Input Stage. V IC(max) V BE1. V CE 5(sat ) V IC(min) = V CC +V BE 3 = V EE. + V CE1(sat )


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1 BJT OPAMPs
2 Input Stage The input stage is similar to MOS design. Take a pnp input stage (Q1 Q2) with npn current mirror load (Q3 Q4) and a pnp tail current source (Q5). Then, V IC(max) = V CC V BE1 V CE 5(sat ) V IC(min) = V EE +V BE 3 V BE1 + V CE1(sat )
3 Input Stage If it is desired to bring the minimum input common mode voltage to ground, the current mirror can be replaced by resistors. In this case, gain will be low. To improve this situalon, a folded cascode stage can be used.
4 Input Stage VCC Q5 Q6 Q7 Bias1 + Vod  Vin Q1 Q2 Vin+ Q3 Q4 BiasCM + + VEE=0
5 Input Stage With this configuralon, v od v id = g m1 ρ 1 R o1 ρ 1 = R 3 R 3 + r e3 R o1 [ r 06 ( 1+ g m6 R 6 )] [ r 03 ( 1+ g m3 R 3 )] The assumplon here is that g m6 R 6 << β pnp and g m3 R 3 << β npn. For rail to rail operalon, add another stage.
6 Input Stage VCC Q8 Q9 Q10 Bias1 Vin Q1 Q2 Vin+ + Vod  Vin Q1 Q2 Vin+ Q6 Q7 BiasCM Q VEE=0
7 Input Stage This stage has rail to rail operalon for a supply of 2V or more. However, it suffers from the problem of non constant gm discussed earlier. This varialon of g m causes a varialon in open loop gain and frequency response of the OPAMP, compromising its performance when designed to be stable in closed loop configuralon.
8 The NE5234 OPAMP
9 NE5234 Biasing Circuit
10 NE5234 Biasing Circuit Q49, Q60, and R60 form a Widlar Current Mirror. The bases of Q49 and Q60 are driven by a unity gain buffer (Q50 Q53) to follow the collector of Q49. This is simply a beta helper to reduce the beta error. Here, a complementary emi\er follower is ullized instead of using a single transistor to reduce the supply voltage.
11 NE5234 Biasing Circuit Similarly, Q54 Q57 act to bias the bases of Q47 and Q58. Finally, Q48 and Q59 are just cascode transistors to reduce the dependence of currents on V CC. To concentrate on the core of the circuit, let us remove the beta helpers and cascodes.
12 NE5234 Biasing Circuit
13 NE5234 Biasing Circuit This is just a self biased current source using thermal voltage as we studied earlier. The emi\er area of Q60 is twice the area of Q49. The current can easily be calculated as I C 47 = I C 49 = I C 58 = I C 60 = V T R 60 ln2 = 6µA
14 NE5234 Biasing Circuit If R60 is constant, these currents are proporlonal to absolute temperature (PTAT). Resistor R57 is used to prevent zero current through the circuit.
15 NE5234 Input Stage
16 NE5234 Input Stage Node Bias1 comes from the previous bias circuit. Thus, I C11 is 6µA, I C12 3µA, and I C13, I C14 6µA. I C12 flows into Q8 and R8, sebng the voltage from the base of Q5 to ground to approx. 0.8V. If the common mode input is much less than 0.8V, the current of Q11 flows through Q3 Q4, turning OFF Q5, Q7, Q6, and thus Q1 Q2.
17 NE5234 Input Stage If the input common mode signal is much larger than 0.8V, the current of Q11 flows through Q5 Q7 and thus Q1 Q2, turning Q3 Q4 OFF. If the input common mode voltage is around 0.8V, the current of Q11 is split between Q1 Q2 pair and Q3 Q4 pair. Thus, the total transconductance remains the same.
18 NE5234 Input Stage Q9 Q10 and Q13 Q14 operate in the aclve mode with 6µA of current. However, the output of the first stage is very sensilve to matching. Thus, a CM signal is created in the second stage and fed back to the bases of Q9 and Q10.
19 NE5234 Second Stage
20 NE5234 Second Stage Nodes 9 and 10 are coming from the first stage. C21 and C22 are for frequency compensalon. Q21 and Q22 are for buffering so that the second stage does not load the first one. Ignore Q23 and Q24 as they are normally OFF. Transistors Q25 Q28 form a differenlal pair.
21 NE5234 Second Stage I C15 = 3µA, I C16 = I C19 = 4µA, I C17 = I C18 = 21µA, and I C20 = 6.6µA. The current I C15 flows through the Scho\ky diode D1 and creates a voltage drop of 0.4V. Q29, R29, and Q30 form a Widlar Current Mirror. Q29 is 7 Lmes larger than Q30, giving I C29 as 42µA.
22 NE5234 Second Stage Using this value, the current of the differenlal pair is 42µA 3µA = 39µA. Therefore, the individual currents of the four differenlal pair transistors are around 10µA. Define V cmout1 = 1 ( 2 V 9 +V 10 ) For the first stage, V cmout1 is an output and V biascm is an input.
