Whereas the diode was a 1junction device, the transistor contains two junctions. This leads to two possibilities:


 Randell Shelton
 2 years ago
 Views:
Transcription
1 Part Recall: two types of charge carriers in semiconductors: electrons & holes two types of doped semiconductors: ntype (favor e), ptype (favor holes) for conduction Whereas the diode was a junction device, the transistor contains two junctions. This leads to two possibilities: i) arrow indicates direction of current flow ii) BC, BE have diode drops. But transistors are NOT equivalent to two backtoback diodes. (BC, BE junctions are also asymmetric, larger drop across BC junction) Notation: i) V B, V C, V E = voltage at base, collector, emitter w.r.t. ground ii) V BE = voltage at base w.r.t emitter iii)v CE = voltage at collector w.r.t emitter iv)v CC  usually the positive power supply voltage v) V EE  usually the negative power supply voltage If V CE > V BE BE junction is forward biased therefore current should flow... CE junction is reversed biased. If CE formed a normal diode, no current would flow through the collector. The difference is that the base region is very thin and a strong E field exists at the CB junction which collects any electrons that come nearby. ) Electrons are injected into base from emitter (from the diode equation I E ~I 0 exp qv BE /kt ) 2) e's arriving at the base have two possibilities i) recombine with holes and flow out of the base ii) if an electron drifts near the CB junction it will be sucked across the junction due to the large E field If I E is large enough (ii) usually dominates. A 'logjam' of electrons looking for holes in the thin base material make the electrons likely to jump into the collector. In this case I C ~ I E where
2 Early transistors had ~0.9 today typical transistors have values ~0.99 More convenient notation uses the parameter to distinguish transistors. I C = I B And typical values of range from ~ I C I B is the current gain of the transistor at (DC). A small base current cause a much larger current to flow through the collector/emitter. = note: =00=.99 =50=.98 is a much more sensitive measure of transistor properties Common Emitter Configuration There are 4 variables of interest: V BE, I B,V CE, I C The most useful relationships among these are: Input characterisic V BE = f I B,V CE Output characteristic I C = f 2 I B,V CE So long as V CE >~ 0.6V, V BE and I C depend only weakly on V CE i.e. V BE ~ f ' I B ~fairly constant I C ~ f 2 ' I B 2
3 Three regions of operation ) Active Region V BE ~0.6V V CE >~V BE (I C > 0) defined as: I C /I B is large ~ 00 This mode is used in linear amplifiers 2) Saturation V BE ~0.8V V CE < V BE Still have I C > 0, but is small ~ 0 Transistor is used as a switch in this mode 3) Cutoff V BE <~0.5V therefore I E,I C = 0 Transistor does not conduct Comments: for pnp reverse the above signs and inequalities I C is never < 0 in normal operation (large reverse voltages can cause breakdown and frequently ruin a transistor) Simple model of transistor in active region V CE >V BE (or for pnp type V CE <V BE ) ) 'current gain' picture I C / I B =~00 since I E = I B +I C, I C = I E ~I E 2) Assume simple VI diode characteristics for BE junction I E 0 V BE ~0.6V I E =0if V BE 0.6V or for pnp I E 0 V BE ~ 0.6V I E =0if V BE 0.6V 3
4 Emitter follower (aka common collector) Used to buffer a higher output impedance source to drive a lower input impedance load. V L =V S R L R L R S V S if R L R S Black Box emitter follower R in = V S L R in R R L R L R S V S~V S if R in R S and R out R L Constructing an emitter follower V KVL: in V BE =V out V out =V in 0.6V V in V BE I E =V EE The output simply follows the input when the transistor is in the active region. RB, C are optional (they spoil Q of resonant circuits formed by stray L,C in circuit. (greater importance at high frequencies) Emitter follower in cutoff region at cutoff V in < V EE 0.6V The transistor can do no more that shut off current flow, allowing the base to sit at V EE (very large negative biases can cause breakdown and damage to transistor) 4
5 Emitter follower at saturation If V in > V C, then the V CB junction is forward biased I in ~ V in V CC 0.6V V B ~V CC 0.6V V sat CE typically ~0.2V Analysis of input and output impedance of the emitter follower V in I B V BE =V out V in I B V BE I E =V EE nonlinear relationship for I E,V BE I E e qv BE /kt makes it difficult to define impedance by using V/I Instead we use small signal analysis to define the impedances for time varying signals. Express V,I as a time invariant part + a time varying part then define Z = V I = (for AC signals) V V DC V t V DC I I DC I t I DC In the simple transistor model V BE =0 and V EE =V CC =0 (they are power supplies!) therefore for the small signal analysis we can write in b = out in b e =0 Note: we now adopt a notation where lower case letters will generally refer to small signal equivalents in a circuit analysis 5
6 Small signal equivalent analysis of the emitter follower power supplies are replaced by AC shorts to ground Small signal current gain h FE c / b h FE need not equal, but in practice they tend to agree w/in ~ 20% and are both large (>>). We'll generally assume ~h FE Input impedance Z in = in b = out b b then use e = b Z in = (usually << ) This is the input impedance looking into the base note: Z in To find the output impedance, we replace the circuit by its small signal Thevenin equivalent. O.C. thev = out Z thev = O.C. out S.C. out finding the open circuit output: O.C. out = in b = in / Z in = in in finding the short circuit current: S.C. out = S.C. b = in / Z out = O.C. out = S.C. out = R E if ~ note: Z out What is in this case? 6
7 Limitations of the emitter follower Consider a typical application: driving a 50 Ohm terminated coaxial cable. Assume t RC so that we are in the DC blocking regime for the high pass filter. For the transistor to remain in the active region we need I E >0 positive input increases I E, no problems as long as V in < V CC negative input pulses will decrease I E and may cause cutoff KVL: V in V BE I S =V SS V in V BE V cap I L R L =0 assume V BE =0 and the capacitor is a short for our signal frequencies small signal form of KVL: in in s =0 in L R L =0 then using e= s cap e = R L The transistor cutsoff when e = I E D.C. or (equivalently) when in I E D.C. R L Note: The cutoff voltage depends strongly on the load if R L <~ Circuit simulation illustrating cutoff effect 7
