EEE225: Analogue and Digital Electronics Lecture II James E. Green Department of Electronic Engineering University of Sheffield j.e.green@sheffield.ac.uk
This Lecture 1 One Transistor Circuits Continued... Emitter Follower or Common Collector Emitter Follower Voltage Gain Emitter Follower Input Resistance Emitter Follower Output Resistance Common Base Common Base Voltage Gain Common Base Input Resistance 2 Inside the Opamp Feedback System Simplified Schematic of an Opamp Opamp Circuit DC Conditions Differential Amplifier 3 Review 4 Bear 2/ 17
One Transistor Circuits Continued... Emitter Follower or Common Collector Emitter Follower / Common Collector A kind of voltage follower or buffer Approximately unity voltage gain pnp or npn versions possible High current gain May be thought of as impedance transformer (so can all transistor circuits...) v s R S +V s V S In this figure the biasing circuitry is contained as an effective resistance within R S v o 3/ 17
One Transistor Circuits Continued... Emitter Follower Voltage Gain v b R S i b v be r be g m v be or β i b v e v s = i b R S + v be + v o (3) ( = v be 1 + R ) S + v o (4) r b e v s v o i e using the result in (2) to eliminate v be, ( ) 1 v o = v be + g m r be (1) v be g m (2) and a relation between v be, v s and v o is given by summing voltages around the input loop. v o v s = = r be g m (5) r be g m + R S + r be 1 g m + R S β + (6) 1 The gain is non-inverting 2 Gain 1 if >> R S /β and >> 1/g m 4/ 17
One Transistor Circuits Continued... Emitter Follower Input Resistance The input resistance is given by considering v b /i b, recall (1) ( ) 1 v e = v be + g m (7) r be and summing up the voltages... v b = v be + v e (8) ( ) 1 = v be + v be + g m r be (9) ( ( )) 1 = v be 1 + + g m r be (10) since v be = i b r be and g m r be = β we can write, r i = v b i b = r be +(β + 1) (11) Generally (β + 1) >> r be so the input resistance is dominated by the (β + 1) term. By comparing this result with the input resistance of the non-degenerated common emitter amplifier we could show negative feedback can be used to increase the input resistance of a transistor stage. 5/ 17
One Transistor Circuits Continued... Emitter Follower Output Resistance To obtain the output resistance inject a test current i t with the input grounded and find v o /i t. Summing currents at v e (1 + β) i b + i t = v e (12) and summing up the voltages in the base loop R S i b v b v be r be g m v be or β i b v e i e i t v e = i b (R S + r be ) (13) substituting (13) into (12) and solving for v e /i t, r o = 1 + 1 (14) 1+β R R S +r E be r o 1 + R S g m β (15) If β >> 1, the first term becomes R S +r be β and if is large, we can ignore the 1 term. 6/ 17
One Transistor Circuits Continued... Common Base Common Base Connection Generally used in conjunction with other transistors in circuit blocks, but sometimes alone 1. i e is the input current (flowing from v s ), since i e = i o + i b the current gain (i o /i e ) is slightly less than 1 (actually it s = α). R S R L I C +V S v s I E I B v o summing currents, -V S i e + i b + g m v be = 0 (16) 1 http://dx.doi.org/10.1088/0957-0233/23/12/125901 7/ 17
One Transistor Circuits Continued... Common Base Voltage Gain v s v e R S + v be rbe +g m v be = 0 (17) v e + v be = 0 so v e = v be therefore (17) can be solved for v be v be = v s R S ( 1 R S + 1 r be + g m ) v s (18) (19) 1 + g m R S approximation is because 1/r be = g m /β and β >> 1 At the output, v o = i o R L = g m v be R L (20) combining this with (19) to eliminate v be v o v s = g m R L 1 + g m R S = R L r e + R S(21) where r e = 1/g m. The gain is non-inverting Gain R L If R S >> r e gain controlled by ratio R L /R S 8/ 17
One Transistor Circuits Continued... Common Base Input Resistance Common Base Input Resistance The resistance looking into the emitter, r i = v e i e = v e v be r be g m v be (22) Since v e = v be and g m >> 1/r be this reduces to r i 1 g m = r e The value is small 10s - 100s Ω There is another model of the transistor called T Model in which r e plays a much bigger role. However hybid-π is the only model we will use. The original π paper is by Giacolletto 2. 2 http://dx.doi.org/10.1109/jssc.1969.1049963 9/ 17
