t ) vc2 (t ) ig (t ) dt R2 (15) Under-damped values of Rs and Cs Fall 2007 ECE 53A N=4 R1

vc2 (t ) i1 + + L1 C1 - C2 vc1 (t ) - iL RL Second-Order RC-Network ECE 53A Fall 2007 (or its dual: RL-Network) v1 R3 + R1 vc1 (t ) v2 + C1 vc2 (t ) - C2 R2 - Figure 1. Second-order RC-network By inspection, nodal (or KVL) equations are given by Node v1

C1 dv1 v1 v1 v2 0 dt R1 R3 (1) Node v1 C2 dv2 v2 v2 v1 0 dt R2 R3 (2) 1 1 R R C1s 3 1 1 R3 1 R3 v (t ) 0 1 1 1 0 v (t ) C2 s 2 R2 R3 Eq.(3) is in the form of the (3) nodal admittance matrix and can be obtained from the circuit in Figure 1. Fall 2007

ECE 53A To treat Equation (1) and (2) (or(3)) simultaneously, we first use (1) to determine v2 in terms of v1 , and then substitute the result into (2) to obtain a second-order differential equation in v1 . We have dv R v2 R3C1 1 1 3 v1 (4) dt R1 d 2 v1 1 1 1 1 dv1 dt 2 R1C1 R2C2 R3C1 R3C2 dt 1 1 1 v1 0 R R C C R R C C R R C C 2 3 1 2 2 3 1 2 1 2 1 2 (5) Note that Eq.(5) is of the generic form d 2 v1 dv1 2

n v1 0 n dt 2 dt 1 1 1 1 2n R1C1 R2C2 R3C1 R3C2 1 1 1 n 2 R1 R2C1C2 R1 R3C1C2 R2 R3C1C2 (6) (7) (8) Fall 2007 ECE 53A Therefore 1 1 1 1 1 Damping Factor 2 R1C1 R2C2 R3C1 R3C2 n Undamped Resonant Frequency in rad/sec 1 1 1

R R C C R R C C R R C C 1 3 1 2 2 3 1 2 1 2 1 2 1 (9) 2 (10) Also, note that Eq.(1) and (2) are already in the form of state-variable formation. Thus, rewriting (1) and (2) as 1 1 v 1 dv1 1 (11) v2 dt R R C R C 3 1 1 3 1 dv2 1 v1 dt R3C2 1 1 v2 R2 R3 C2 (12)

Since x1 v1 vc and x2 v2 vc , x Ax, 1 where A 2 1 1 R C R C 3 1 1 1 1 R3C1 1 R3C1 1 1 R C 2 2 R3C2 (13) Fall 2007 ECE 53A For the special case of RC-network, the nodal equations are indeed the state equations using state variables xk vck . The roots of the characteristic equation are constrained to be only the negative -axis as shown below

j 2 1 v1 (t ) A1e 1t A2e 2t Where 1 0, 2 0 and the constant A1 and A2 are determined by the I.C. x1 (0) vc1 (0) and x1 (0) vc1 (0). ECE 53A High-Order (n=4) RLC-Network N=4 (2Ls and 2Cs) + vg vc2 (t ) i1 Rg + - + L1 (1) Fall 2007 C1 - C2 vc1 (t ) iL L2 (2) RL - Can use Loop (KVL) analysis (3 loops) or use Nodal (KCL) analysis (2 node-pairs) Better get use state-variable analysis x x1 x2 x3

Let x1 vc1 x2 vc 2 x3 iL1 x4 T Set up equations is straight forward use a mixture of KCL equations (tree branch) and KVL equations (Cuts). Solve by computing state and output equations. x4 iL 2 to obtain 4 coupled 1st-rder differential equations. Fall 2007 ECE 53A First obtain directed graph: Rg L1 C2 L2 Vg1 C1 RL Tree: Connect all nodes No closed path Tree branches: vg1 , C1 , C2 and Rg 4 KCL equations Loops: L1 , L2 and RL 3 KVL equations Use a selected mixture of KCL and KVL equations No integration only 1st-order derivatives. To obtain 4 coupled 1st-order differential equations which we can put into a compact matrix from I.C. x(0) vc1 (0) vc 2 (0) iL1 (0) iL 2 (0) Question: what kind of network is this? Look at behavior of element (Ls and Cs) at s 0 (DC) and s ECE 53A Example E-1 Fall 2007

K R iL + V0 C Vc i (t ) iL (t ) L i(t) - Figure E-1. RC-network used to illustrate the finding the particular solution from the general solution Let the capacitor C be charged to a voltage V0 and at t=0 let the switch be closed. The value of the resistance R will determine whether the system is (a) over-damped (b) critically damped, (c) under-damped Figure E2-2. Network response for the three cases: (a) over-damped (b) critically damped and under-damped Fall 2007 ECE 53A General solutions in terms of and n 1 or Q 1 2 i (t ) e nt K1ent 1 1 or Q 2 i (t ) K1 K 2 e nt 1 or Q

2 1 K1e nt 21 (1) (2) 1 2 i (t ) e nt K1e jnt 1 2 K1e jnt 1 2 (3) In Eq.(1-3), K1 and K2 are arbitrary constant of integration. Eq.(3) can be rewritten equivalently as i (t ) Ke nt sin nt 1 2 K K12 K 22 K tan 1 1 K2 ECE 53A Two Cascaded Inverters RL RL Vout1 Vin1

Fall 2007 Vout2 Vin2 Cascaded inverters with parasitic wiring inductance and gate capacitance RL Vin1 RL Vout1 Vin2 L1 Cgs Vout2 ECE 53A Circuit model of the cascaded inverters when the input at Vin1 is low RL RL Vout1 Vin1 RON Series-RLC circuit: Ron L1 Cgs2 L1 Vin2 Cgs2 RON Vout2 Fall 2007 ECE 53A Fall 2007 DUALITY and DUAL NETWORKS R and G 1

Dual Quantities: R L and C Loop current iR and node-pair voltage vR Loop and node pair Short circuit and open circuit Example 1: KCL and KVL + R Vs L i(t) + Vs Dual is R L C Network: - Example 2: C V(t) Loop current i (t ) node pair v(t ) R L 1 C 2 -

1 Dual is R C 3 3 L 2 ECE 53A State-variable method I. First-order circuits and systems x(t ) ax(t ) bu (t ) y (t ) cx(t ) du (t ) For LTI systems, a,b,c,d are constants. II. Second-order circuits and systems x(t ) Ax(t ) Bu (t ) y (t ) Cx(t ) Du (t ) For continuous-time LTI systems, A,B,C and D are constant matrices. For LTI analog lumped circuits containing R, L, C and transformers. A 2 2 constant matrix, known as state matrix T x(t ) x1 x2 state vector; x1 and x2 are state variables vc (t ) and/or iL (t ) u (t ) Input vector y (t ) output vector I.C needed to solve the state equations: x(0)