Exact Analysis of a Common-Source MOSFET Amplifier

Transcription

1 Exact Analysis of a Common-Source MOSFET Amplifier Consider the common-source MOSFET amplifier driven from signal source v s with Thévenin equivalent resistance R S and a load consisting of a parallel resistor R L and capacitor C L as shown below: V DD R L v S R S i S M C L v out Calculate the voltage gain, A V (= v out /v s ), for this amplifier.

2 We start with the small-signal equivalent circuit: R S C gd C gs v S i S v gs C L g m v gs r O R L v out R L,eq = r O R L We can combine the parallel combination of R L and r o as indicated. The branch currents are indicated in the figure on the next slide. 2

3 sc gd (v gs v out ) R S i S C gd v S i S C gs v gs sc gs v gs g m v gs v out r O R L sc L v out r O R L C L v out Equating the current through C gd to the three shunt currents at the output: Now solving for v gs : v sc v v g v sc v out gd gs out m gs L out ro RL v gs At the input loop: v s( CL Cgd ) ( ro RL ) out gm( ro RL ) Cgd s gm v i R v S S S gs 3

4 sc gd (v gs v out ) R S i S C gd v S i S C gs v gs sc gs v gs g m v gs v out r O R L sc L v out r O R L C L v out Using KCL at the gate node: i sc v sc ( v v ) S gs gs gd gs out and then we can write v v s( C C ) R sc R v S gs gs gd S gd S out After a lot of algebra we can write the final expression (on next page). 4

5 Cgd gm( ro RL ) s g vout m v S s CgsCgd( gm( ro RL )) RS ( Cgd CL ) ( ro RL ) 2 s ( Cgd CL ) Cgs CLC gd RS( ro RL ) Now look at the denominator D(s) We can write in the form, s s D s s s 2 ( ) p p2 p p2 p p2 So we have two poles which are written on the next page: p and p2. 5

6 p Cgs Cgd gm ro RL RS Cgd CL ro RL Miller capacitance term ( ( )) ( )( ) p2 Cgs Cgd( gm( ro RL )) RS ( Cgd CL )( ro RL ) ( Cgd CL ) Cgs CgdC L RS( ro RL ) Often we have a dominant pole meaning the other pole is 4 times greater in magnitude than the dominant pole p. Returning to the definition of D(s): s s D s s s 2 ( ) p p2 p p2 p p2 D s s s 2 ( ) where p is the dominant pole. p p p2 6

7 2 s s Ds ( ), where p is the dominant pole. p p p2 j p2 p Because p p2 7

8 DIGRESSION: If you have a quadratic polynomial of the form: as 2 bs c = 0 Then the solution can be found from s,2 2 b b 2ac 2a You may have two real roots, or you may have a complex conjugate pair of roots of the form s,2 = j j H Inductance involved f 8

9 9 Miller s Theorem (from Razavi) If A v is the voltage gain from node to 2, then a floating impedance Z F can be converted to two grounded impedances Z and Z 2 : v F F F A Z V V V Z Z Z V Z V V F F F v V V V V Z Z Z Z Z V V A

10 Miller s Theorem (continued) Applying Miller s theorem, we can convert a floating capacitance between the input and output nodes of an amplifier into two grounded capacitances. The capacitance at the input node is larger than the original floating capacitance. ZF jcf Z2 j C A F v A v Av Z ZF jcf A A j A C v v v F 0

11 0 C C gs gd 0 0 Application of Miller s Theorem - I Miller s Theorem A Capacitor C in is given by g R V m D C C ( ( g R )) C ( g R ) in F m D F m D Capacitor C out is given by Cout CF CF gmrd gmrd

12 0 C C gs gd 0 0 Application of Miller s Theorem - II Miller s Theorem A g R V m D Pole at Input: in R g G m R D C F Pole at Output: out R D g m R D C F 2

13 Back to our common-source MOSFET amplifier R S C gd C gs v S i S v gs C L g m v gs r O R L v out R L,eq = r O R L What is the midband voltage gain A V? Set all capacitors to open circuits for midband voltage gain calculation. R S v S v gs g m v gs r O R L v out R L,eq = r O R L A g ( r R ) V m O L 3

14 Applying Miller s Theorem to CS MOSFET Amplifier R S v S i S C in v gs g m v gs r O R L C out v out C C C g ( r R in gs gd m O L Cout CL Cgd g ( r R m O L Now we have only an input loop and an output loop to deal with. This means we will have an input pole and an output pole to contend with. 4

15 Applying Miller s Theorem to CS MOSFET Amplifier If the voltage gain is large, the input pole frequency will be lower than the output pole frequency. The input pole is then the dominant pole in this amplifier. vout g r m O RL v sr C s( r R ) C S S in O L out v gm( ro RL ) out v ( sr C )( s( r R ) C ) S S in O L out Input pole Output pole The input pole is given by s in C C g ( r R R gs gd m O L S 5

16 Comparing dominant pole terms from exact analysis to that from Miller s Theorem From our exact analysis : p Cgs Cgd gm ro RL RS Cgd CL ro RL ( ( )) ( )( ) From our analysis using Miller s theorem: s in p Cgs Cgd gm( ro RL RS 6

17 Putting some numbers to our CS MOSFET Amplifier C gs = 25 ff; C gd = 5 ff; C L = 2 ff g m = 2 ma/v (mmho) R S = 5,000 ; r O = 20,000 ; R L = 5,000 r O R L = 4,000 and A V = - g m (r O R L ) = - 8 [ g m (r O R L ) ] = 9 in Cgs Cgd gm ro RL RS in fin 454 MHz 2 ( 25 ff 5 ff( 0.002(4, 000))]5, 000 7

18 Putting some numbers to our CS MOSFET Amplifier For the output pole we have f out 2 CL Cgd ( ro RL ) gm( ro RL 2,266 MHz Clearly the input pole is dominant at 454 MHz. 8

19 f p Dominant pole from exact analysis 2 Cgs Cgd( gm( ro RL )) RS ( Cgd CL )( ro RL ) Compared to the Miller capacitance. Additional term f p 2 C C g ( r R R gs gd m O L S The added term lowers the break frequency from 454 MHz down to 38 MHz. 9

20 NEXT: Open-Circuit Time Constant Analysis H() s K 2 m as a2s ams 2 n bs b2s bns When the poles and zeros are easily found, then it is relatively easy to determine the dominant pole. But often it is not easy to determine the dominant pole. The coefficient b is especially important because b p p2 p3 pn How do we determine the pi values? We will consider all capacitors in the overall circuit one at a time. 20

21 Open-Circuit Time Constant Analysis We consider each capacitor in the overall circuit one at a time by setting every other small capacitor to an open circuit and letting independent voltage sources be short circuits. The value of b is computed by summing the individual time constants, called open-circuit time constants. and b H n RioC i n b RC i i io Time constant = RC i 2

22 Open-Circuit Time Constant Rules For each small capacitor in the circuit: Open-circuit all other small capacitors Short circuit all big capacitors Turn off all independent sources Replace cap under question with current or voltage source Find equivalent input impedance seen by capacitor Form RC time constant This procedure is best illustrated with an example 22