Presentation on theme: "Differential Amplifiers and Integrated Circuit (IC) Amplifiers"— Presentation transcript:
1 Differential Amplifiers and Integrated Circuit (IC) Amplifiers Chapter 7Differential Amplifiers and Integrated Circuit (IC) Amplifiers
2 Discrete and Integrated Circuits A discrete circuit is constructed of components that are manufactured separately. Later, these components are connected together, by conductors like wires, in a circuit board or a printed circuit board (PCB).On the other hand, in an integrated circuit, the components and their inter-connections are manufactured concurrently by a sequence of processing steps.The types of components that are available and their practical values depend heavily on the approach taken for implementation. (for example, the capacitors in discrete circuits can be in the range of 1pF to 1F, but only 1pF to 100pF in ICs. Also, inductors are almost impractical in ICs. But in ICs, matching of components is much easier)See table 7.1 in page 412 for detail.Applications of discrete circuits will persist especially for some special circuits that are to be mass produced, but today the bulk of electronic systems are based on ICs.Processing steps in manufacturing ICs incur cost and failures and are usually different for different technologies.BJT technology are used more for high-quality analog circuits, while MOS are more for general analog circuits and digital circuits. Today, semiconductor industry can manufacture both BJT and MOS on the same chip, called BiCMOS technology.
3 Figure 7.1 The current mirror. DC biasing for Integrated CircuitsDifferent from biasing of discrete circuits, resistors and capacitors are “expensive” in terms of cost and chip area, are therefore avoided whenever possible.For amplifier circuits, the BJT transistors operate in active region.The following circuits show how matched transistors, when combined with a few resistors, can act as current sources that are useful in biasing IC amplifiers.Collector of Q1 is connected to its base. Thus , and Q1 is in the active region. If is larger than 0.2V, Q2 is also in the active region.See page , it can be shown thatFigure 7.1 The current mirror.
4 Figure 7.1 The current mirror. DC biasing for Integrated Circuits IITo a first-order approximation, the base current of Q2 is independent of the output voltage , therefore the output characteristics is almost identical to one of the collector characteristic curves for Q2.An important specification of a current source is the range of output voltage for which the output current is approximately constant, which is called compliance range.Another important specification of a current source is its dynamic output resistance, which is the ratio of the incremental voltage divided by the incremental output current (ideally it should be infinite).In small-signal equivalent circuit, the current source is replaced by its dynamic resistance.Figure 7.1 The current mirror.
5 Figure 7.2 Emitter follower with bias current source. Biasing an Emitter FollowerAn example of how the current mirror can help establish the bias point of an IC amplifier is shown below.The current source is formed by R, Q1 and Q2, while Q3 is an emitter follower amplifying the input signal and delivering it to the load.Often, we can simplify the circuit diagram as in Figure7.2(b).Note:(1) the amplifier is direct-coupled compared to AC-coupling in discrete amplifiers(2) Output voltage is -0.7V for input voltage of zero. In this case, the circuit displays a DC offset, which is not desirable. This problem can be solved or reduced by the circuit shown in the next slide.Figure 7.2 Emitter follower with bias current source.
6 Biasing an Emitter Follower: reducing offset A simple way to reduce offset for this follower is to cascade a second stage consisting of a pnp emitter follower as shown in the figure below.Note that in discrete circuits, offset is not an issue as a coupling capacitor is used.Figure 7.3 The offset voltage can be reduced by cascading a complementary (pnp) emitter follower.
7 Effects of transistor area on current mirror Doubling the area of a transistor is the same as connecting two of the original transistors in parallel, as shown in the Figure 7.4.The output current of a current mirror for which the relative junction areas of the transistors are A1 and A2 is given byStudy example 7.1 in page 418.Figure 7.4 Doubling the junctionarea of a BJT is equivalent to connectingtwo of the original BJTs in parallel.Figure 7.5 Current mirror for Examples 7.1.Figure 7.9 Collector characteristic of Q2, illustrating the Early voltage.
8 Figure 7.7 Output characteristic for the current mirror of Figure 7.5. Figure 7.8 Dynamic output resistance of thecurrent mirror of Figure 7.5.
