ECE 202, Spring 18

Lecture 12: Switched Capacitor Networks 1. Introduction (a) What are switched capacitor circuits/networks?

ANSWER: Resistorless and inductorless—only capacitors, switches, sources, and op amps. (b) Can they be ALL that other RLC circuits can be?

ANSWER: Only approximately. (c) Yeah right. How does a capacitor approximate a resistor?

ANSWER: A capacitor and a switch or two.

(d) Same Question. How does a capacitor which does not dissipate any energy approximate a resistor? Did you forget about those little properties Professor Ray?

ANSWER: NOT. Indeed, that is precisely the reason for approximating a resistor with a capacitor and a switch—very little energy dissipation. Some researchers found that resistor behavior could be approximated by two MOS switches and a capacitor. AND resistors burn energy released as heat, while capacitors only store and transfer energy (ideally). What a great way to reduce heat and power losses and still have a “pretend” resistor circuit! WOW, this is Sofa King (a Saturday Night Live Trademark) interesting.

(e) Light switches? Ha ha. ANSWER: No, electronic switches. We build clocks that use switches

whose on-off behavior is faster than a college administrator running from (or covering up) a difficult situation.

(g) Can you prove all this for us prof ray?

ANSWER: Actually, I can’t and don’t have the time to do that. The proofs (in papers that I have not looked at in over 30 years) use discrete time state space models. But we will do examples, using either Laplace transform ideas and/or conservation of charge. Yawl get a charge out of that. ☺

ECE 202, Spring 18

(h) So are you telling us that switched capacitor stuff is simple?

ANSWER: “It’s complicated,” like in the Count of Monte Christo movie. (i) What is so complicated?

ANSWER: What you learned in 201 complicates the whole lesson. No doubt some extinguished professor told you about the continuity of the voltage across a capacitor. And way down at the nano-level, that is true as Professor Furgason used to reminded me again and again and again. But I can’t SEE MUCH down there at the nano-level, and things happen Sofa King fast that it is not worth figuring out for a circuits course. So I just say, capacitors want to “jump for your love” or capacitors “just want to have jumps.”

(j) CAN capacitors really have instantaneous jumps in voltage levels? ANSWER: Approximately, for all practical purposes. If a capacitor is

connected to a voltage source, by KVL, it takes on that voltage. If two capacitors are connected to each other in parallel, then, the voltage across both will change “instantaneously” in accordance with conservation of charge!!!! Nano, no more nano, no nano foorrrr me.

ECE 202, Spring 18

2. Examples Example 1: A simple switched capacitor circuit. Circuit operation: C1 = 10 F, C2 = 40 , vin(t) = 20u(t +10) V. (i) Switch in position A t < 0 . (ii) Switch moves to B at t = 0 . (iii) Switch moves to A at t = 1 s. (iv) Switch moves back to B at t = 2 s, where it remains forever and ever.

Step 1. Compute vC1(0− ) . By KVL, vC1(0− ) = 20 V. Step 2. Label s-domain equivalent for 0 ≤ t < 1. Find Vc1(s) , vc1(t) , and

vc2(t) .

(a) C1vC1(0− ) = 200

ECE 202, Spring 18

(b) By Ohm’s law,

VC1(s) =VC2(s) = C1vC1(0− )× 1

C1s+C2s=

C1C1 +C2( )s vC1(0− ) = 200

50s= 4

s

(c) Inverse Transform:

vC1(t) = vC2(t) = 4 V for 0 ≤ t <1 s. Step 3. What happens for 1≤ t < 2 s? (a) 4 = vC1(1− ) ≠ vC1(1+ ) = 20 = vC1(t) for 1≤ t < 2 s. (b) vC2(t) = 4 for 1≤ t < 2 s. Step 4: Label s-domain equiv. for 2 ≤ t . Find Vc1(s) , Vc2(s) , vc1(t) , and

vc2(t) .

(a)

VC1 =VC2 =

1(C1 +C2)s

C1vC1(2− )+C2vC2(2− )⎡⎣

⎤⎦e−2s

ECE 202, Spring 18

= e−2s

50s200+160⎡⎣ ⎤⎦ =

7.2s

e−2s

(b) Take inverse transform. Thus for 2 ≤ t

vC1(t) = vC2(t) = 7.2u(t − 2) V

ECE 202, Spring 18

EXAMPLE 2. Switched Op Amp Worksheet OBJECTIVE. Imitate an op amp integrator behavior by a switched capacitor circuit.

Preliminaries: R = 2 Ω and C = 2 F. Thus H (s) =

Vout (s)Vin(s)

= −0.52s

.

Hence, if vin(t) = 2u(t) V, then Vin(s) = 2

s and

Vout (s) = −0.5

s2 in which case

vout (t) = −0.5tu(t)V.

ECE 202, Spring 18

Switched Op Amp Circuit. Find vout (t) for 0 ≤ t ≤ 4 for the RELAXED

circuit below. Relaxed means zero initial conditions. Switching Action: (i) position A at t = 0 and moves to position B at t = 1. (ii) Returns to A at t = 2 (iii) Subsequently moves to B at all odd integer valued time instants and back to A at all even integer time instants.

Step 1. Find vc1(t) for n ≤ t < n+1, n = 0, 2, 4, 6, …., i.e., when switch is in position A. The answer is obvious, says Mr. Obvious!

vC1(t) = 2 V for n ≤ t < n+1, n = 0, 2, 4, 6, …. Step 2. Find vout (t) for 0 ≤ t < 1. Mr Obvious says: vout (t) = 0 for 0 ≤ t < 1. Step 3: Find Vout (s) and vout (t) for 1 ≤ t < 2. Label the s-domain equivalent circuit below.

ECE 202, Spring 18

H (s) =

Vout (s)Vin(s)

= −Yin(s)Yf (s)

which implies that Vout (s) = − Cs

2Cs× 2

s= −1

s

Therefore vout (t) = −1 V for 1 ≤ t < 2. Step 4: From step 1, vout (t) = −1 V for 2 ≤ t < 3 since there is no resistance to sap energy. Step 5: Using the result of step 4, label the equivalent s-domain circuit below valid for 3 ≤ t < 4.

Strategy: Superposition to obtain Vout (s) .

(a) Contribution to Vout (s) due to the source on input branch is: −1s

(b) Contribution to Vout (s) due to feedback branch source is: −1s

(c) By superposition, Vout (s) = −2

s

ECE 202, Spring 18

Therefore vout (t) = −2 V for 3 ≤ t < 4. Step 6: True-False:

vout (t) =

0 0 ≤ t <1−0.5(n+1) n ≤ t < n+ 2, n = odd #

⎧⎨⎪

⎩⎪

"Odd number" means positive odd integer. Plot vout(t).

vc1(t)