8 The problem of reproducing negative pulses can solved by using a pnp transistor In this case a negative pulse increase I E, keeping the transistor active (but a positive pulse can cause it to cutoff in the same way illustrated above) The general solution for bidirectional pulses is to use a complementary follower This simple approach has an unfortunate drawback. It does not work for small signals in 0.6V Large cross over distortion, output only follows input if in 0.6V, otherwise V out = 0 A better design that compensates much of the cross over distortion In this design: D, D 2 keep V BE ~ ±0.6V (the downside is that Z in ~ R/2!) 8
9 Strategy for biasing an emitter follower ) choose quiescent current I Q, where I Q is the current drawn by the largest pulse which would tend to shut off the transistor. I Q max in max in (if is symmetric) in R L R L 2) choose (assuming AC coupled load otherwise load would affect I Q ) V B 0.6V I Q =VEE = V B 0.6V I Q I Q For maximum symmetric voltage swing at output, V B is set to ~midpoint between the two power supplies V B ~ 2 V CCV EE = 2 V CC V EE.2V I Q 3) Choose the appropriate resistor divider to set the base voltage V B ~ 2 V CC V EE for our example R R 2 and V CC V EE R R 2 ~0. I Q this is the current through the divider Solving for R: R V CC V EE I Q Note: ~00 therefore above relation guarantees that I divider ~0 I B so the base voltage is roughly independent of the base current 4) Finally choose the appropriate AC coupling capacitors for the input signal Z in = R R 2 R L R in choose C in, C out by the necessary low frequency signal cutoff C in ~ 3dB R in C out ~ 3dB R L 9
10 Building a simple current source I in = I E~I E I E = V BB 0.6 V EE for negligible I B This circuit is a good constant current source a long as V C > V BB (small rise of I C w/ V BB ) Otherwise, decreases as we approach saturation. Compliance A current source can only provide constant current for a limited range of voltage at the collector. This output voltage range is called the output compliance of the current source. The compliance is set by the requirement that the transistor stay in the active region. A current source such at the one to the right has limited compliance related to : Compliance ~ V CC V sat I E V EE 0
11 Inverting amplifier (aka common emitter amplifier) Assume the transistor is in the active region, then the analysis of this circuit is identical to the emitter follower e = e = in using c ~ e out = c R C = R C in GainG= R C Z in is identical to the emitter follower Z out is different The collector looks like a very large resistance Z out =R C R C cross checking with the Thevenin equivalent O.C. out = R C R in E S.C. out = c = in here we have use the small signal short: This circuit cannot maintain a large gain while driving a low impedance load G RL = R C R L R C The solution to allow driving a heavier load is to add an emitter follower G= R C R L ~ R C
12 Hybrid Parameters Part 2 In our simple analysis of transistor circuits thus far we have assumed the applicability of a simple model V BE ~0.6V,V BE = BE =0 in the active region. These assumptions lead to unphysical results in some cases ) G= R C implies infinite gain of inverting amplifier 2) Z out = 0 implies zero (or at least arbitrarily small) output impedance We can refine the simple model for small signal analysis by making a Taylor expansion of the input and output characteristics that we saw earlier. Using V=IR V BE = f I B,V CE BE = f I B V CE b f V CE I B ce h ie is a resistance h re is unit less =h ie b h re ce h fe is unit less h oe is /resistance I C = f 2 I B,V CE C = f 2 I B V CE b f 2 V CE I B ce =h fe b h oe ce 2
13 The transistor model circuit described by these equations is shown in the figure below: Equivalent circuit for a transistor In practice V BE, I C depend only weakly on V CE, in this limit h re and h eo ~ 0 With this approximation we get be =h ie b c =h fe b this should look somewhat familiar, recalling that h fe ~ The transistor model the simplifies to An alternate and more convenient model commonly used is here r e h ie This is called the T equivalent circuit 3
14 EbersMoll Approximation Above we have abandoned simple assumptions V BE ~0.6V,V BE = BE =0 to fully describe a transistor in the active region. More precisely I E and V BE are related by (the diode equation): I E ~I 0 e V /V BE T ~I C where V T kt /q 25 mv at room temperature then I E = I E D.C. V V BE T e =g m BE and BE = e r e where r e = = V T g m I = 25 D.C. in ma at E I room temperature Small signal T equivalent circuit of the transistor looks like:  a current source controlled by b voltage drop across BE junction depends on current flowing through the device EbersMoll illustrated. Logarithmic increase in V BE with increasing I E while in active region. The lower plot zooms in on the effect of I B on the saturation voltage. 4
15 Small signal analysis of the inverting amplifier Assume: R 3 C 2, R R 2 C 2 time intervals of interest i.e. DC blocking (refer to Lab manual pp. 56 for a review of methods to establish proper DC biasing) Equivalent circuit (for small signal analysis): in e r e ' =0 Gain: out = c R C Gain= out = R C R C ' in r e r e R 3 if R 3 c ~ e ) Gain < infinity even when R 3 = 0 2) When R 3 = 0, G = R C /r e Since r e =V T / I E =V T / I E DC e the gain will depend on the amplitude of the signal. One can get a well defined G by setting R 3 >> r e for the entire signal range. But then the maximum gain is limited by the R C constraint namely, V CC I DC C R C /2V CC (for max. symmetric voltage swings at output) 5
