Example: High-frequency response of a follower

Transcription

1 Example: Hih-requency response o a ollower o When body eects are cluded, db actually appears between dra and round. ce both termals o db are rounded, it does not aect the circuit. o d is also between the dra and the round and is parallel to. o s is between put (ate and output (source.

2 Example: Hih-requency response o a ollower Use Miller s Theorem to replace capacitor between put & output ( s with two capacitors at the put and output. A = s, i v v s A = + = m( ro = + ( r s m ( m ( r o o A = + m ( r o s * Both s,i and s,o are small as A = d + s, i = = + ( + db s,o

3 Example: Hih-requency response o a ollower p = 2π [ si G ]. onsider : 2. Fd resistance between apacitor termals

4 Example: Hih-requency response o a ollower p2 = 2 π [ (/ m ]. onsider : 2. Fd resistance between apacitor termals

5 Hih-requency response o a ollower A M / / / p p2 H = + [ = = d = / G ( + p G + si ( s, o / si m p2 m( ro + ( r + ( r o db G, (/ s m o m ] Note that both poles are at very hih requencies as both ( / p and / m ( / p2 are small.

6 esponse o an active-loaded ampliier Miller s Between put & round Between output & round = + s d, i = d, o d 2 db db2

7 esponse o an active-loaded ampliier r o v d A = = m( ro ro m 2 v d, i = d( A = d( + m md / p si / p ( ro ro 2 2 d, o d ( / d( / m d = = = + s d, i A = + d, o d 2 db db2

8 Hih-requency response o multi-stae ampliiers Note: some capacitors may not be present. ompute mid-band a o each stae o overall mid-band a (you need i and i2 2. Use Miller s Theorem to replace Miller s capacitors. 3. ombe all parallel capacitors. You will end up with capacitors at put, output, and between staes. 4. ompute circuit time constants rom: / / / p p2 p3 ( ( ( si o i2 i

9 Hih-requency response o ascode ampliiers

10 Hih-requency response o ascode ampliiers Between output & round = + + d 2 db2 Miller s Between output & round = + + d, o s2 db Between put & round = + s d, i

11 i2 r o Hih-requency response o ascode ampliiers A A v2 i2 v = = r m2 o ( r ro 2 + = + r = = + r m m2 ( r o + A m2 i2 ( d, i d v = r o m2 ( / A d, o d v r + r m2 = Miller s apacitors Gas and resistances ombe parallel capacitors = + + d 2 db2 = + + d, o s2 db = s + d, i ircuit time constants / / / p p2 p3 ( ( si o i2

12 Domant Pole ompensation. Very oten we need to purposely troduce an additional pole the circuit order to stabilize it (e.., provid phase mar at zeros. This is called the domant pole compensation. 2. This pole has to be the domant pole (several octave below any zero or pole. 3. As such, we can ore transistor ternal capacitances the analysis as poles troduced by these capacitances would be at hiher requencies and does not impact the domant pole. 4. Domant pole is troduce by the addition o a capacitor either between output & round or between put & output o one stae.

13 Domant Pole ompensation by Miller s eect / A p M si. Because o the Miller s eect, M appears as a much larer capacitor, A M and is desirable or an I. 2. Unortunately, p is proportional to si 3. For the irst-stae, si is typically small, so this is NOT the preerred method or domant pole compensation or the irst stae 4. For second or latter staes, si is o o previous stae and can be lare. As such, this is the preerred method or clud a capacitor side the chip.

14 Domant Pole ompensation by / p ( o. Because the required is lare, is always located outside the chip (i.e., between output termal and round.