9 The Wilson current source An improved circuit, called Wilson current source, with higher output impedance that the previous current mirror is shown in the Figure.For the Wilson current source, the following holds:A1, A3 are the relative junction areas of the Q1 and Q3 respectively. (see page 421)Figure The Wilson current source,which has a high output resistance.
10 The Widlar current source When the desired current is small, the Widlar current source may be a better alternative, as shown in the Figure.For Widlar current source, the following holds (see page 422):See example 7.3 in page 423.Figure The Widlar current source,which is useful for small currents.
11 Figure 7.12 Typical biasing circuit for a bipolar IC. The combined current sourcesIn an Integrated Circuit amplifier, several current sources use the same reference current, as shown below.The current through R1 is the reference current for all four current sources. Q1, Q2 forms a current mirror, and Q1, Q3 forms a Widlar source. Notice the pnp current source by Q4, Q5 and Q6.Figure Typical biasing circuit for a bipolar IC.
12 Figure 7.15 NMOS current mirror. IC biasing with MOSFETThe BJT current sources have counterparts constructed with MOSFET.The shown MOSFET current mirror is very similar to the BJT current mirror.In typically cases, the MOSFET M1 operates in saturation region, as drain-to-gate voltage is zero.Assuming the transistors are identical and that the output voltage is large enough so that M2 is in saturation as well. The currentBy using devices with different W/L ratios, circuits having output current equal to a predetermined constant times reference current can be designed.Figure NMOS current mirror.
13 Figure 7.16 NMOS Wilson current source. IC biasing with MOSFETAn improved current source is shown below, which has higher output resistance than the simple current mirror.The output current is related to the reference current by the equation below as well (assuming the transistors operating in saturation region):The reference current I1 may be approximated byFigure NMOS Wilson current source.
14 Figure 7.22 Basic BJT differentiial amplifier. Emitter-coupled differential pairThe emitter-coupled differential pair is a very important circuit that is used many bipolar analog integrate circuits.The circuit is shown in the figure and the two transistors are assumed identical. The current source IEE is typically implemented as a current source circuit discussed before (eg. Current mirror, wilson current source).The input voltages vi1 and vi2 can be considered to be composed of a differential signal vid and a common mode signal vicm defined below:Differential output voltage is defined asFigure Basic BJT differentiial amplifier.
15 Figure 7.23a Basic BJT differential amplifier with waveforms. Emitter-coupled differential pair IIFirst, consider the two input signal vi1 and vi2 are equal. Then the differential input voltage vid is 0 and we have a pure common-mode input signal.In this case, the current IEE splits equally between the Q1 and Q2, therefore vod=0.In other words, the circuit does not respond to the common-mode component of the input.Figure 7.23a Basic BJT differential amplifier with waveforms.
16 Figure 7.23b Basic BJT differential amplifier with waveforms. Emitter-coupled differential pair IIIFor a pure differential input (when vicm=0), it can be shown the a non-zero differential output voltage vod is resulted, as a differential input signal steers IEE twoard one side or the other.In summary, the circuits rejects common-mode input and responds to the differential input. In amplifiers, a small differential input signal is amplified to a differential output signal.Figure 7.23b Basic BJT differential amplifier with waveforms.
18 Figure 7.25 Collector currents versus differential input voltage. Signal transfer characteristics ISee page 437 for derivation of the collector current for the emitter-coupled differential pair.The following collector current versus differential input voltage can be obtained.Note that in the plot, when vid=0, ic1=ic2. For vid>5VT, the current is steered almost entirely through Q1 and similarly when vid<-5VT, the current is entirely through Q2.Figure Collector currents versus differential input voltage.
19 Signal transfer characteristics II Using the previous equation of , one can findA plot of this transfer characteristics is shown in the following figure. The curvatureShows that the differential amplifier can distort a signal if the amplitude is too large.For input voltage less VT, the characteristics is quite straight giving linear gain.Figure Voltage transfer characteristic of the BJT differential amplifier.