16 Grounded emitter amplifier consider the following circuit G= R C /r e Z in =R R 2 r e Z out R C We obtain a large gain but pay the cost with: small Z in large Z out But the more serious problem with this design is in the stability of the DC operating point. Consider the effect of temperature variations: At constant I E ~I C V BE = 2.mV / o C (see text) 2 I At constant V C BE I 0T 2 T /30 o C C As a consequence, a G.E. Initially biased so that V C = 2 V cc will saturate (i.e. V C ~0.6V ) when T =8 o C Also V BE necessary to give a specific I C will vary ~ 00mV for different transistors of the same type The above circuit is a lousy design for these reasons! A stable operating point can be obtained by adding an Choose R, R 2 so I E ~V i.e. V BE C E so r e C e f 3dB The DC bias is now stable and the large gain for AC signals is unaffected. But we may experience significant signal distortion due to variations in r e with emitter current. 6
17 Analysis of grounded CE amplifier with a triangle waveform as input. Output distortions are obvious for changing currents in the transistor. The bottom plot shows how the gain of the circuit changes as the input wave form varies G= out R C R C ma. The gain varies a in r e 25/ I E factor of two over the input signal shape from ~ 300 to almost
18 Simple current source I in = I E~I E I E = V BB 0.6 V EE for negligible I B This circuit is a good current source a long as V C > V BB (small rise I C w/ V CE ) Otherwise, decreases as we approach saturation. of A current source such at the one to the right has very limited compliance due to : Compliance ~ V CC V sat I E V EE A current mirror can increase compliance range to within a few 0ths of a volt of the power supply and also improve Z out. A simple current mirror No to limit compliance of output transistor in this design. V 2 BE =V BE I,2 I 0,2 e qv BE /kt,2 (V BE < 0 for pnp transistor in active region) substituting V BE =kt ln I I 0 I 2 =I 2 I /T 2 0 T I 2 0 I 0 I 0 I { T in the expression for I_2 T 2 ln I I } 0 I 2 'mirrors' or is 'programmed by' I 8
19 Comments on the current mirror: i) I =V CC 0.6/ R P (nearly constant in the active region) then 2 ii) if T =T 2 I 2 = I 0 I I (area scaling current is 'mirrored') 0 iii)thermal dependence: if T 2 T I 2 0 Saturation current grows w/ temp, leading to a I 0 net INCREASE in I C with temperature (for fixed V BE ), net DECREASE in V BE with temperature of ~ 2.mV/ o C for fixed I C. That's not obvious from looking at EbersMoll/diode equation. Ideally Z out should be ~ infinity for a current source. I =0= V R This is not the case. Use the Early Effect to understand the finite Z out. The Early Effect The Early Effect describes I C dependence on V CE It is parameterized as I C =I 0 e V BE /V T V BC V A (for V CB > 0) ~% effect / volt In the mirror above V BE will break the symmetry due to the Early Effect. Z out =[ r o I C V CE] [ I C V BC 2 remains constant with changes in V CB ] =[ I C V A] is related to the hybrid parameter r o Recall: in our T equivalent model we took h oe ~00 ma,2, but differences in V CE I C V CE goes as h oe ~0 as one of our simplifying assumptions. 9 r o
20 The two following methods can be used to increase Z out for the current source, however the cost is a slight reduction in compliance. ()Emitter resistors (neglecting base currents) V CC I C V BE =V B =I C R P I C I P V CC 0.6V R P To find Z out V BE =V B V CC I T C subs. I C =I 0 e V BE/V V BC V A find changes in V BE so be = be I c r e I C V BE be I C V BC bc V A bc E= R be r e r o bc noting that be = c (V B2 is fixed by Q) = r bc Z out = bc =r o o R E c r e c r e The increase above r o seems plausible since any increase in I C would increase V BE thereby decreasing I C. The emitter resistors work as a negative feedback system to help keep the current stable. But compliance is reduced somewhat by addition of to the output transistor. 20
21 (2) Wilson Mirror Both V CE(2) and V CE(3) are fixed (to fight Early effect) Say I C wants to increase due to an increase in V CE(). This will also increase V BE(2) thus increasing I P thus increasing V BE() thus decreasing I C I C is held stable against Early Effect For this circuit Z out =r 0 R P r e Compliance is reduced somewhat by addition of a second diode drop. 2
22 Differential Amplifier Part 3 i) basis of most op amps and comparators ii) good for sensing low level signals in a noisy environment DC analysis of a basic differential amplifier Assume: I C = I E V BE = 0.6V (operate in active region: I C >0, V CB >=0) To find V out in terms of V,V 2 it is sufficient to find I,I 2 KVL (from base to base 2): V V BE I I 2 V BE V 2 =0 I 2 = V V I T where I T =I I 2 V out =V CC I 2 R C To find I T use KVL from base (base2) to V EE V V BE I I T R T =V EE V 2 V BE I 2 I T R T =V EE V I T = V therefore I V 2 2 V V 2 0.6V V EE V 2.2V 2V 2,= EE 2 2 R T 2 R T differential mode common mode define V Dif =V V 2 V CM = 2 V V 2 so I,2 = V dif 2 V CM 0.6V V EE 2 R T large V Dif = large I 2 small V Dif current suppressed by 2 R T Therefore if << R T, Gain is much larger for differential voltages than for common voltages at the two inputs. 22
23 Small signal analysis of a differential amplifier Assumptions: V DC DC =V 2 I DC DC =I 2 V T r e =r e2 = /2 I T Using KVL from base to base 2 b r e b2 r e 2 =0 use b c and 2 = T 2 = 2 2r e 2 T therefore o = 2 R C = 2 Dif R C 2r e Differential Gain G Dif 2 2 CM R C r e 2 R T Common Mode Gain G CM Common Mode Rejection Ratio: CMRR= G Dif = 2 R T r e G CM 2R e r e r e CMRR is large in a good differential amplifier. R T is often replaced by a current source to give maximum common mode rejection. R T 23
24 The CMMR of the differential amplifier may be significantly improved by adding a current source to the tail. Assuming a perfect current source, we again solve for I,2 as a function of V,2 KVL: V V BE I I 2 V BE2 V 2 =0 use I I 2 =I T and I I S e V BE/V T to write: V V 2 I T 2 I 2 I T V T ln I T I 2 =0 this relates I 2 to V dif for the perfect current source (constant I T ), the output ( I 2 ) only depends on V dif. Consider two limiting cases: i) if I T V T (25 mv at room temperature) then the current reduces to: I 2 = V V I Differential gain is similar to above case. T ii) if =0 then = I T e V V 2 /V T I 2 =I T for small V V 2 I 2 Amplifier quickly saturates to one of two states I 2 =0 or 24
