Part 2 of 2
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Electronic Circuits Revision on Basic Transistor Amplifiers Contents • Biasing • Amplification principles • Small-signal model development for BJT Aim of this chapter To show how transistors can be used to amplify a signal. amplifier 2 Basic idea Step 1: Set the transistor at a certain DC level — biasing Step 2: Inject a small signal to the input and get a bigger output — coupling 0.6V amplifier 7V 3 Biasing the transistor To set the transistor to a certain DC level = To set VCE and IC VCC=10V IC RL RB IB + VBE – Transistor: β = 100 + VCE – Suppose we want the following biasing condition: IC = 10 mA and VCE = 5 V Find RB and RL Start with VBE ≈ 0.7 V. Then, IB = (10 – VBE )/ RB = (10 – 0.7)/ RB IC = βIB = 100 (10 – 0.7)/ RB = 10 mA Also, VCE = 10 – RL IC Hence, 5 = 10 – 10RL So, RB = 94kΩ So, RL = 0.5kΩ 4 β dependent biasing — bad biasing VCC=10V Now, let’s go to the lab and try using RB = 94kΩ and RL = 0.5kΩ, and see if we get what we want. IC RL RB IB + VBE – Transistor: β = 100 …totally wrong! We don’t get IC = 10mA and VCE = 5V + VCE – This is not a good biasing circuit! because it relies on the accuracy of β, but β can be ±50% different from what is given in the databook. 5 A slightly better biasing method Again, our objective is to find the resistors such that IC = 10mA and VCE = 5V. VCC=10V IB + + RB2 IC RL RB1 VBE – VCE – First, if IB is sm all, we can approximately write ⇒ Suppose we get IC = 10mA. Then RL = 0.5kΩ. We can start with RB1 = 940Ω and RB2 = 60Ω. Such resistors will make sure IB is m uch sm aller than the current flowing down RB1 and RB2, which is consistent with the assumption. What we need in practice is to fine tune RB1 or RB2 such that VCE is exactly 5V. 6 A much better biasing method — emitter degeneration Again, our objective is to find the resistors such that IC = 10mA and VCE = 5V. VCC=10V RL RB1 IB RB2 IC + Surely, RL = 0.3kΩ in order to get VCE = 5V. – Finally, we have VB = VE + 0.6. Therefore, if IB is small compared to IRB1 and IRB2, we have VCE + VE – Set VE = 2V, say. Then, RE = 2V/10mA = 0.2kΩ. RE Hence, RB1 = 740Ω and RB1 = 260Ω. NOTE: β is never used in calculation!! 7 Stable (good) biasing Summary of biasing with emitter degeneration: VCC=10V RB2 IC RL RB1 VB Choose VE , IC and VCE . IB + RE VCE + VE – – RE RL Use VBE ≈ 0.6 to get VB. Then use to choose RB1 and RB2 such that IB is much smaller the current flowing in RB1 and RB2. 8 Terminology The following are the same: Biasing point Quiescent point Operating point (OP) DC point 9 Alternative view of biasing IC VCC IC RL + + VBE – VCE – RL + – VCC IC + VR – Load line Slope=–1/RL + VCE – operating point VCC VCE 10 What controls the operating point? IC VCC IC RL Load line Slope=–1/RL a bigger VBE + + VBE – VCE – operating point a smaller RL VCC VCE CONCLUSION: VBE or IB controls the OP RL also controls the OP 11 What happens if VBE dances up and down? IC VCC IC RL + + VBE VCE Load line Slope=–1/RL The OP also dances up and down along the load line. a bigger VBE = 0.65 a smaller VBE = 0.6 VCE also moves up and down. – – VCC VCE Typically, when VBE moves a little bit, VCE moves a lot! THIS IS CALLED AMPLICATION. 12 Derivation of voltage gain Question: what is VCC Clearly, Ohm’s law says that IC RL + + VBE – ? VCE – Then, what relates ∆IC and ∆VBE ? Last lecture: transconductance Hence, 13 Common-emitter amplifier The one we have just studied is called COMMON-EMITTER amplifier. VCC SUMMARY: RL RB1 IC + RB2 + vBE – vCE – Small-signal voltage gain = –gmRL That means we can increase the gain by increasing gm and/or RL. Output waveform is antiphase. 