5.7 Single Stage BJT Amplifiers

4/15/2011 section 5 7 Single Stage BJT Amplifiers 1/1 5.7 Single Stage BJT Amplifiers Reading Assignment: 460-485 Small signal BJT amplifiers typically can be classified as one of three types. Each type has its own specific characteristics, and thus each type has its own specific uses! First, we consider the common-emitter amplifier: HO: THE EMITTER CAPACITOR: WHAT’S UP WITH THAT? Next, the common collector amplifier—otherwise known as the emitter follower. HO: THE COMMON-COLLECTOR AMPLIFIER Finally, the common-base amplifier: HO: THE COMMON-BASE AMPLIFIER Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 1/18 The Emitter Capacitor: What’s up with that? Note that in a previous amplifier example, there is a mysterious capacitor attached to the emitter: 15 V 15 V 5.0 mA 1K 3.7 K vO (t ) COUS β 100 vi (t ) - 2.3K 1K Q: Why is this big capacitor here? Is it really required? COUS A: Let’s do a small-signal analysis and see why we place this large capacitor at the emitter. Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 2/18 Let’s analyze this amplifier! Step 1 – DC Analysis This is already completed! Recall that we designed the single supply DC bias circuit such that: IC 5 mA and VCE 5.0 0.7 Step 2 – Calculate the BJT small-signal parameters If we apply the hybrid-π model, we will require the small signal parameters: gm rπ Jim Stiles IC 5 mA 200 mA V VT 0.025V VT βV 100(0.025) T 0.5 K IB IC 5.0 The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 3/18 This is step 3 Steps 3 and 4 – Replace the BJT with its small-signal equivalent circuit , and turn off all DC sources. Tuning off the DC sources, and replacing the Capacitors Of Unusual Size with short circuits, we find that the circuit becomes: 1K 3.7 K vo (t ) β 100 vi (t ) Jim Stiles - 2.3K 1K The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 4/18 and this is step 4 Now carefully replace the BJT with its small-signal model: B C - 3.7 K vbe 2.3 K vi (t ) - rπ 0.5 K vo (t ) RC 1 K 200 v be E Note that 3.7 2.3 0.5 370 Ω , therefore our small-signal circuit is equivalently: vo (t ) vi (t ) - vbe - Jim Stiles 0.37 K RC 1 K 200 v be The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 5/18 A hefty gain Step 5 – Analyze the small-signal circuit. Since for this circuit vbe vi and vo (1)200vbe , the open-circuit, small-signal voltage gain of this amplifier is: Avo vo 200vbe 200 vi vbe Likewise, we can find that the small-signal input and output resistances are: and Rin 370Ω Rout 1.0 K Note that the gain in this case is fairly large—46 dB. Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 6/18 Still, what’s up with the capacitor? Q: I still don’t understand why the emitter capacitor is required. Sure, our amplifier has large voltage gain, but I don’t see how a capacitor could be responsible for that. A: To see why the emitter capacitor is important, we need to compare these results to those obtained if the emitter capacitor is removed. Note that if we remove the emitter capacitor, the first two steps of the smallsignal analysis remains the same—the DC operating point is the same, and thus the small-signal parameters remain unchanged. However, this does not mean that our resulting small signal circuit is left unchanged! Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 7/18 The emitter resistor is not “shorted out”! * Recall that large capacitors (COUS) are approximated as AC shorts in the small-signal circuit. * The emitter capacitor thus “shorts out” the emitter resistor in the small-signal circuit—the BJT emitter is connected to small-signal ground. * If we remove the emitter capacitor, the emitter resistor is no longer shorted, and thus the BJT emitter is no longer connected to ground! Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 8/18 A horse of an entirely different color The small-signal circuit in this case is: B C - 3.7 K 2.3 K vi (t ) vbe - rπ 0.5 K vo (t ) RC 1 K 200 v be E RE 1 K Note that the resistors R1 3.7 K and R2 2.3 K are no longer in parallel with base resistance rπ 0.5 K ! As are result, we find that small signal voltage vbe is not equal to small signal input voltage vi. Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 9/18 This circuit—it’s harder Note also that the collector resistor is not connected in parallel with the dependent current source! ib ic B vi (t ) - 2.3 3.7 vbe C rπ 0.5 K - ie RE 1 K vo (t ) RC 1 K 200 v be E ve - Analyzing this small-signal circuit is not so easy! We first need to determine the small signal base-emitter voltage vbe in terms of input voltage vi . Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 10/18 Start with KCL From KCL, we know that: ie ib ic Where: ie ve v e ve RE 1 ib vbe vbe 2.0 vbe rπ 0.5 ic 200vbe Therefore: Jim Stiles ve 2.0 vbe 200 vbe 202 vbe The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 11/18 And now for KVL ib C B vi (t ) - 2.3 3.7 vbe rπ 0.5 K - ie ic vo (t ) RC 1 K 200 v be E RE 1 K Likewise, from KVL: ve - 0 vi vbe ve 0 vbe vi ve Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 12/18 This is NOT voltage division! Inserting this into the first KCL result: ve 202 vbe 202 vi 202ve And now solving for small-signal emitter voltage: ve 202 vi 203 Note that the small-signal base voltage is not related to the small signal input voltage by voltage division, i.e.: ve Jim Stiles RE rπ RE vi 1 vi 1.5 The Univ. of Kansas !!! Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 13/18 vbe is really small! Therefore, we can finally determine vbe in terms of input voltage vi : vbe vi ve vi vi 202 202 vi 1 v i 203 203 203 Note then that not only is vBE vi , the small-signal base-emitter voltage is much smaller than input voltage vi ! This of course is evident from the relationship: ve 202 202 vb v 203 203 i which states that the emitter voltage is approximately equal to the base (input) voltage vb (vi ) . Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 14/18 Now for the output voltage This result will have a profound impact on amplifier performance! To determine the output voltage, we begin with KCL: ir ic 200vbe ib B C vi (t ) - 2.3 3.7 vbe ic rπ 0.5 K - ie vo (t ) ir RC 1 K 200 v be E RE 1 K Jim Stiles ve - The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 15/18 What a wimpy gain Now applying Ohm’s Law to RC: vo 0 vo ir 200vbe RC 1 But recall that: vbe vo 200vbe vi 203 so we find that the small-signal output voltage is: vo 200 vbe 200 v 203 i And thus the open-circuit voltage gain of this amplifier is: Avo Jim Stiles vo 200 1.0 vi 203 The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 16/18 See, the emitter capacitor is important Yikes! Removing the emitter capacitor cause the voltage gain to change from – 200 (i.e., 46 dB) to approximately –1.0 (i.e., 0dB)—a 46 dB reduction! That emitter capacitor makes a big difference! We can likewise finish the analysis and find that the small-signal input and output resistances are: Rin R1 R2 3.7 2.3 1.42 K Rout 1.0K Note that input resistance actually improved in this case, increasing in value from 370 Ω to 1.42 K Ω. However, the decrease in voltage gain makes this amplifier (without a emitter capacitor) almost completely useless. Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 17/18 He only knows this because your TA explained it to him The amplifier in this case (with the emitter capacitor) is an example of a design known as a common-emitter amplifier. There are an infinite number of common-emitter designs, but they all share one thing in common—the emitter of the BJT is always connected directly to smallsignal ground. Common-emitter amplifier, such as the one examined here, typically result in large small-signal voltage gain (this is good!). However, another characteristic of common emitter amplifiers is a typically low small-signal input resistance and high small-signal output resistance(this is bad!). Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Emitter Capacitor lecture 18/18 Make sure you can answer this question One way to construct a common-emitter amplifier without using an emitter capacitor is simply to connect the BJT emitter directly to ground: R1 In this case, the emitter is at both AC (small-signal) ground and DC ground! 15 V 15 V RC vO (t ) β 100 COUS vi (t ) - R2 Q: Why is this common-emitter design seldom used? A: Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 1/14 The Common-Collector Amplifier The common-collector amplifier: the BJT collector is at small-signal ground! Examples of this type of amplifier include: Do you see why each VCC VCC V CC COUS COUS - vi (t ) VBB COUS (c) COUS of these four circuits is a gul-durn common-collector amplifier? vO (t ) vi (t ) vO (t ) - (a) VCC VCC - (b) Jim Stiles - vO (t ) vi (t ) Make dang sure that you do! vO (t ) vi (t ) (d) VEE VEE The Univ. of Kansas Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 2/14 We’ll use the T-model Let’s consider circuit (a). It turns out that for common-collector amplifiers, the T-model (as opposed to the hybrid-π) typically provides the easiest small-signal analysis. Using the T-model, we find that the small-signal circuit for amplifier (a) is: C gmvbe αie ib - RB vi (t ) B re vbe - E RE Jim Stiles vo (t ) ie The Univ. of Kansas Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 3/14 Let’s analyze this amplifier! Let’s determine the open-circuit voltage gain of this small-signal amplifier: vooc Avo vi We therefore must determine the output voltage vo in terms of input voltage vi ic From KVL, we find that: C 0 vi RB ib vbe vo gmvbe αie ib - RB vi (t ) B re vbe - E RE Jim Stiles The Univ. of Kansas vo (t ) ie Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 4/14 Let’s apply KCL! And from KCL, we find: C ib ie ic ie gm vbe Where from Ohm’s Law: ie vo 0 vo RE RE gmvbe αie ib - RB vi (t ) B re vbe - E So: ib ie gm vbe vo gm vbe RE RE vo (t ) ie Inserting this into the KVL equation above: vo vi RB ib vbe v vi RB o gmvbe vbe RE R vi B vo gm RB vbe RE Jim Stiles The Univ. of Kansas Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 5/14 Let’s apply Ohm’s Law! Likewise using KCL and Ohm’s Law: C ie Or rearranging: vbe vbe vo re RE r e vo RE - Inserting this result in the solution above: R vo vi B RE R vi B RE R vi B RE vo gm RB vbe re v g R m B o vo R E RB v g r m e o vo R E R vi ( gm re 1 ) B vo RE Jim Stiles The Univ. of Kansas gmvbe αie ib RB vi (t ) B re vbe - E RE vo (t ) ie Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 6/14 It’s the gain—but look closer! From this result we can determine the small-signal output voltage: 1 R vo 1 (1 gm re ) B vi RE And so the open-circuit voltage gain is: R v Avo o 1 (1 gm re ) B RE vi We now note that: gm re VT IC IC α IE VT IE Therefore: 1 gm re 1 α 1 Jim Stiles 1 β β 1 The Univ. of Kansas 1 β 1 Dept. of EECS

4/15/2011 The Common Collector Amplifier lecture 7/14 The output is no bigger than the input! And so the gain becomes: v 1 RB Avo o 1 β 1 RE vi We note here that: 1 1 1 β 1 We find therefore, that the small-signal gain of this common-collector amplifier is approximately: 1 RB Avo 1 β 1 RE The gain is approximately one! Jim Stiles 1 ( 1 0 ) 1 1.0 The Univ. of Kansas Dept. of EECS