25 The Miller Effect Consider a high gain CE amplifier. o b and o 0 (Large dv/dt across C BC) b An intrinsic capacitance connects the base to the collector. At high frequencies current can be diverted from the base to the collector via this capacitive coupling. (This current flow is not related to the DC case of forward biasing the BC junction.) The effects of this are two fold: i) Z in looking into the base is reduced at large frequencies ii) The gain is reduced because of the voltage divider effect of R S and Z in Millers Theorem Consider a 3terminal network: In the equivalent circuit: Z = Z K Z 2 = ZK K K 2 proof: = 2 = Z K Z = Z 2 = 2 = Z 2 / K Z = 2 Z 2 How much current flows through C BC? dv I =C Cap BC =C dt BC b o C CB G b C CB looks like a capacitor that is G times larger (remember G is negative...) 25
26 Analysis of the Miller effect in an amplifier circuit Assume we have an ideal inverting amplifier (Z in =, Z out =0, Gain = A) Applying Miller's theorem, we can redraw the circuit as: K = 2 = A Z C = Z C A = = j C j C A C =C A The negative gain effectively increases the input capacitance. Note: we can ignore C 2 in this analysis, because it's tied to the low Z output of the amplifier and thus has no effect on the signal. Next we compute the overall gain of the amplifier circuit which is the product of the attenuation due to the voltage divider (R S C ) and the gain of the ideal amplifier (A). 2 = A = A s j R S C = s A Gain= j R S C A A A j R S C A j R S C i) In the limit of low source impedance, the gain is simply A as per the design of the amplifier subcircuit. ii) For finite source impedance, the device becomes a perfect integrator as the gain approaches infinity. i.e. for in =e j t out j e j t 26
27 Miller Effect in the Grounded CE Amplifier This is more complex that the ideal case above,because of finite input and output impedances. However, we can simplify matters for nottoolarge values of the input frequency. Begin by calculating K: (KCL) c = L f (Trans. Eqn.) c e = /r e where: L = 2 / R C f = 2 j C CB Then (KCL) becomes r e = 2 R C 2 j C CB K 2 = R C r e j R C C CB j R C C CB K R C r e K K for R C C CB and r e R C i.e. current flow through C CB is small Z C CB R C =Z out output not loaded Therefore the equivalent circuit can be modeled as below (using C =C CB R C /r e C 2 =C CB ) 27
28 Since the impedance looking into the base of the transistor is r e the input signal sees Z in at the base to be Z C r e Z = in Z C S = r e r Z in R S = e S Z C r e R S S R S r e[ j C R S r e ] R 2 = C from our assumption that r e R C C CB Solving for the circuit's gain G= 2 = R C e[ S R S r j R C r e ] C R r CB S e This has the form of a low pass filter with 3dB = R C r C R r CB S e e Notice that for large R S r e 3dB 2 R R C C C C CB CB then (and above derivations are still valid) This means that the Miller Effect becomes the dominant cause of gain loss at high frequencies. Summary: The Miller Effect arises when the following conditions are present: ) large inverting gain 2) nonnegligible output impedance of signal source 3) capacitance between input and output of amplifier circuit 28
29 Designs that avoid the Miller Effect Differential amp w/ one input grounded Cascode connection Input and output have the same phase At base of Q2 Z out =0 (grounded) preventing local Miller effect at Q2 collector resistor for Q is r e2, thus G () ~  (not large) Z out =0 at the base of Q2 29
30 The Bootstrap Technique On the flipside, the Miller Technique may also be used to raise the input impedance of a circuit. In order to provide a stiff DC bias voltage at the base of the transistor in an emitter follower, we required R R 2 0 for a single supply follower like the one show above. From the small signal equiv. circuit we calculated Z in =R Div R Div where R Div 20 This is much less that what is possible for an emitter follower, namely Z in The circuit below has a much larger input impedance Z in ~. Almost no (A.C.) current flows through because both ends are at nearly the same (A.C.) voltage. R L = R div R' B = R L R' ' B = R L r e K 2 = R L R L r e K = r e R L R e r e R L K K = R L r e (large and negative) ' ' = R L r e and R L ' ' R L therefore Z in r e R L 30
Chapter 2.  DC Biasing  BJTs
Chapter 2.  DC Biasing  BJTs Objectives To Understand : Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented
More informationElectronic Circuits Summary
Electronic Circuits Summary Andreas Biri, DITET 6.06.4 Constants (@300K) ε 0 = 8.854 0 F m m 0 = 9. 0 3 kg k =.38 0 3 J K = 8.67 0 5 ev/k kt q = 0.059 V, q kt = 38.6, kt = 5.9 mev V Small Signal Equivalent
More informationChapter 2  DC Biasing  BJTs
Objectives Chapter 2  DC Biasing  BJTs To Understand: Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented
More informationCHAPTER.4: Transistor at low frequencies
CHAPTER.4: Transistor at low frequencies Introduction Amplification in the AC domain BJT transistor modeling The re Transistor Model The Hybrid equivalent Model Introduction There are three models commonly
More informationBipolar Junction Transistor (BJT)  Introduction
Bipolar Junction Transistor (BJT)  Introduction It was found in 1948 at the Bell Telephone Laboratories. It is a three terminal device and has three semiconductor regions. It can be used in signal amplification
More informationBiasing the CE Amplifier
Biasing the CE Amplifier Graphical approach: plot I C as a function of the DC baseemitter voltage (note: normally plot vs. base current, so we must return to EbersMoll): I C I S e V BE V th I S e V th
More informationMod. Sim. Dyn. Sys. Amplifiers page 1
AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance
More informationKOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II )
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II ) Most of the content is from the textbook: Electronic devices and circuit theory,
More informationLecture 7: Transistors and Amplifiers
Lecture 7: Transistors and Amplifiers Hybrid Transistor Model for small AC : The previous model for a transistor used one parameter (β, the current gain) to describe the transistor. doesn't explain many
More informationMod. Sim. Dyn. Sys. Amplifiers page 1
AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance
More informationBJT Biasing Cont. & Small Signal Model
BJT Biasing Cont. & Small Signal Model Conservative Bias Design Bias Design Example Small Signal BJT Models Small Signal Analysis 1 Emitter Feedback Bias Design Voltage bias circuit Single power supply
More informationBiasing BJTs CHAPTER OBJECTIVES 4.1 INTRODUCTION
4 DC Biasing BJTs CHAPTER OBJECTIVES Be able to determine the dc levels for the variety of important BJT configurations. Understand how to measure the important voltage levels of a BJT transistor configuration
More informationDC Biasing. Dr. U. Sezen & Dr. D. Gökçen (Hacettepe Uni.) ELE230 Electronics I 15Mar / 59
Contents Three States of Operation BJT DC Analysis FixedBias Circuit EmitterStabilized Bias Circuit Voltage Divider Bias Circuit DC Bias with Voltage Feedback Various Dierent Bias Circuits pnp Transistors
More informationChapter 13 SmallSignal Modeling and Linear Amplification
Chapter 13 SmallSignal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 131 Chapter Goals Understanding of concepts related to: Transistors
More informationBJT Biasing Cont. & Small Signal Model
BJT Biasing Cont. & Small Signal Model Conservative Bias Design (1/3, 1/3, 1/3 Rule) Bias Design Example SmallSignal BJT Models SmallSignal Analysis 1 Emitter Feedback Bias Design R B R C V CC R 1 R
More informationESE319 Introduction to Microelectronics. Output Stages
Output Stages Power amplifier classification Class A amplifier circuits Class A Power conversion efficiency Class B amplifier circuits Class B Power conversion efficiency Class AB amplifier circuits Class
More informationTransistor amplifiers: Biasing and Small Signal Model
Transistor amplifiers: iasing and Small Signal Model Transistor amplifiers utilizing JT or FT are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly. Then, similar FT
More informationECE342 Test 2 Solutions, Nov 4, :008:00pm, Closed Book (one page of notes allowed)
ECE342 Test 2 Solutions, Nov 4, 2008 6:008:00pm, Closed Book (one page of notes allowed) Please use the following physical constants in your calculations: Boltzmann s Constant: Electron Charge: Free
More informationDevice Physics: The Bipolar Transistor
Monolithic Amplifier Circuits: Device Physics: The Bipolar Transistor Chapter 4 Jón Tómas Guðmundsson tumi@hi.is 2. Week Fall 2010 1 Introduction In analog design the transistors are not simply switches
More informationESE319 Introduction to Microelectronics Common Emitter BJT Amplifier
Common Emitter BJT Amplifier 1 Adding a signal source to the single power supply bias amplifier R C R 1 R C V CC V CC V B R E R 2 R E Desired effect addition of bias and signal sources Starting point 
More informationID # NAME. EE255 EXAM 3 April 7, Instructor (circle one) Ogborn Lundstrom
ID # NAME EE255 EXAM 3 April 7, 1998 Instructor (circle one) Ogborn Lundstrom This exam consists of 20 multiple choice questions. Record all answers on this page, but you must turn in the entire exam.
More informationTutorial #4: Bias Point Analysis in Multisim
SCHOOL OF ENGINEERING AND APPLIED SCIENCE DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING ECE 2115: ENGINEERING ELECTRONICS LABORATORY Tutorial #4: Bias Point Analysis in Multisim INTRODUCTION When BJTs
More informationJunction Bipolar Transistor. Characteristics Models Datasheet
Junction Bipolar Transistor Characteristics Models Datasheet Characteristics (1) The BJT is a threeterminal device, terminals are named emitter, base and collector. Small signals, applied to the base,
More informationHomework Assignment 08
Homework Assignment 08 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance
More informationAC Circuits. The Capacitor
The Capacitor Two conductors in close proximity (and electrically isolated from one another) form a capacitor. An electric field is produced by charge differences between the conductors. The capacitance
More informationOPAMPs I: The Ideal Case
I: The Ideal Case The basic composition of an operational amplifier (OPAMP) includes a high gain differential amplifier, followed by a second high gain amplifier, followed by a unity gain, low impedance,
More informationChapter 5. BJT AC Analysis
Chapter 5. Outline: The r e transistor model CB, CE & CC AC analysis through r e model commonemitter fixedbias voltagedivider bias emitterbias & emitterfollower commonbase configuration Transistor
More informationVI. Transistor amplifiers: Biasing and Small Signal Model
VI. Transistor amplifiers: iasing and Small Signal Model 6.1 Introduction Transistor amplifiers utilizing JT or FET are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly.
More informationCARLETON UNIVERSITY. FINAL EXAMINATION December DURATION 3 HOURS No. of Students 130
ALETON UNIVESITY FINAL EXAMINATION December 005 DUATION 3 HOUS No. of Students 130 Department Name & ourse Number: Electronics ELE 3509 ourse Instructor(s): Prof. John W. M. ogers and alvin Plett AUTHOIZED
More informationEE105 Fall 2015 Microelectronic Devices and Circuits Frequency Response. Prof. Ming C. Wu 511 Sutardja Dai Hall (SDH)
EE05 Fall 205 Microelectronic Devices and Circuits Frequency Response Prof. Ming C. Wu wu@eecs.berkeley.edu 5 Sutardja Dai Hall (SDH) Amplifier Frequency Response: Lower and Upper Cutoff Frequency Midband
More informationEngineering 1620 Spring 2011 Answers to Homework # 4 Biasing and Small Signal Properties
Engineering 60 Spring 0 Answers to Homework # 4 Biasing and Small Signal Properties.).) The inband Thevenin equivalent source impedance is the parallel combination of R, R, and R3. ( Inband implies the
More informationESE319 Introduction to Microelectronics. BJT Biasing Cont.
BJT Biasing Cont. Biasing for DC Operating Point Stability BJT Bias Using Emitter Negative Feedback Single Supply BJT Bias Scheme Constant Current BJT Bias Scheme Rule of Thumb BJT Bias Design 1 Simple
More informationLecture 17. The Bipolar Junction Transistor (II) Regimes of Operation. Outline
Lecture 17 The Bipolar Junction Transistor (II) Regimes of Operation Outline Regimes of operation Largesignal equivalent circuit model Output characteristics Reading Assignment: Howe and Sodini; Chapter
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output.