14 How do we inject signal into the amplifier? VCC RL RB1 ∆vin + ? ~ vin ±20mV IC RB2 + VBE – ~ vCE = VCE + vCE – or ∆vCE 15 Note on symbols = + DC point A ~ vCE = VCE + vCE total signal (large signal) operating point or DC value or quiescent point Total signal a small signal or ac signal Small signal ~ a or ∆a 16 Solution: Add the same biasing DC level Exactly the same biasing VBE VCC RL RB1 IC + + – RB2 ~ vin ±20mV + vBE – vCE – But, it is impossible to find a voltage source which is equal to the exact biasing voltage across B-E. VBE could actually be 0.621234V, which is determined by the network RB1, RB2 and the transistor characteristic!! How to apply the exact VBE? 17 The wonderful voltage source: capacitor VCC The capacitor voltage is exactly equal to VBE because DC current must be zero + VC – RL RB1 0A IC + RB2 + VBE VCE – – 18 Solution — insert coupling capacitor VCC DC voltage equal to exactly the same biasing VBE This is called a coupling capacitor ~ vin ±20mV – RL RB1 + IC + RB2 + vBE vCE – – 19 Complete common emitter amplifier VCC RL RB1 – + ~ vin + IC + RB2 + vBE vCE + – – + ~ vo – – – coupling capacitors (large enough so that they become short-circuit at signal frequencies) 20 Can we simplify the analysis? We are mainly interested in the ac signals. The DC bias does not matter! Can we create a simple circuit just to look at ac signals? + ~ vin – common emitter amplifier + ~ vo – 21 Small-signal model Two basic questions: + ~ vin ? common emitter amplifier – 1 What is the loading (resistance) seen here? 2 ? + ~ vo – What is the Thévenin or Norton equivalent circuit seen here? 22 Small-signal model of BJT: objectives To find: rin rin Ro Gm Gmvin Norton form rin Ro Am or Ro + – rin Ro Amvin Thévenin form 23 Derivation of the small-signal model Input side: iB + vBE For small-signal, rπ – rπ β/gm where gm is the BJT’s transconductance 24 Derivation of the small-signal model Output side: VCC RL IC + vCE = VCC – ICRL For small-signal, ~ gmvBE RL ~ vCE – + vCE – where gm is the BJT’s transconductance 25 Derivation of the small-signal model Output side: VCC + Including BJT’s Early effect ~ RL IC ro ~ RL vCE gmvBE – + vCE – where ro is the Early resistor of the BJT. Recall: ro = VA/IC , where VA is typically about 100V. A very rough approx. is ro = ∞. 26 Initial small-signal model for BJT “MUST” REMEMBER + ~ vBE – B C rπ ~ gmvBE E ro Small-signal BJT parameters: + ~ vCE – BJT model 27 Initial small-signal model for FET Similar to BJT, but input resistance is ∞. G + ~ vGS – D ~ gmvGS S + ~ Small-signal FET parameters: ro vDS – FET model All amplifier configurations using BJT can be likewise constructed using FET. 28 Example: common-emitter amplifier VCC is ac 0V. Assume the coupling caps are large enough to be considered as short-circuit at signal frequency VCC RL RB1 + vin – RB2 B + vBE RB1 ||RB1 + vo rπ E C gmv~BE RL ro E – – 29 Complete model for common-emitter amplifier Complete model: + vin – + RB1 ||RB1 ~ rπ – gmv~BE vBE ro + RL vo – vin – RB1 ||RB1 || rπ Rin = RB1 ||RB1 || rπ Total output resistance Simplified model: + Total input resistance + ~ vBE – gmv~BE RL||ro + vo – Ro = RL||ro Voltage gain 30 Alternative model for common-emitter amplifier Output in Thévenin form: Total input resistance RL||ro + vin – RB1 ||RB1 || rπ + ~ vBE – + – + vo – ~ gm(RL||ro) vBE Rin = RB1 ||RB1 || rπ Total output resistance Ro = RL||ro Voltage gain 31 More about common-emitter amplifier Because the output resistance is quite large (equal to RL||ro ≈ RL), the common-emitter amplifier is a POOR voltage driver. That means, it is not a good idea to use such an amplifier for loads which are smaller than RL. This makes it not suitable to deliver current to load. RL||ro + vin – RB1 ||RB1 || rπ + ~ vBE – 1kΩ, for example + – + vo – practically no output!! 10Ω ~ gm(RL||ro) vBE 32 Bad idea — wrong use of common-emitter amplifier Transconductance gm = IC /(25mV) = 5/25 = 0.2 A/V +10V RB1 5mA 1kΩ Expected gain = gmRL = (0.2)(1k) = 200 or 46dB But the output circuit is: + vin – RB2 + vBE – + speaker vo 10Ω – 200vin – + 1kΩ 10Ω + vo – The effective gain drops to 33 Proper use of common-emitter amplifier The load must be much larger than RL. +10V RB1 + vin – RB2 5mA + vBE – 1kΩ Now the output circuit is: nearly open circuit + 10MΩ vo – 200vin – + + 1kΩ 10MΩ vo – The effective gain is 34 How can we use the amplifier in practice? +10V RB1 5mA How to connect the output to load? 1kΩ ? + vin – RB2 + vBE – + vo – speaker 10Ω 35 Emitter follower Biasing conditions: +10V RB1 IC biased to 10mA VCE biased to 5V + vin – RB2 RE + vo – Base voltage ≈ 5.6V Emitter voltage ≈ 5V Collector current ≈ 10mA RE = 500Ω RB1:RB2 ≈ 44:56 Say, RB1 = 440kΩ RB2 = 560kΩ VE = VB – 0.6 Thus, for small signal, ∆V E = ∆V B or vo = vin Gain = vo / vin =1 36 Small-signal model of emitter follower +10V B RB1 rπ RB1 || RB2 + vin – C RB2 RE + vo – gmv~BE E ro E RE + vo – 37 Small-signal model of emitter follower vin iB r RB1 || RB2 B Input resistance is C rπ gmv~BE E ro E RE + vo – which is quite large (good)!! 