TRANSISTORS: DYNAMIC CIRCUITS. Introduction
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1 TRANSISTORS: DYNAMIC CIRCUITS Introdution The point of biasing a iruit orretly is that the iruit operate in a desirable fashion on signals that enter the iruit. These signals are perturbations about the bias point, for instane, you might bias your input port at 2V, and then add a 50 mv peak-to-peak sine wave to this bias voltage. Ideally, amplifiers are linear devies, meaning the output signal is some multiple of the input signal, independent of the input amplitude. Transistors are non-linear devies (reall their I-V harateristis), so the output amplitude does depend upon the input amplitude. However, by suitably restriting the input swing, the resultant output will show very little urvature, meaning that the non-linear iruit ats approximately linear for small-signal deviations about the bias point. In this lab, the small-signal transistor model is introdued, and this forms the basis for understanding the dynami performane of several ommonly enountered iruits.
2 Theory Small-Signal Model for the Bipolar Transistor The small-signal model for the bipolar transistor is as follows: r b C µ r base olletor ontat i ontat + βi = v r π C π g m v r o _ r e emitter ontat Fig. 1 Small-signal model for the bipolar transistor. The key element is the ontrolled urrent soure, whih an be shown as depending on the internal base urrent i or the internal base-emitter voltage v. The quantity g m is defined as 1 ƒi ƒ v be, signifying how responsive the olletor urrent is to hanges in the driving voltage v. Resistor r e is the small resistane assoiated with the highly-doped emitter, and is usually negleted in analysis when one is foused on dominant effets. Resistor r b is a distributed, non-linear resistane, and thus hard to haraterize with a single value, but it orresponds to the resistane between the base ontat and that region of the base material lying underneath the emitter. Resistor r likewise is hard to haraterize with a single value, but represents the net resistane between the olletor ontat and the bottom portion of the base material. Resistor r π is not a physial resistane like the preeding three, but rather a dynami quantity defined as 1 ƒi ƒ b v be, representing how resistant the input base urrent is to hanges in the internal base-emitter voltage (i.e., that voltage not inluding the voltage drop aross rb, represented as v in Fig. 1). The ontrolled soure indiates how muh the olletor urrent hanges for a hange in base urrent (or equivalently, base-emitter voltage). Like r π, resistor r o is a dynami resistane; it represents the influene of hanges in olletor-emitter voltage on olletor urrent, and thus is given by 1 ƒi ƒ v e. For high Early voltages, this resistane is negligible, and thus the olletor voltage has a negligible impat on the urrent flowing out of the olletor ontat. C µ represents the depletion apaitane of the base-olletor juntion.
3 C π is omposed of two parts: 1) a diffusion apaitane given by ƒqb = ƒv ƒi τ F = τ F ƒv be g m and 2) a depletion apaitane, whih is usually negligible ompared to the diffusion apaitane sine the baseemitter juntion is forward-biased. To develop numerial values for the symbols in the small-signals model, the defining derivatives must be evaluated symbolially, and ast in terms of bias quantities. With the bias quantities speified, numerial values may be assigned to eah small-signal parameter. Deriving the symboli expressions for the smallsignal parameters is a straightforward exerise, and left as a prelab exerise. Common-Emitter Amplifier In the previous lab, the ommon-emitter amplifier was biased, but no mention was made of why it is alled an amplifier. To answer this, we analyze the iruit in the previous lab, with a few modifiations. First, we need to feed our input signal into the base, without the bias points of the signal soure and the ommonemitter amplifier interfering with eah other. This is aomplished by adding an AC oupling apaitor to the input port, large enough so that it will at like an AC short at the frequenies at whih we operate. Seond, the amplifier needs to drive a resistive load whih we don t want to upset our bias point, so we append another AC oupling apaitor to the output port. Third, we replae the transistor symbol used in the previous lab with the small-signal model, leading to the following iruit: C in r b C µ r C out v out i v s R b1 R b2 r π C π βi r o R R load r e + R ee Fig. 2 Common-emitter amplifier for AC analysis. Assuming that we are at low-enough frequenies that the parasiti apaitanes of the BJT don t affet our results, and further negleting the Early resistane and internal emitter resistane, analysis of our model leads to: v v out s = r b β + ( Rload R ) rπ + ( β + 1) R e (1)
4 With large beta, this redues to R R > R long as ( load ) e R load R R e, a rather simple independent of transistor parameters. As, the transfer funtion has a magnitude greater than 1, explaining why the ommon-emitter is alled an amplifier. For the values derived in the previous lab, this requires that R load be at least 375Ω for gain. Wilson Current Soure Another type of urrent soure is the Wilson urrent soure, shown below: V R I ref I out Q 3 Q 1 Q 2 Fig. 3 Wilson urrent soure. This urrent soure has two advantages, namely, it gives very good mathing between output and referene urrents, and it provides very high output resistane. Analysis of both the mathing and output resistane are left as prelab exerises. The advantage of high output resistane is that the output urrent is insensitive to whatever load is attahed to it. Thought of another way, if you onsider the urrent soure as a parallel ombination of a Norton urrent and Norton resistane, and your load as a pure resistane, then the resultant urrent divider is suh that nearly all of the Norton urrent flows into the load resistane, with very little lost to the internal resistane of the urrent soure.