More informationA twoport network is an electrical network with two separate ports
5.1 Introduction A twoport network is an electrical network with two separate ports for input and output. Fig(a) Single Port Network Fig(b) Two Port Network There are several reasons why we should study
More informationSwitching circuits: basics and switching speed
ECE137B notes; copyright 2018 Switching circuits: basics and switching speed Mark Rodwell, University of California, Santa Barbara Amplifiers vs. switching circuits Some transistor circuit might have V
More informationCHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE
CHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE To understand Decibels, log scale, general frequency considerations of an amplifier. low frequency analysis  Bode plot low frequency response BJT amplifier Miller
More information(Refer Slide Time: 1:41)
Analog Electronic Circuits Professor S. C. Dutta Roy Department of Electrical Engineering Indian Institute of Technology Delhi Lecture no 13 Module no 01 Midband Analysis of CB and CC Amplifiers We are
More informationECE342 Test 3: Nov 30, :008:00, Closed Book. Name : Solution
ECE342 Test 3: Nov 30, 2010 6:008:00, Closed Book Name : Solution All solutions must provide units as appropriate. Unless otherwise stated, assume T = 300 K. 1. (25 pts) Consider the amplifier shown
More informationSOME USEFUL NETWORK THEOREMS
APPENDIX D SOME USEFUL NETWORK THEOREMS Introduction In this appendix we review three network theorems that are useful in simplifying the analysis of electronic circuits: Thévenin s theorem Norton s theorem
More informationForwardActive Terminal Currents
ForwardActive Terminal Currents Collector current: (electron diffusion current density) x (emitter area) diff J n AE qd n n po A E V E V th  e W (why minus sign? is by def.
More informationfigure shows a pnp transistor biased to operate in the active mode
Lecture 10b EE215 Electronic Devices and Circuits Asst Prof Muhammad Anis Chaudhary BJT: Device Structure and Physical Operation The pnp Transistor figure shows a pnp transistor biased to operate in the
More informationElectronic Circuits 1. Transistor Devices. Contents BJT and FET Characteristics Operations. Prof. C.K. Tse: Transistor devices
Electronic Circuits 1 Transistor Devices Contents BJT and FET Characteristics Operations 1 What is a transistor? Threeterminal device whose voltagecurrent relationship is controlled by a third voltage
More informationSemiconductor Physics Problems 2015
Semiconductor Physics Problems 2015 Page and figure numbers refer to Semiconductor Devices Physics and Technology, 3rd edition, by SM Sze and MK Lee 1. The purest semiconductor crystals it is possible
More informationEE105  Fall 2006 Microelectronic Devices and Circuits
EE105  Fall 2006 Microelectronic Devices and Circuits Prof. Jan M. Rabaey (jan@eecs) Lecture 21: Bipolar Junction Transistor Administrative Midterm Th 6:308pm in Sibley Auditorium Covering everything
More informationPHYS225 Lecture 9. Electronic Circuits
PHYS225 Lecture 9 Electronic Circuits Last lecture Field Effect Transistors Voltage controlled resistor Various FET circuits Switch Source follower Current source Similar to BJT Draws no input current
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) Recall the basic twoport model for an amplifier. It has three components: input resistance, Ri, output resistance, Ro, and the voltage gain, A. v R o R i v d Av d v Also
More informationQuick Review. ESE319 Introduction to Microelectronics. and Q1 = Q2, what is the value of V Odm. If R C1 = R C2. s.t. R C1. Let Q1 = Q2 and R C1
Quick Review If R C1 = R C2 and Q1 = Q2, what is the value of V Odm? Let Q1 = Q2 and R C1 R C2 s.t. R C1 > R C2, express R C1 & R C2 in terms R C and ΔR C. If V Odm is the differential output offset
More information3. Basic building blocks. Analog Design for CMOS VLSI Systems Franco Maloberti
Inverter with active load It is the simplest gain stage. The dc gain is given by the slope of the transfer characteristics. Small signal analysis C = C gs + C gs,ov C 2 = C gd + C gd,ov + C 3 = C db +
More informationUNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences
UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EE 105: Microelectronic Devices and Circuits Spring 2008 MIDTERM EXAMINATION #1 Time
More informationE40M Review  Part 1
E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs,
More informationProf. Paolo Colantonio a.a
Prof. Paolo olantonio a.a. 2011 12 The D bias point is affected by thermal issue due to the active device parameter variations with temperature I 1 I I 0 I [ma] V R } I 5 } I 4 } I 3 Q 2 } I 2 Q 1 } I
More informationThe BJT Differential Amplifier. Basic Circuit. DC Solution
c Copyright 010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The BJT Differential Amplifier Basic Circuit Figure 1 shows the circuit
More information13. Bipolar transistors
Technische Universität Graz Institute of Solid State Physics 13. Bipolar transistors Jan. 16, 2019 Technische Universität Graz Institute of Solid State Physics bipolar transistors npn transistor collector
More informationVidyalankar S.E. Sem. III [EXTC] Analog Electronics  I Prelim Question Paper Solution
. (a) S.E. Sem. [EXTC] Analog Electronics  Prelim Question Paper Solution Comparison between BJT and JFET BJT JFET ) BJT is a bipolar device, both majority JFET is an unipolar device, electron and minority
More informationCommon Drain Stage (Source Follower) Claudio Talarico, Gonzaga University
Common Drain Stage (Source Follower) Claudio Talarico, Gonzaga University Common Drain Stage v gs v i  v o V DD v bs  v o R S Vv IN i v i G C gd C+C gd gb B&D v s vv OUT o + V S I B R L C L v gs  C
More informationLecture 37: Frequency response. Context
EECS 05 Spring 004, Lecture 37 Lecture 37: Frequency response Prof J. S. Smith EECS 05 Spring 004, Lecture 37 Context We will figure out more of the design parameters for the amplifier we looked at in
More informationEE 330 Lecture 22. Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits
EE 330 Lecture 22 Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits Exam 2 Friday March 9 Exam 3 Friday April 13 Review Session for Exam 2: 6:00
More information0 t < 0 1 t 1. u(t) =
A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 13 p. 22/33 Step Response A unit step function is described by u(t) = ( 0 t < 0 1 t 1 While the waveform has an artificial jump (difficult
More informationBasic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati
Basic Electronics Prof. Dr. Chitralekha Mahanta Department of Electronics and Communication Engineering Indian Institute of Technology, Guwahati Module: 2 Bipolar Junction Transistors Lecture4 Biasing
More informationNotes for course EE1.1 Circuit Analysis TOPIC 10 2PORT CIRCUITS
Objectives: Introduction Notes for course EE1.1 Circuit Analysis 45 Reexamination of 1port subcircuits Admittance parameters for port circuits TOPIC 1 PORT CIRCUITS Gain and port impedance from port
More informationLecture 35  Bipolar Junction Transistor (cont.) November 27, Currentvoltage characteristics of ideal BJT (cont.)