20 Emitter degenerationSometimes it is advantageous to add emitter generation resistor REF to the circuit, as shown in the Figure.There resistors have the disadvantage of reducing the differential voltage gain of the circuit. However, two reasons for this is to increase input impedance and to reduce distortion due to the nonlinearity of the BJTs.The right figure shows the transfer characteristic of the differential amplifier (REF=40VT/IEE).Figure Differential amplifier with emitterdegeneration resistors.Figure Voltage transfer characteristic withemitter degeneration resistors. REF = 40(VT/IEE).
21 Balanced versus single-ended outputs The output of a differential amplifier can be balanced, in which case the output voltages from both collectors are connected to the inputs of another differential amplifier.On the other hand, the output can be taken from one collector, in which case we say the output is single-ended. If a single-ended output is desired, there is no need for a resistor in the collector of the other resistor. (resistor at collector of Q1 omitted as shown).Figure Either a balanced or single-ended output is available\break from the differential amplifier.
22 The current mirror as a load The following figure shows a variation of the emitter-coupled pair in which the collector resistors are replaced by a current mirror.This circuit is particularly favored in ICs, as transistors are much cheaper than resistors.A simple analysis by assuming large so that base currents of Q3 and Q4 are neglected, results in the equation as follows:For is approximately proportional to vid. Notice furthermore that the common-mode input component does not affect the output current.Figure Emitter-coupled pairwith current-mirror load.
23 Small-signal analysis of the Emitter-coupled differential pairc Using small-signal analysis, we can derive expressions for voltage gain, input impedance and output impedance of the emitter-coupled differential pair.The small-signal equivalent circuit for the differential pair is shown below by replacing the transistors by their small-signal models.Note that power supply has been shorted to GND in small-signal circuit.Also note that the IEE current source is replaced by a resistance REB in the small-signal circuit, as practical current sources has a finite output impedance.Figure Small-signal equivalent circuitfor the differential amplifier of Figure 7.27.(REB is the output impedance of the current source IEE.)
24 Figure 7.34 Half-circuit for a differential input signal. Small-signal analysis: differential inputFirst, we analyze the circuit for a pure differential input signal. Therefore the input voltage are vi1=-vi2=vid/2.The analysis can be simplified by observing that the equivalent circuit is symmetrical. Due to this symmetry and opposite polarity of the independent sources, the voltage at point J is zero. The circuit behavior would not change by shorting point J to Ground.We can then consider only the left-hand side circuit as shown in the Figure. We need to analyze only this half circuit as the right half is the same except different polarity.Figure Half-circuit for a differential input signal.
25 Figure 7.34 Half-circuit for a differential input signal. Small-signal analysis: differential input IIThe half circuit, we then find out the gain and input impedance:Notice that we have defined Rid as the ratio of the entire differential voltage vid to the input current. Thus, Rid is the input impedance between the input terminals of the complete circuit.The voltage gain is:For output impedance, we have:Figure Half-circuit for a differential input signal.
26 Small-signal analysis: common-mode input When the input voltage are vi1=vi2=vicm, the equivalent circuit is depicted in the figure. We have shown the output impedance of the current source as the parallel combination of two resistors.The equivalent circuit is symmetrical with respect to the dashed line including the polarities of the signal sources. Therefore, we conclude that current iJ must be zero.As such, we can open the connection and consider only left or right hand half circuit.Figure Small-signal equivalent circuit with a pure common-mode input signal.
27 Figure 7.36 Half-circuit for a pure common-mode input signal. Small-signal analysis: common-mode inputFrom the half circuit, we can then compute the gain, input impedance and output impedance.Note that we have defined the common-mode input impedance to be the voltage divided by the total current the source must deliver to both terminals.The gain from a single-ended load to common-mode input is:As vo1=vo2=vocm.For output impedance, we have:Figure Half-circuit for a pure common-mode input signal.
28 Small-signal analysis: CMRR In amplifier circuits, it is often desirable to reject common-mode signal while amplifying the differential signal.A measure of how well the amplifier rejects the common-mode signal relative to the differential signal is the common-mode rejection ratio (CMRR). By definition, the CMRR is ratio of the gain for the differential signal to the gain for common-mode signal.From results of previous results, the CMRR for the single-ended output and balanced output can be defined respectively as follows:It can be seen that CMRR is nearly independent of To increase CMRR, it is desired to select a larger value for REB and a small REF.