Inside the Opamp Feedback System Feedback Systems (Quick reminder) In EEE118 we discussed the opamp in terms of a general feedback system. v i + (v i H v o ) G v o H v o H So v o = G (v i H v o ) (23) If G H >> 1, or v o (1 + G H) = G v i (24) v o = G v o G v i G H = 1 (26) H = (25) v i 1 + G H System dependent on H, designer controls H with ratio of resistors. 10/ 17
Inside the Opamp Simplified Schematic of an Opamp v + v i v I C1 I E Q 1 Q 2 I C3 R VA Q 3 R 1 v o1 Input stage subtracts inputs v + + v I C2 Q4 + V S V A Q 5 R L Voltage amplifier stage G v a V S Ouput Stage allows power gain O/P v o4 Input stage: differential amplifier or long tailed pair. Subtracts the inputs. Voltage amplifier stage (VAS): common emitter amplifier. Provides majority of voltage gain. Output stage: emitter follower. Increases current capability of VAS (voltage current = power... hence power gain. 11/ 17
Inside the Opamp Opamp Circuit DC Conditions Opamp will not work properly without feedback. Feedback controls the gain of the circuit but also helps define the DC conditions. Feedback adjusts v i in order to achieve the internal voltage drops required for proper operation. If v o = 0, v i will be at the value it needs to be in order to make v o = 0. Feedback is not shown on prior slide. If v + v 0, V E1 and V E2 0.7 so I E (+V S 0.7)/. I E splits between Q 1 and Q 2 to form I C1 and I C2. I C1 has two functions 1) create a voltage drop of 0.7 V across R 1 in order to bias Q 3 into conduction. 2) Provide the base current for Q 3. I C1 will be 0.7/R 1 + I C3 /h FE3. The value of I C3 varies with V A and hence with V o4 but assuming V A = 0, I C3 = +V S /R VA. I C2 is returned directly to the negative supply. In the case where v + v 0, there is a common mode input voltage, v cm, and I E (+V S v cm )/. 12/ 17
Inside the Opamp Differential Amplifier V + i V i + V V I 2 I R 1 +V S I Q 1 Q 2 V o = I 2 I R 1 I 2 + I v o = I R 1 If v + increases by v i and v decreases by v i, the average of v + and v is unchanged so I E is unchanged because V be is unchanged. If v + and v increase or decrease by v i, v i is called a common mode signal ideally the differential amplifier will not amplify any common mode component of the input. -V S 13/ 17
Inside the Opamp Differential Amplifier v + +0.7 V I C1 R 1 +V S I Q 1 Q 2 V S I C2 v o1 Q 3 We must consider the effects of three transistors. Q 1 and Q 2 are the input differential pair. Q 3 must also be considered now because its input resistance forms part of Q 1 s collector load resistance. If the input signal is regarded as v + with respect to ground, Q 2 looks like a common base connection and can be represented by its common base input resistance 1/g m2. The collector current of Q 1 sees two resistors in parallel, R 1 and the input resistance of Q 3. Q 3 is a common emitter amplifier without degeneration. Its input resistance is r be3. 14/ 17
Inside the Opamp Differential Amplifier i b1 r be1 v be g m1 v be1 or β 1 i b1 v i i e1 v e1 r e2 R 1 v o1 r be3 A small signal equivalent circuit describes the three transistor circuit block according to our simplifications. i 1 This small signal model is very similar to the common emitter with degeneration from Lecture 1. In this case R S = 0 and and R L are parallel combinations //r e2 and R 1 //r be3. Since >> r e2, r e2 dominates. The gain expression for the circuit is (based on the degenerated CE analysis) v o1 v i = R 1//r be3 r e1 + r e2 (27) 15/ 17
Review Review Considered the emitter follower circuit (voltage gain, current gain, input and output resistances). Considered the common base circuit (voltage gain, current gain, input resistance). Recapped the idea of the opamp as a feedback system. Introduced a simplified schematic of an opamp. Developed some ideas around the DC conditions of the simplified opamp Looked at the combination of three transistors into a differential amplifier + common emitter stage and considered their combined effect. 16/ 17
Bear 17/ 17