6.720J/3.43J  Integrated Microelectronic Devices  Fall 2002 Lecture 351 Lecture 35  Bipolar Junction Transistor (cont.) November 27, 2002 Contents: 1. Currentvoltage characteristics of ideal BJT (cont.)
More informationOPERATIONAL AMPLIFIER APPLICATIONS
OPERATIONAL AMPLIFIER APPLICATIONS 2.1 The Ideal Op Amp (Chapter 2.1) Amplifier Applications 2.2 The Inverting Configuration (Chapter 2.2) 2.3 The Noninverting Configuration (Chapter 2.3) 2.4 Difference
More informationECE343 Test 2: Mar 21, :008:00, Closed Book. Name : SOLUTION
ECE343 Test 2: Mar 21, 2012 6:008:00, Closed Book Name : SOLUTION 1. (25 pts) (a) Draw a circuit diagram for a differential amplifier designed under the following constraints: Use only BJTs. (You may
More informationIntroduction to Transistors. Semiconductors Diodes Transistors
Introduction to Transistors Semiconductors Diodes Transistors 1 Semiconductors Typical semiconductors, like silicon and germanium, have four valence electrons which form atomic bonds with neighboring atoms
More informationElectronic Circuits. Bipolar Junction Transistors. Manar Mohaisen Office: F208 Department of EECE
Electronic Circuits Bipolar Junction Transistors Manar Mohaisen Office: F208 Email: manar.subhi@kut.ac.kr Department of EECE Review of Precedent Class Explain the Operation of the Zener Diode Explain Applications
More informationELEC 3908, Physical Electronics, Lecture 18. The Early Effect, Breakdown and SelfHeating
ELEC 3908, Physical Electronics, Lecture 18 The Early Effect, Breakdown and SelfHeating Lecture Outline Previous 2 lectures analyzed fundamental static (dc) carrier transport in the bipolar transistor
More informationAnalog Integrated Circuit Design Prof. Nagendra Krishnapura Department of Electrical Engineering Indian Institute of Technology, Madras
Analog Integrated Circuit Design Prof. Nagendra Krishnapura Department of Electrical Engineering Indian Institute of Technology, Madras Lecture No  42 Fully Differential Single Stage Opamp Hello and welcome
More information6.012 Electronic Devices and Circuits
Page 1 of 1 YOUR NAME Department of Electrical Engineering and Computer Science Massachusetts Institute of Technology 6.12 Electronic Devices and Circuits Exam No. 1 Wednesday, October 7, 29 7:3 to 9:3
More informationCircle the one best answer for each question. Five points per question.
ID # NAME EE255 EXAM 3 November 8, 2001 Instructor (circle one) Talavage Gray This exam consists of 16 multiple choice questions and one workout problem. Record all answers to the multiple choice questions
More informationHomework Assignment 09
Homework Assignment 09 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. What is the 3dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L =
More informationAt point G V = = = = = = RB B B. IN RB f
Common Emitter At point G CE RC 0. 4 12 0. 4 116. I C RC 116. R 1k C 116. ma I IC 116. ma β 100 F 116µ A I R ( 116µ A)( 20kΩ) 2. 3 R + 2. 3 + 0. 7 30. IN R f Gain in Constant Current Region I I I C F
More informationELECTRONICS IA 2017 SCHEME
ELECTRONICS IA 2017 SCHEME CONTENTS 1 [ 5 marks ]...4 2...5 a. [ 2 marks ]...5 b. [ 2 marks ]...5 c. [ 5 marks ]...5 d. [ 2 marks ]...5 3...6 a. [ 3 marks ]...6 b. [ 3 marks ]...6 4 [ 7 marks ]...7 5...8
More informationSpring Semester 2012 Final Exam
Spring Semester 2012 Final Exam Note: Show your work, underline results, and always show units. Official exam time: 2.0 hours; an extension of at least 1.0 hour will be granted to anyone. Materials parameters
More informationElectronics Prof. D C Dube Department of Physics Indian Institute of Technology Delhi
Electronics Prof. D C Dube Department of Physics Indian Institute of Technology Delhi Module No. 07 Differential and Operational Amplifiers Lecture No. 39 Summing, Scaling and Averaging Amplifiers (Refer
More informationCHAPTER 7  CD COMPANION
Chapter 7  CD companion 1 CHAPTER 7  CD COMPANION CD7.2 Biasing of SingleStage Amplifiers This companion section to the text contains detailed treatments of biasing circuits for both bipolar and fieldeffect
More informationDigital Integrated CircuitDesign
Digital Integrated CircuitDesign Lecture 5a Bipolar Transistor Dep. Region Neutral Base n(0) b B C n b0 P C0 P e0 P C xn 0 xp 0 x n(w) b W B Adib Abrishamifar EE Department IUST Contents Bipolar Transistor
More informationSmallSignal Midfrequency BJT Amplifiers
SmallSignal Midfrequency JT Amplifiers 6.. INTRODUTION For sufficiently small emittercollector voltage and current excursions about the quiescent point (small signals), the JT is considered linear; it
More information6.301 SolidState Circuits Recitation 14: OpAmps and Assorted Other Topics Prof. Joel L. Dawson
First, let s take a moment to further explore device matching for current mirrors: I R I 0 Q 1 Q 2 and ask what happens when Q 1 and Q 2 operate at different temperatures. It turns out that grinding through
More informationChapter 9 Bipolar Junction Transistor
hapter 9 ipolar Junction Transistor hapter 9  JT ipolar Junction Transistor JT haracteristics NPN, PNP JT D iasing ollector haracteristic and Load Line ipolar Junction Transistor (JT) JT is a threeterminal
More information(e V BC/V T. α F I SE = α R I SC = I S (3)
Experiment #8 BJT witching Characteristics Introduction pring 2015 Be sure to print a copy of Experiment #8 and bring it with you to lab. There will not be any experiment copies available in the lab. Also
