Lecture 36: MOSFET Common Drain (Source Follower) Amplifier.

Transcription

1 Whites, EE 320 Lecture 36 Pae 1 of 10 Lecture 36: MOSFET Coon Drain (Source Follower) Aplifier. The third, and last, discrete-for MOSFET aplifier we ll consider in this course is the coon drain aplifier. This type of aplifier has the input sinal fed at the ate siilar to the CS aplifier but the sinal output is taken at the source terinal, as shown in Fi. 4.46(a): (Fi. 4.46a) Sall-Sinal Aplifier Characteristics We ll calculate the followin sall-sinal quantities for this MOSFET coon ate aplifier: R in, A, A o, G, G i, A is, and R out. To bein, we construct the sall-sinal equialent circuit: 2009 Keith W. Whites

2 Whites, EE 320 Lecture 36 Pae 2 of 10 D R si i i G s i = si i - - R G s - S 1/ o R in R out R L r o i o (Fi. 4.46b) Because the drain terinal is an AC round, we shifted one end of the output resistance r o so it appears in parallel with R L. This akes the T odel particularly well suited for the CD aplifier since r o appears in series with 1. Input resistance, R in. With si = 0 and i = 0, we can see directly fro this sall-sinal equialent circuit that R = R (4.99),(1) in Partial sall-sinal oltae ains, A and A o. At the output side of the sall-sinal circuit with i = 0 o = s( ro) (2) At the input, usin oltae diision 1 s = i (3) 1 + R r G L o

3 Whites, EE 320 Lecture 36 Pae 3 of 10 Substitutin (3) into (2), ies the partial sall-sinal AC oltae ain to be o ro A = (4.102),(4) i ro + 1 Notice that if r o R L and 1 then A 1 In the case of an open circuit load ( ), the sall-sinal partial oltae ain becoes ro Ao A = (4.103),(5) r + 1 Oerall sall-sinal oltae ain, G. Usin oltae diision at the input to the sall-sinal equialent circuit Rin i = Rin + Rsi si (6) Substitutin this into o i o i G = = si si i si A (7) and usin (1) and (4) ies the oerall sall-sinal oltae ain of this coon drain aplifier to be G o RG ro = (4.104),(8) R si G + R R si L r o + 1 Aain, notice that if r o R L and 1, as well as R R, then G si = A o

4 Whites, EE 320 Lecture 36 Pae 4 of 10 G 1 (9) Consequently, this coon drain aplifier is often called the source follower aplifier. Oerall sall-sinal current ain, G i. Applyin current diision at the output and notin that i = 0 then ro io = s (10) ro + while at the input 1 + ( r i o) ii = = s (11) RG ( 3) RG Substitutin (11) into (10) ies the oerall sall-sinal AC current ain to be G io ro RG i = (12) i r + R 1 + R r ( ) i o L L o With a little anipulation, this can be expressed as ( ro) RG Gi = (13) 1 + ( ro) If r o R L and 1, then RG Gi (14) R L which likely is quite lare.

5 Whites, EE 320 Lecture 36 Pae 5 of 10 Short-circuit sall-sinal current ain, A is. The short circuit sall-sinal AC current ain can be easily deterined fro (12) with R L = 0 as A G = R (15) is i R = 0 G L Output resistance, R out. To deterine R out fro the sallsinal circuit aboe we set si = 0 and apply a fictitious AC oltae source x at the output as shown: s (Fi. 1) Notice that the ate terinal has zero oltae because si = 0 and i si = 0. By definition R out i x (16) x

6 Whites, EE 320 Lecture 36 Pae 6 of 10 We can see that with x attached, the oltae s will not usually be zero. This eans the current in the dependent current source is also not zero. In such instances, we would norally need to analyze this circuit to find the oltae x in ters of i x, and then apply (16) to deterine the output resistance of this aplifier. Howeer, in this case both terinals of the dependent current source are rounded so it akes no contribution to the output resistance. By inspection, the output resistance is siply 1 Rout = ro (17) Suary and Coparison with the CE Aplifier In suary, we find for the CG sall-sinal aplifier: o A non-inertin aplifier. o Potentially ery lare input resistance [see (1)]. o Sall-sinal oltae ain less than one, and potentially close to one [see (8) and (9)]. o Potentially ery lare sall-sinal current ain [see (13) and (14)]. o Relatiely sall output resistance [see (19)].

7 Whites, EE 320 Lecture 36 Pae 7 of 10 Siilar to the BJT coon collector (eitter follower) aplifier we discussed in Lecture 21, the coon drain (source follower) aplifier finds use in applications that require a unityain oltae bufferin function. That is, in applications where a oltae sinal source has sufficient aplitude, for exaple, but it has a lare internal resistance while the sinal needs to be supplied to a load with a uch saller resistance. Other applications of oltae bufferin aplifiers are: The output stae of a ulti-stae aplifier chain to proide a low resistance output. To separate a filter circuit fro a subsequent aplifier circuit that loads the filter with a aryin ipedance load, which will likely adersely affect the filter behaior. Exaple N36.1 (based on text exercise 4.35). Use the circuit of Fi. E4.30 to desin a coon drain aplifier. Assue R si = 1 MΩ, R L = 15 kω, and r o = 150 kω. Coputer R in, A o, A, G, G i, and R out both with and without considerin r o. This is the sae DC biasin circuit we used in Exaple N34.1 for the desin of a coon ate aplifier. Here we re oin to use it as the basis for a coon drain aplifier. The DC analysis results are shown in Fi. E4.30:

8 Whites, EE 320 Lecture 36 Pae 8 of 10 (Fi. E4.30) Usin (4.71) 2I D = = = 1 V OV S Based on this DC biasin, the correspondin coon drain aplifier circuit is: Notice the addition of the bypass capacitor on the drain terinal of the MOSFET. This was added so that R D will affect only the

9 Whites, EE 320 Lecture 36 Pae 9 of 10 DC functionality of the circuit. In the AC operation, the drain terinal will be an AC round, which fits the analysis presented in this lecture. The sall-sinal equialent circuit for this aplifier is then: s Fro (1), R in = R G = 4.7 MΩ (with and without r o ). Fro (17), R out R out 1 = = 1 kω (w/o r o ). 1 = r = = kω (w/ r o ), or 3 o 150k ro 150k V Fro (5), Ao = = = ro k V V or A o = 1 (w/o r o ). V (w/ r o ),

10 Whites, EE 320 Lecture 36 Pae 10 of 10 o 15k 150k V Fro (4), A = r ro + 1 = 15k 150k + = 10 V ro 15k V (w/ r o ), or A = = = (w/o r 3 o ). R r k + 10 V L o Fro (8), RG ro 4.7M 15k 150k G = RG + Rsi ro + 1 = 4.7M + 1M 15k 150k + 10 V RG = (w/ r o ), or G = V RG + Rsi + 1 = 4.7M 15k V M 1M 15k 1 = (w/o r o ) V Fro (13), 3 ( ro ) R 10 ( 15k 150k G ) 4.7M A Gi = = = ( ro) ( 15k 150k) 15k A 3 RG 10 15k 4.7M A (w/ r o ), or Gi = = = k 15k A (w/o r o ). 3