5 MOS inverter Another ommonly enountered struture is the MOS inverter, shown below: M 2 V in V out M 1 Fig. 4 MOS inverter. The idea here is very simple, namely, that when the input voltage is high (i.e., lose to the supply voltage), the output voltage is low (i.e., near ground) and vie-versa. The analysis of this involves several steps. Start by onsidering the input at 0 volts. Between 0 volts and the threshold voltage of M 1, M 1 is utoff; this implies that the urrent is zero, whih happens when M 2 is in triode, with its drain-soure voltage equal to zero. Thus, the output is at the supply voltage for 0 < V in < V t1. One V in goes beyond V t1, M 1 enters saturation, while M 2 remains in triode. As the input voltage inreases further, both transistors beome saturated, while further inreases ause M 1 to enter triode while M 2 remains saturated, and ultimately, M 2 is utoff when the input voltage exeeds V t2. A formal analysis involving the non-linear MOS I-V harateristi is left as a prelab exerise.
6 Prelab Exerises 1) Given the definitions of the small-signal parameters for the bipolar transistor, express r π, g m, and r o in terms of bias parameters (e.g., olletor urrent, V e, et.). For a 1mA urrent and V e of 3.2V (just as in the preeding lab) as well as an Early voltage of 200 and a Beta of 100, what are the values of r π, g m and r o? 2) Use SPICE to simulate the low-frequeny gain of the ommon-emitter amplifier in Fig. 2 (don t enter the small-signal model use an atual transistor!). If you have no SPICE.model ard for a bipolar transistor, use the default SPICE model. Use the.op ommand to verify the bias urrents and voltages as well as the transistor g m for your iruit, then perform an.a analysis to observe the frequeny response. Plae a large (e.g., 0.1 uf) apaitor aross resistor R e. What happens to the gain? Do a hand analysis and SPICE simulation to onfirm your response. 3) Express the output urrent of the Wilson urrent soure in terms of the referene urrent. Do not neglet base urrents in your analysis, and assume eah transistor has the same β. Your final answer should be in terms of the referene urrent and β. (Hint: your final answer should look like 2 I out I ref ( 1 error), where the error depends inversely on β.) Do a small signal analysis to determine a symboli expression for the output resistane of the Wilson urrent soure. Assume every transistor has idential small-signal parameters. Assume you desire you an output urrent of 1 ma, and your supply is 5V. What is the approximate size of the required referene resistor? 4) Determine symboli V out vs. V in equations for the MOS inverter for eah of the five regions disussed. Assume the supply is 5V, the NMOS devie has V tn = 1.4V, K n (W/L) = A/V 2 and λ = 0.02V -1, and the PMOS devie has V tp = -1.2V, K p (W/L) = A/V 2 and λ = 0.02 V -1 and determine the values of V in and V out at eah of the four transitions between the five regions of operation. Plot the V out vs. V in harateristi based on your equations. Verify the hand analysis by plotting the V out vs. V in harateristi in SPICE using the same power supply and MOS parameters.
7 Lab Exerises 1) Build the ommon-emitter amplifier of Fig. 2. Determine an appropriate input signal level so that the output signal is large enough to measure, but small enough so that the transistor ats linearly. Determine the low-frequeny gain at 10 khz. Does it math reasonably well with your hand results? Change the input signal level by +/- 6-dB. Does the iruit behave linearly? Plae a 0.1µF apaitor aross Re and re-measure the AC gain (you may need to adjust your input signal amplitude one more to ensure linear behavior). Does this math your predition as well? At really low frequenies, the gain rolls off. Why is this? At high frequenies, the gain diminishes as well. What auses this derease? 2) Build the Wilson urrent soure disussed in this lab, using the referene resistor value you alulated in the prelab. Determine the urrent through this resistor (it may differ from 1 ma depending on the V be values of the transistors), and then measure the output urrent (use a DVM with several digits auray). What is the perentage error? How does this ompare to the urrent soures of the previous lab? Based on the expression you determined for the error, determine a value for the β of the transistors. 3) Build the CMOS inverter. Step the input DC voltage from 0 to 5V, and reord the orresponding output voltage. At what voltage are the input and output equal. Why might this voltage not be exatly 2.5V, i.e., halfway between ground and the power supply? Change from the ground-to-5v system to a +/-2.5V system, and apply a square-wave input of 5V peak-to-peak. Observe the input and output waveform s on the osillosope, and ompare the two as you inrease the input frequeny.
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