29 Figure 7.38 Addition of emitter followers to increase input impedance. Amplifier design: how to increase input impedance?Figure Addition of emitter followers to increase input impedance.
30 An design example for high CMRR Figure Differential amplifier ofExample7.4 using the Wilson current source.Figure First attempt in Example 7.4.
31 the differential amplifier of Example 7.4. Figure 7.42a Waveforms forthe differential amplifier of Example 7.4.
32 Figure 7.43 Source-coupled differential amplifier. The source-coupled differential pairUsing MOSFET, we can construct an source-coupled differential pair, which is a counterpart of the emitter-coupled differential pair using BJTs.The main advantage of using MOSFET for a differential pair compared to BJTs is the nearly infinite input impedance, while the disadvantage is lower gain magnitude.Figure Source-coupled differential amplifier.
33 The source-coupled differential pair II Assuming the two MOSFETs are the same.The analysis of the source-coupled differential pair proceeds in the same way as the emitter-coupled differential pair for both common-mode signal and differential input signal.The transfer characteristics for drain current Id1 and Id2 are shown in the figure.Figure Differential output voltage versusnormalized input voltage.Figure Drain currents versusnormalized input voltage.
34 (Note: RSB is the output resistance of the bias current source I.) The source-coupled differential pair IIIThe small-signal equivalent circuit fir the source coupled differential pair is shown in the figure.The power supply is replaced by a short circuit and the resistance RSB represents the output impedance of the bias current source.The circuit can be analyzed for differential and common-mode input signal in almost the same way as the emitter-coupled differential pair discussed before.Refer to results in Table 7.3 in page 462.Figure Small-signal equivalent circuit for the source-coupled amplifier of Figure 7.43.(Note: RSB is the output resistance of the bias current source I.)
35 An example of IC amplifier: MOS The advantage of the amplifier is that it can be fabricated on the same chip as CMOS logic circuits.The PMOS M8, M1 and M2 form a dual current mirror that supplies bias currents to the amplifier stages.The resistor Rset is selected to yield the desired reference current.The input stage consists of transistors M3 and M4, which forms a source-coupled differential pair. Note that Q-point currents on M3 and M4 are approximately equal to Iset.Transistors M5 and M6 form a current mirror load for the input stage.Transistor M7 is a common-source amplifier and M2 acts like a active load.Figure CMOS op amp.
36 An example of IC amplifier II: MOS The small-signal equivalent circuit for the output stage is shown below.Note that the in the model we treated the Ccomp as a open circuit and included the drain-source small-signal resistance rd.Transistor M2 forms the output device of the a current mirror and its gate-to-source voltage is pure DC with no signal component. In other words, the small signal vgs2=0, therefore the current source id2=gm2vgs2 is zero as well. Then we obtainFigure Small-signal equivalent circuit for the output stage consisting of M7 and M2.
37 An example of IC amplifier III: MOS Next, we need to analyze the source-coupled pair to find its differential voltage gain. The small-signal equivalent circuit is shown below.Note that source terminals of M3 and M4 are connected to ground as the circuit is symmetrical.Assuming the current on the small-signal rd resistors are small compared to the current source, we can derive the following: (refer to page )Figure Small-signal equivalent circuit for the output stage consisting of M7 and M2.
38 Figure 7.53 Open-loop gain versus frequency for the CMOS op amp.
39 An example of IC amplifier: BJT Q1, Q2 form a differential amplifier balanced outputs, Q3, Q4 form a differential amplifier with single-ended outputs, Q5 is a pnp emitter amplifier an emitter resistance R6, Q6 is an emitter follower, and finally Q7, Q8, Q9 form a double current mirror.Figure A BJT op amp.
40 Equivalent circuit for the first stage of Figure 7.55. Figure REB represents the output impedance of current sink Q8.Ri = is the differential input impedance of the second stage.