23 NE5234 Second Stage For small values of V biascm, the transistors Q9 and Q10 are off and V cmout1 is approx. V CC For large values of V biascm, V cmout1 drops to about 0.3V. The threshold is around 0.9V. Please see your book for details. For the second stage, V biascm is an output and V cmout1 is an input.
24 NE5234 Second Stage In this case, V biascm = V cmout V The solulon to these two characterislcs is when V biascm = 0.9V and V cmout1 = 0.4V. This structure is also an example of CMFB.
25 NE5234 Output Stage
26 NE5234 Output Stage The NE5234 does not use an emi\er follower configuralon for the output due to the low supply voltage limitalons. Nodes 25 and 26 are the inputs coming from the second stage. Capacitors C25 and C26 as well as resistors R25 and R26 are for frequency compensalon. The output is driven by Q74 and Q75.
27 NE5234 Output Stage The high current drive requirement for the output transistors result in high base currents. The base of Q75 is driven by an emi\er follower Q68. The base of Q74 also has to be driven. However, the β of pnp transistors are lower. Thus, a complementary pair Q64 Q65 is used.
28 NE5234 Output Stage The voltages Bias1 and Bias5 are coming from the bias circuit discussed earlier. They set I C61 to 6µA and I C63 and I C64 to 33µA. Transistors Q70, Q72, and Q73 are normally OFF. All other biasing is set by the output bias circuit. PBASE and NBASE are set by the output stage and are inputs to the output bias circuit. The next slide shows a simplified diagram of the output bias circuit and the output stage together.
29 NE5234 Output Stage
30 NE5234 Output Stage In this diagram, transistors which are normally OFF are omi\ed and some current mirrors are shown as ideal current sources. In classical class AB common collector output stages, the product of currents is a constant. In theory, these transistors never turn off. However, due to voltage drops in the base and emi\er resistances, they do turn off. This causes extra delays, worsening the crossover distorlon even further.
31 NE5234 Output Stage In the NE5234, the transistors can never turn OFF. The collector currents of the output transistors are observed by observing the base emi\er voltages. These voltages are eventually sent to Q45 Q46 for comparison.
32 NE5234 Output Stage This can be wri\en analylcally as, V B 46 = V BE 75 = V T ln I C 75 V B 45 = V BE 42 + I C 43 R 42 I C 43 = V EB 74 V EB 43 R 43 Assuming R 42 = R 43, I S 75 V B 45 = V BE 42 +V EB 74 V EB 43 = V EB 74 +V T ln I S 43 I S 42
33 NE5234 Output Stage This yields V B 45 = V T ln I C 74 I S 75 if I S 75 = I I S 74 S 42 I S 43 If the difference between V B45 and V B46 exceeds 3V T in either direclon, the transistor with the higher base voltage turns OFF. Then, the emi\er voltage is controlled by the other transistor. This is the input to Q40 of the other differenlal pair.
34 NE5234 Output Stage The other side is a constant voltage created by I REF across the two diode connected transistors Q37 and Q38. This forms a feedback loop. Assume that Q75 conducts a large current to pull the OPAMP output low. Now assume that the voltage at the base of Q40 rises.
35 NE5234 Output Stage Then, I C40 is increased and I C39 is reduced. Then, the node voltage of 25 is increased, thus increasing the voltage on Pbase. This change reduces V B45 and hence V B40. This is opposite to what happened inilally. In normal operalon, V B39 and V B40 are about equal.
36 NE5234 Output Stage Now, let us outline the analylcal calculalon. I C 45 + I C 46 = I C 44 V BE 75 +V EB 46 V EB 45 V B 45 = 0 V BE 75 +V EB 46 V BE 40 I 40 R 40 + I 39 R 39 +V BE 39 V BE 37 V EB 38 = 0 Using the concept of feedback to equalize the base voltages of Q39 and Q40, V BE 40 + I R 40 R 40 = V BE 39 + I R 39 R 39 V BE 75 +V EB 46 V BE 37 V EB 38 = 0 V T ln I C 75 +V I T ln I C 46 V S 75 I T ln I C 37 V S 46 I T ln I C 38 = 0 S 37 I S 38
37 NE5234 Output Stage Sebng I C37 = I C38 = I REF = I C36  I C35, I C 75 I REF I S37 = I S75 I REF I C 46 I S 46 I S 38 Also, from the previous equalons, V T ln I C 75 +V I T ln I C 46 V S 75 I T ln I C 45 V S 46 I T ln I C 74 S45 I S 75 Assuming Q45 and Q46 are idenlcal, I C 75 I C 74 = I C 45 I C 46 = I C 44 I C 46 I C 46 = 0
38 NE5234 Output Stage Solving for I C46 and subsltulng it to one of the earlier equalons yields, I C 46 = I C 44 I C 74 I C 75 + I C 74 I C 75 I C 74 I C 75 + I C 74 = I REF 2 I C 44 I S 75 I S 37 I S 46 I S 38 In the NE5234, I REF = 7.4µA, I C44 = 6µA, I S75 /I S37 = 10, and I S46 / I S38 = 2.