38 Small-signal model of emitter follower B Output resistance is C rπ gmv~BE E ro im E RE rout + – vm which is quite small (good)!! 39 Small-signal model of emitter follower Thevenin form: very large + vin – rπ+(1+β)RE Large input resistance Small output resistance Voltage gain = 1 RE||(1/gm) + – very small + 1 vin vo – Draw no current from previous stage Good for any load 40 A better “emitter follower” +10V Input resistance is very LARGE because RE = ∞. RB1 Output resistance is 1/gm. Gain = 1. + vin – RB2 ∞ + IE 1/gm vo – This circuit is also called CLASS A output stage. Details to be studied in second year EC2. 41 Common-emitter amplifier with emitter follower as buffer +10V RB1 + vin – RB2 RL + vBE – common-emitter amplifier (high gain) emitter follower (unit gain) ∞ IE 1 gm + speaker vo 10Ω – 42 FET amplifiers (similar to BJT amplifiers) +10V RG1 + vin – RG2 RL + vGS – common-source amplifier (high gain = –g m R L ) source follower (unit gain) ∞ IS 1 gm + speaker vo 10Ω – 43 Further thoughts Will the biasing resistors affect the gain? Rbias Seems not, because RL Gain = –gmRL + + vin – vo – which does not depend on Rbias . However, a realistic voltage source has finite internal resistance. This will affect the gain. 44 Input source with finite resistance Rs The input has a voltage divider network. RL Rbias + + Therefore, the gain decreases to vo vin – – assuming ro very large. Rs + vin – Rbias + vBE – rπ gmvBE ro RL + vo – 45 Example 10V 94kΩ 50Ω – 1kΩ + + vin 5mA vo 600Ω – By how much does the gain drop? 50Ω + Rbias vin – + vBE – rπ 94k||600 = 596Ω gm = 5mA/25mV = 0.2A/V rπ = β/gm = 100/0.2 = 500Ω Voltage divider attenuation = Hence, the gain is reduced to 0.845(gmRL) = 169 46 Further thoughts Recall that the best biasing scheme should be β independent. 10V 84kΩ 5mA 1kΩ + vo + vin – 16kΩ – 200Ω One good scheme is emitter degeneration, i.e., using RE to fix biasing current directly. Here, since VB is about 1.6V, as fixed by the base resistor divider, VE is about 1V. Therefore, IC ≈ VE/RE = 5mA (no β needed!) Question: Will this biasing scheme affect the gain? 47 Common-emitter amplifier with emitter degeneration Exercise: Find the small-signal gain of this amplifier. VCC RB1 RL Answer: + vo + vin – RB2 – RE The gain is MUCH smaller. We have a good biasing, but a poor gain! Can we improve the gain? 48 Common-emitter amplifier with emitter by-pass Add CE such that the effective emitter resistance becomes zero at signal frequency. VCC RB1 RL + vo + vin – RB2 – RE CE So, this circuit has good biasing, and the gain is still very high! Gain = – gmRL which is unaffected by RE because effectively RE is shorted at signal frequency. CE is called bypass capacitor. 49 Summary Basic BJT model (small-signal ac model): C B + vBE – E rπ gmvBE ro E 50 Summary Basic FET model (small-signal ac model): Similar to the BJT model, but with infinite input resistance. Therefore, the FET can be used in the same way as amplifiers. D G + vGS ∞ – S gmvGS ro S 51 Summary Common-emitter (CE) amplifier small-signal ac model: Rbias + vin RL – Rbias + vBE – rπ gmvBE ro RL + vo – + + vin – vo – Gain = –gmRL Input resistance = Rbias || rπ Output resistance = RL ||ro ≈ RL (quite large — desirable) (large — undesirable) 52 Summary Emitter follower (EF) small-signal ac model: Rbias + vin + vin – + RE + – vBE Rbias – rπ vo – gmvBE RE + vo – Gain = 1 Input resistance = Rbias || [rπ +(1+β)RE ] Output resistance = RE || (1/gm) (quite large — desirable) (small — desirable) 53
