Page1 Name Solutions ES 330 Electronics Homework # 6 Soltuions (Fall 016 ue Wednesday, October 6, 016) Problem 1 (18 points) You are given a common-emitter BJT and a common-source MOSFET (n-channel). Fill in the table below. Assume the BJT to be in the forward active mode and the n- channel MOSFET to be in the saturation region of operation. [The purpose of this exercise is to compare BJT and MOSFET parameters. Remember that A 0 is the voltage gain without a separate load resistance R L.] NPN BJT Cell = 100, A = 100 and TH = kt/q = 5 millivolt N-channel MOSFET Cell n C OX = 00 A/, (W/L) = 40 and A = 10 Bias Current is C = 0.1 ma C = 1 ma = 0.1 ma = 1 ma g m (ma/) 4 40 1.6 4 r 0 (k) 1000 100 100 10 A 0 (/) 4000 4000 16 40 R N (k) 5.5 For the BJT: g ; r ; A g r ; R r C A m 0 O m 0 in TH C gm 1 W For the MOSFET: g C 4 L mi n OX i i i ' L A 40 r0i ; AO gmi r0i ; Rin i i ma ;
Page Problem (0 points) You are given the circuit drawn below. t is fabricated in a CMOS process for which n C OX = p C OX = 00 A/, An = Ap = 0 /m, tn = - tp = 0.5 volt and =.5 volts. The two transitor types have L = 0.5 m and are to be operated at = 0.1 ma and O = 0.3 volt. Find the required gate node voltage G, and the (W/L) ratios for both the n-channel and p-channel MOSFETs to meet the required conditions. Note: 0.1 ma 100 μa is specified as the operating current. 0.5 0.3 0.8 and.5 0.8 1.7 SG tp O G SG 1 W 1 W W 1 1 nc OX O 100 00 (0.3) 11.1 L L L 1 W C L p OX O p n n n 1 W W 100 100 (0.3). L L 1 0.1 A gm1 r01 r0 and gm1 0.67 0.3 r ' A L 0 0.5 r 100 k ; therefore, 0.1 01 0 A 0.67 100 100 33.5 O 1 p ma p G = _1.7_ volts (W/L) n = _11.1_ (W/L) p = _._
Page3 Problem 3 (1 points) The schematic below shows a cascoded n-channel MOSFET pair used to achieve higher output resistance R OUT. We want to achieve R OUT = 00 k using this cascode pair operating at drain current = 0.5 ma. Assuming identical geometrical device layouts so that (W/L) 1 = (W/L) = (W/L), n C OX = 0.1 ma/volt, and A = 10 volts; what is the required gate width-to-length ratio (W/L) for this circuit? R g r r OUT m1 01 0 must equal 00 k is given. A 10 r0 0, 000 ohms (0.0005) We know that r r r 01 0 0 00, 000 g (0, 000) ; hence, g.0005 m1 m1 ma W W Using gm1 nc OX ; 0.0005 (0.0001) 0.0005 L L Solving fo r W/L gives, W L.5
Page4 Problem 4 (0 points) George Wilson (was an C designer at Tektronix who lead a highly productive bipolar C design group that made many contributions to Tektronix oscilloscope products in the 1980 s and 1990 s) proposed an improved current mirror over the basic current mirror using only two BJT transistors. What is now known as the Wilson current mirror adds transistor Q 3 to the basic current mirror as is illustrated in the schematic below. (a) erive an expression for the ratio of ( REF / O ). The above figure shows the current components in all circuit branches assuming the transistor current gains () are identical. Simply sum the currents at each node. Therefore,
Page5 REF O 11 C ( 1) 1 1 ( ) C 1 ( 1) (b) Compare the ( REF / O ) ratio for the Wilson current mirror to the ( REF / O ) ratio of the basic current mirror, that is, a current mirror without transistor Q 3. What improvement in ( REF / O ) does the Wilson current mirror over the basic current mirror? The basic current mirror has a ( REF / O ) ratio of 1 REF Therefore, if = 100, for the basic current mirror cell ( REF / O ) = 1.000, but for the Wilson current mirror ( REF / O ) = 1.000. Another way to look at this is to examine relative to REF which allows us to compute the % deviation from the ideal. Then for the basic current mirror is about % deviation, but it is 0.0% for the Wilson current mirror. Problem 5 (15 points) Starting with the basic current mirror (with transistor Q 1 and Q only), we place a small resistance R E in series with the emitter of transistor Q as shown in the schematic below. For this problem assume you can neglect base currents (that is, assume infinite ), the NPN transistor saturation current S is 1 10-15 A, and both transistors are identical. f resistor R E = 0, then REF = C. But the presence of R E reduces C. Let REF = 1 ma. f C = 0.75 REF, what is the value of R E? O
Page6 We use the relationship: BE sat exp 15 BE 1 At 1 ma we have C 1 0.001 1 10 exp, 0.05 Solving for gives 0.69078 volt, and BE 1 BE 1 At 0.75 ma we have C 0.00075 1 10 exp Solving for gives 0.68358 volt BE BE 0.0070 volt BE 1 BE C TH 15 BE 0.05, R = 0.00075 R 0.0070 volt; R 9.960 ohms C E E E Problem 6 (15 points) Current mirrors are often used for C biasing in integrated circuits. But they can also be used as signal-current amplifiers. Consider the circuit schematic below for an example of a signal-current amplifier. Transistor Q 1 acts as the input current source (driven by voltage N ). erive an expression for the small-signal current gain ( 0 / 1 ) and the small-signal voltage gain (v out /v in ) with (W /L) and (W 3 /L) being the primary variable parameters?
Page7 Solution: n this circuit when W = W 3, then O = 1. However, we can scale the output current O (= 3 ) by (W 3 /W ) if we wish. This allows for the function of a signal current amplifier with gain O / 1. Note that O is equal to 3. MOSFET Q 1 establishes current. The current in MOSFET Q 1 (and also MOSFET Q ) is Large-signal: 1 = = ½ n C OX (W/L)( in ) where in = ( GS t ) = O Small-signal: i 1 = i = - g m1 v in Signal current gain = ( O / 1 ) = ( 3 / 1 ) = (W 3 /W ) Then v out = i 3 R L and i 3 = (W 3 /W )(- g m1 )v in Hence, the voltage gain A = - (W 3 /W )(g m1 )R L