More informationElectronic Circuits for Mechatronics ELCT 609 Lecture 2: PN Junctions (1)
Electronic Circuits for Mechatronics ELCT 609 Lecture 2: PN Junctions (1) Assistant Professor Office: C3.315 Email: eman.azab@guc.edu.eg 1 Electronic (Semiconductor) Devices PN Junctions (Diodes): Physical
More informationE2.2 Analogue Electronics
E2.2 Analogue Electronics Instructor : Christos Papavassiliou Office, email : EE 915, c.papavas@imperial.ac.uk Lectures : Monday 2pm, room 408 (weeks 211) Thursday 3pm, room 509 (weeks 411) Problem,
More informationChapter 10 Instructor Notes
G. izzoni, Principles and Applications of lectrical ngineering Problem solutions, hapter 10 hapter 10 nstructor Notes hapter 10 introduces bipolar junction transistors. The material on transistors has
More informationITT Technical Institute ET215 Devices I Unit 1
ITT Technical Institute ET215 Devices I Unit 1 Chapter 1 Chapter 2, Sections 2.12.4 Chapter 1 Basic Concepts of Analog Circuits Recall ET115 & ET145 Ohms Law I = V/R If voltage across a resistor increases
More information(Refer Slide Time: 03:41)
Solid State Devices Dr. S. Karmalkar Department of Electronics and Communication Engineering Indian Institute of Technology, Madras Lecture  25 PN Junction (Contd ) This is the 25th lecture of this course
More informationUniversity of Pennsylvania Department of Electrical and Systems Engineering ESE 319 Microelectronic Circuits. Final Exam 10Dec08 SOLUTIONS
University of Pennsylvania Department of Electrical and Systems Engineering ESE 319 Microelectronic Circuits Final Exam 10Dec08 SOLUTIONS This exam is a closed book exam. Students are allowed to use a
More informationAs light level increases, resistance decreases. As temperature increases, resistance decreases. Voltage across capacitor increases with time LDR
LDR As light level increases, resistance decreases thermistor As temperature increases, resistance decreases capacitor Voltage across capacitor increases with time Potential divider basics: R 1 1. Both
More informationL03: pn Junctions, Diodes
8/30/2012 Page 1 of 5 Reference:C:\Users\Bernhard Boser\Documents\Files\Lib\MathCAD\Default\defaults.mcd L03: pn Junctions, Diodes Intrinsic Si Q: What are n, p? Q: Is the Si charged? Q: How could we make
More informationBJT  Mode of Operations
JT  Mode of Operations JTs can be modeled by two backtoback diodes. N+ P N N+ JTs are operated in four modes. HO #6: LN 251  JT M Models Page 1 1) Forward active / normal junction forward biased junction
More informationTransistors  a primer
ransistors  a primer What is a transistor? Solidstate triode  threeterminal device, with voltage (or current) at third terminal used to control current between other two terminals. wo types: bipolar
More informationTime Varying Circuit Analysis
MAS.836 Sensor Systems for Interactive Environments th Distributed: Tuesday February 16, 2010 Due: Tuesday February 23, 2010 Problem Set # 2 Time Varying Circuit Analysis The purpose of this problem set
More informationECE PN Junctions and Diodes
ECE 342 2. PN Junctions and iodes Jose E. SchuttAine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu ECE 342 Jose Schutt Aine 1 B: material dependent parameter = 5.4 10
More informationLecture 15: MOS Transistor models: Body effects, SPICE models. Context. In the last lecture, we discussed the modes of operation of a MOS FET:
Lecture 15: MOS Transistor models: Body effects, SPICE models Context In the last lecture, we discussed the modes of operation of a MOS FET: oltage controlled resistor model I curve (SquareLaw Model)
More informationFinal Examination EE 130 December 16, 1997 Time allotted: 180 minutes
Final Examination EE 130 December 16, 1997 Time allotted: 180 minutes Problem 1: Semiconductor Fundamentals [30 points] A uniformly doped silicon sample of length 100µm and crosssectional area 100µm 2
More informationElectronic Circuits. Prof. Dr. Qiuting Huang Integrated Systems Laboratory
Electronic Circuits Prof. Dr. Qiuting Huang 6. Transimpedance Amplifiers, Voltage Regulators, Logarithmic Amplifiers, AntiLogarithmic Amplifiers Transimpedance Amplifiers Sensing an input current ii in
More informationPhysics 364, Fall 2012, reading due your answers to by 11pm on Thursday
Physics 364, Fall 2012, reading due 20120920. Email your answers to ashmansk@hep.upenn.edu by 11pm on Thursday Course materials and schedule are at http://positron.hep.upenn.edu/p364 Assignment: This
More informationProf. Anyes Taffard. Physics 120/220. Voltage Divider Capacitor RC circuits
Prof. Anyes Taffard Physics 120/220 Voltage Divider Capacitor RC circuits Voltage Divider The figure is called a voltage divider. It s one of the most useful and important circuit elements we will encounter.
More informationI. Frequency Response of Voltage Amplifiers
I. Frequency Response of Voltage Amplifiers A. CommonEmitter Amplifier: V i SUP i OUT R S V BIAS R L v OUT V Operating Point analysis: 0, R s 0, r o >, r oc >, R L > Find V BIAS such that I C
More informationDEPARTMENT OF ECE UNIT VII BIASING & STABILIZATION AMPLIFIER:
UNIT VII IASING & STAILIZATION AMPLIFIE:  A circuit that increases the amplitude of given signal is an amplifier  Small ac signal applied to an amplifier is obtained as large a.c. signal of same frequency
More information