(in Hz) at this bias condition. (in Hz) if the bias current is doubled. C in. 50k

Transcription

1 Proble bipolar transistor is biased so that I 05, and at this bias and pf 2 F 75 0 a) Fd the frequency The transistor s low-frequency alue of is 00 It is also gien that ( Hz) at this bias condition 5pF b) Expla the eang of general For ore space, use the back of this page c) Fd the frequency ( Hz) if the bias current is doubled Proble 2 onsider the circuit of Figure P-2 a) Fd the capacitance lookg to the put of the circuit seen b) Estiate the 3dB frequency of s S 50k 50k 20pF 20 Figure P- 2 Proble 3 The followg questions ay be answered fro eorized characteristics of a first-order low-pass response n aplifier is described as hag a D ga of 0 and a sgle-pole high frequency roll-off The upper 3dB frequency is khz a) What is the approxiate agnitude of the aplifier's ga R at 00Hz b) What is the approxiate agnitude of the aplifiers ga at 0KHz BIS Q2 c) What is the approxiate phase of the aplifier's ga at R S 00kHz Proble For the circuit of Figure P-a, assue that I It is gien that RS 0k, and R k lso assue that r 0k For both transistors, pf and S Figure P-a Q

2 5pF BIS is a ery low-ipedance bias oltage that keeps all transistors forward actie a) For Figure P-a, what is the oltage ga R b) For Figure P-a, what is the resistance to ground at the node R S c) For Figure P-a, what is the effectie capacitance to ground at the base of Q S Q d) Expla words how the transistor Q2 helps the circuit of Figure P-a to hae a higher bandwidth copared to the circuit of Figure P-b (Figure P-b is the sae circuit at the sae bias current, but with Q2 reoed) Figure P-b Proble 5 For the circuit of Figure P-, assue that the transistor M is biased so that it has , pf, 02pF It is also gien that R 00k X a) What is the tie constant at the gate of M b) What is the resultg 3dB frequency of gd c) If you could reduce the alue of either capacitor or capacitor gd by 0%, which would you reduce order to ake the biggest iproeent bandwidth State your reasong s s R X g G 2k Figure P-5 DD D M S Proble 6 onsider the circuit of Figure P-6 Gien: R 50k R 2 0k 000pF a) There is one tie constant for this network What is it b) What is the 3dB frequency of s c) What is the agnitude of s d) What is the agnitude of s Hz at low frequencies at ten ties the 3dB frequency of s s R R 2 Figure P-6

3 Solution a) f T 3 g MHz b) f is the frequency at T 2 2 6pF which the agnitude of the short-circuit current ga collector current doubles, g, where je F je je fro the gien conditions, then double only new alue of i i c b is equal to one c) Note that when doubles But the alue of also changes! Recall that is essentially constant with bias current Hence, we should fd g 2 3 5pF je F (because doubles) and calculate the Fd that g 5pF and hence 35pF Hence the alue of je at I is 65pF and MHz 2 75pF Solution 2 a) This proble is a straightforward application of the Miller Effect The capacitance lookg is gien by put, where is the ga ties the capacitance between the put and the If you cannot see how to detere the oltage ga fro the ( 20) 20pF 20pF b) Recognize that the tie constant at gien circuit, please ask! Note that by spection, 20, so F is needed The capacitance to ground at that node has already been found It reas to fd the resistance to ground We see 50k 50k 25k Hence, 25k 20pF=05μs and 56kHz 2 Solution 3 Figure S- shows a Bode plot with the axes labeled accordg to the gien foration for this proble a) The frequency 00Hz is well below the upper -3dB frequency of khz Therefore, the ga at 00Hz will be the 0 20dB sae as the D ga, which is gien as 0dB 20dB 0 The low-frequency ga is the sae as the D ga for a low-pass filter, because the ga does not change significantly with frequency for frequencies ore than one decade below the -3dB frequency b) 0kHz is 0 Hz, and this frequency is one decade aboe the -3dB frequency, so the agnitude response is down by 20dB Note that the agnitude response 0 20dB dB 0dB Lowfrequency ga 3dB 2 0 Figure S-3 Lowfrequency ga 3dB One decade One decade 20dB dB

4 at 0 Hz is 20dB lower than the straight-le approxiation, which is 20dB 0 at khz Hence, the agnitude of the ga at is c) 00kHz is, which is two decades aboe the -3dB frequency Therefore, the phase shift is the ultiate phase shift due to the sgle pole, which is Solution : a) b) t 90 0 Hz B 0dB 3 3 g R Hz, there are two paths to ground Lookg to the collector of Q, we see a relatiely high resistance Lookg to the eitter of Q2, we see just e2 resistance to ground is 25 c) t r 25 Hence the g, there are two capacitances to ground The Miller put capacitance due to B pf ( ) 2pF In parallel, we hae the capacitance 5pF Hence, the capacitance to ground is 7pF equal to d) For the circuit of Figure P-b, the bandwidth is doated by the tie constant at the base of Q The oltage ga is -0, and therefore 2, is is ultiplied by to contribute pf to the capacitance at the base of Q In the circuit of Figure P-a, the oltage ga is kept to - by the low resistance at the collector of Q due to Q2 This decreases the capacitance at the base of Q So it could be said that the effect of Q2 is to lower the ga of the first stage (Q) order to reduce the Miller effect Solution5 a) To fd the tie constant at the gate, fd the resistance and capacitance to ground Because the resistance lookg to the gate of the MOSFET is fite, the only resistance to ground at the gate is The capacitance to ground is gien by ( ) ( ) First, the oltage ga is zero because there is no gate gd gd R X signal at the collector Next, the oltage ga g gd of the stage fro s g2k It follows that ( ) g 2k 3 Then put capacitance at the gate is gien by (0323) () Fally, gate gd pf(00323) 02pF() 0232pF Hence the tie constant at the gate is 2323ns gate b) the -3dB frequency is 685 MHz

5 c) Note that ( ) ( ) It can be seen that the total capacitance at the gate is uch ore sensitie to 0% reduction gate gd gd gd gd because the Miller effect reduces the effect of would crease the bandwidth by ore than a 0% decrease Hence, a Solution 6 a) The tie constant is gien by cr R2h c50k 0k h 000pF 8 33 sec b) f 3dB 9 khz 2 c) R2 0k 07 d) t ten ties the 3dB frequency of a low-pass filter, the s R R2 60k agnitude response is 20dB fro its low frequency alue 20dB represents diision by a factor of 0, so the answer is 0 07