39 NE5234 Output Stage IC IC
40 NE5234 Output Stage When the load current is zero, both currents are equal at about 360µA. When one current goes to infinity, the other saturates at around 180µA. Now, you can calculate all other currents
41 NE5234 Small Signal Analysis Let us break the circuit into three stages, input stage, second stage, and output stage. β npn = 40, β pnp = 10, V A,npn = 30V, and V A,pnp = 20V. The input stage is a fully differenlal circuit with two pairs and the input resistance depends on which of the two are conduclng. Assume V IC << 0.8V. Q1 Q2 is OFF, Q11 biases the pnp pair Q3 Q4.
42 NE5234 Small Signal Analysis + Q13 Q3 + Vid/2 + Q9 +
43 NE5234 Small Signal Analysis From this figure, R id = 2r π 3 =170kΩ R up1 = ( ) r R g m13r 013 β pnp +1 β pnp 1+ g m13 R 13 g m13 r π13 β pnp +1 r g m13 R 13 β pnp 1+ g m13 R 13 β pnp =18MΩ Similarly, R down1 β npn +1 r g m 9 R 9 β npn 1+ g m 9 R 9 β npn r 09 ( 1+ g m 9 R 9 ) 1+ g m 9 R 9 β npn = 27MΩ
44 NE5234 Small Signal Analysis Then, the overall output impedance is R o1 = R up1 R down1 =11MΩ The transconductance of the input stage is, G m1 = g m 3 ρ 3 ρ 3 R 9 R 9 + r e9 = 0.84 G m1 = 97µS
45 NE5234 Small Signal Analysis The input resistance of the second stage is given by, R i2 = 2 r π 21 + ( β pnp +1) ( r π 25 r π 26 ) [ ] =1.3MΩ To find the output resistance, we again divide the resistance into two parallel branches. ( ) r R g r m β pnp +1 β pnp +1 β pnp r g m17 R 17 β pnp R up2 = 1+ g R m g R = 5.0MΩ m17 17 g m17 r π17 β pnp
46 NE5234 Small Signal Analysis Looking down into the collector of Q25 or Q26, R down2 = r 025 ( 1+ g m 25 R E 25 ) β npn R E 25 = r e26 + β npn +1 β npn R E 25 β npn +1 g m26 g m27 R down2 = 4.0MΩ Combining these, R up 2 R ( in 3(26) ) g m26 r 026 r e 27 r e28 g m28 = 0.85kΩ R o2 = R up 2 R down2 = 2.2MΩ
47 NE5234 Small Signal Analysis Q25 Q26 Q27 Q28 + av21v9 + av22v10
48 NE5234 Small Signal Analysis a v21 and a v22 represent the small signal gains of the emi\er followers Q21 and Q22. Each emi\er follower drives a load of R L where R L = r π 25 r π 26 = r π 27 r π 28 = r π 25 2 a v21 = a v 22 = 1+ 1 r π 21 ( ) r 021 ( β pnp +1) r π 25 2 [ ] = 0.90 The transconductance of the second stage can be defined as i o G m2 = v 9 v 10 v25 =0
49 NE5234 Small Signal Analysis Instead of using a differenlal voltage v 9 v 10 at the input, assume that v 10 is zero and redefine G m2 as G m2 = i o v 9 v 25 =0 Use superposilon to find i o such that i o1 = g m g m25 R E 25 a v21 v 9
50 NE5234 Small Signal Analysis To find the second component, R E 26 = r e 25 r e27 r e 28 = r e25 3 = 0.85kΩ ( ) = 4.0MΩ R C 26 = r g m 26 R E 26 g i c 26 m 26 R a 1+ g m26 R v21 v C 26 9 E 26 R C 26 + R up 2 R in 3(26) i o2 i c 26 3 g m26 1+ g m 26 R E 26 a v 21 v 9 Combining these two components, i o = i o1 + i o2
51 NE5234 Small Signal Analysis We can find G m2 as, g G m2 a m 25 g v 21 m 26 =170µS 1+ g m25 R E g m26 R E 26 The input resistance of the output stage is, R i3(25) = r π 64 + ( β npn +1)R E 64 [ ] r π 65 + β pnp +1 R E 64 = r 063 ( 1+ g m63 R 63 ) R E 65 = R 65 r π 66 r π 74 = 200Ω R E 64 = 3.5kΩ R i3(25) = 404kΩ ( )R E 65 [ ]
52 NE5234 Small Signal Analysis The output resistance is R o3 = r o74 r o75 =15kΩ The transconductance of the output stage is G m3 = a v 64 a v65 g m 74 1 a v64 = r 1+ π 64 ( β npn +1)R E 64 a v65 = 1+ G m3 = 1 96Ω 1 r π 65 β pnp +1 ( )R E 65 = 0.36 = 0.63
53 NE5234 Small Signal Analysis Now, let us combine all stages and calculate the gain for a load of 2kΩ. R a v1 = G m1 R i2 o1 = 60 2 a v2 = 2G ( m2 R o2 R ) i3(25) =120 a v3 = G ( m 3 R o3 R ) L =18 a v = a v1 a v 2 a v3 =130,000
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