What would you like to do?
What are some soures of bias?
Sour grapes aren't ripe and haven't produced the sugar that makes them sweet yet. Also, wine grapes tend to be more sour and table grapes tend to be sweeter. Wine grapes a…ren't necessarily intended to be eaten and table grapes are.
not so much sour, its kind of got a tangy taste. you should just try some.
lemon, lime, sour warheads, and grapefruit
The Bureau of Indian Affairs or the BIA is an organization created to improve the quality of life for Native Americans.
being against a diffrent race that you are.
Some sour fruit could be green grapes, green apples, and mabey a grape fruit that you ate to early.
Because they're body isn't use to sour milk. An if their memory cells had antioxidants to fight off it....then they would'NT be harmed by it. For ex. people in south America a…nd in Africa are more prone to live through malaria...because they are used to it....and they live around it. Their body has adapted to it. But Americans will most likely die....unless you just got here and your raelly from Africa or etc. lol. hope I've anserwed your Q the way you expected!!!!
on the side of your tounge you have taste buds that have hydrogen ions causing acids to relase and give off a sour taste.
Temperature Compensation Of BJT Differential Amplifiers The bipolar junction transistor (BJT) emitter-coupled differential-pair circuit is a familiar amplifier stage in the re…pertoire of analog designers, but has asurprising obscurity that needs to be revealed. This TechNote examines the emitter-circuit current source, I0, of BJT diff-amps and the effects on amplifier gain of different implementations for it. The widespread belief that a BJT current source can temperature-compensate the diff-amp is true, but the conditions for it do not appearto be widely known, based on most designs. The typical circuitis shown below. +V RL1 RL2 vo1 vo2 vi1 RB1 Q1 RB2 Q2 vi2 RE1 RE2 I0 This is a differential-input, differential-output voltageamplifier. Both input and output quantities are differential, and the incremental gain of the circuit is: v v - v A × v - A × v v v v A = o = o 2 o1 = v 2 i2 v1 i1 = o 2 - o1 = vi = -a 2 × vi 2 - vi1 vi 2 - vi1 RL 2 vi 2 v = 0 i1 vi1 v = 0 i2 re1 + re 2 + RE1 + RE 2 + RB1 /(b1 + 1) + RB 2 /(b 2 + 1) R - - a1 × L1 re1 + re 2 + RE1 + RE 2 + RB1 /(b1 + 1) + RB 2 /(b 2 + 1) The condition for differential amplification is that Av1 = Av2. The circuit is made symmetrical for: RE = RE1 = RE2 ; RB = RB1 = RB2 ; RL = RL1 = RL2 ; b = b1 = b2, Q1, Q2 matched Then the gain simply becomes: Av = 2 × Av1 = 2 × Av 2 = -a × re + RE RL + RB /(b + 1) = -a × RL rM A goal of good design is to make Av a fixed value. The choice of resistors with a low temperature coefficient (TC) and sufficiently tight accuracy is one factor. This is usually easy to achieve, though for high-precision design, the change in resistance due to change in ambient temperature is afactor to be considered. Even more so are thermals, changes in resistance due to changes in power dissipation with vi. For very precise designs, the change in resistance with applied voltage must be considered too. Other transistorparameters than thetwo (re and b) of the BJT T model used here - namely, ro - also need to be included for precision design. Wewill assume that theBJTs have a sufficiently high Early voltage that ro need not come into our list of considerations -- at least not here. In practice, this assumption is often valid. BJTs are typically the least ideal elements of the circuit. From the gain formula, it is evident that two BJT parameters affect gain, theincremental emitter resistance, re, and b. For high b -- that is, for b >> 1 -- the gain factor, a = b b + 1 approaches 1. For a typical b value of 200, then a = 0.995, contributing a gainerror of 0.5%. If that is too much error, a-compensationtechniques are required. Usually, this error can be compensated by including it in the gain formula, as we have done. What is more important is how much it drifts with temperature. Typical TC(b) @ 1%/°C, then for large b, a has nearly a zero TC of around 50 ppm; a is not much of a problem. The transresistance expression of Av -- the denominator -- is the resistance across which the input voltage develops the common (emitter) current. Theoutput current is modified bya, which accounts for loss along the way from theemitter circuit. This transresistance, rM, also includes b in the RB term. If RB is kept small, and the inputs are driven by voltage sources, then this b is of no concern. If the sources are high in resistance, then the RB term will affect gain by b variation with temperature. Its 1%/°C variation is scaled down by the extent to which RB/(b + 1) is not dominant in rM. Thus, keepingthis term negligible is another design factor. The most troublesome term in rM is re, for it varies with temperature and emitter current, IE, according to: re = VT | I E | = kT / qe | I E | @ 26 mV , T = 300 K | I E | Thus re varies with the thermal voltage, VT, which varies in proportion with absolute temperature: dVT = VT dT T At 300°K (about 80°F), this is 1/300°K or about 0.33%/°K = 0.33%/°C. For laboratory-quality instrument design, let us suppose the temperature range over which the equipment ought to be able to operate within its specifications is 15°C from room temperature, about 25°C ± 15°C, or 10°Cto 40°C. Over a 15°C change from ambient, VT changes about 5% -- far too much for most precision designs. Therefore, VT variation in gain needs to be compensated. The simplest compensation for re is to make it a negligible term (along with the RB term) in rM. This is accomplished by making RE dominant. ForRE >> re, the drift in re affects gain far less than 5%. In many cases, dominant external emitter resistance solves the drift problem, but at the expense of gain and power dissipation. By increasing I0, then re is reduced proportionally, though circuit power increases. This is undesirable for power-limited equipment and it also exacerbates thermals by increasing DPD(vi) in the BJTs. In some cases, re cannot be made negligible, and some compensation for it is desired. One of the most common schemes is to make I0 track re and cancel its effect, at least approximately. To make I0 have the TC of VT, the simplest approach is to use a BJT current source implementation of I0. The b-ejunction voltage of the current-source BJT then decreases with temperature, I0 increases, and decreases re. Current-Source Circuits The first circuit sourcing I0 that we will consider is simply a resistor, R0, returned to a negative supply. This long-tailed current source approaches an ideal current source as the supply voltage, -V, approaches negative infinity, along with the value of R0. It does nothing to compensate for the TC of re. The second implementation to consider is shown below. I0 Q0 R0 -V This simple circuit has a voltage acrossR0 of V - VBE(Q0). As temperature, T, increases, VBE decreases, but not with the TC of VT. The other major BJT parameter affecting VBE is the saturation current, IS, as found in thep-n junction (b-ejunction) voltage equation: VBE = VT × ln I C , I C ( ) >> I S I S T For a typical BJT, such as a PN3904, IS ≈ 10-14 A. Then 1 mA of current produces a VBE @ 0.65 V. Both VT and IS contribute to the TC(VBE). IS has a greater effect than VT and of opposite polarity on VBE, resulting in a combined effect of about -2 mV/°C for VBE. (For more on BJT TC effects, see the volume, Signal-Processing Circuits of Analog Circuit Design, available at http://www.innovatia.com.) It is therefore more important to cancel IS effects than those of VT. Depending on the relative values of V and VBE, the effect of the TC(VBE) can be scaled by choice of RE and supply voltage, V, which is often constrained bysystem- level design. By adding a resistor network between the emitter and ground, a Thévenin equivalent supply voltage and value of R0 can be independently set. Ifscaled properly, as Tincreases, VBE decreases and I0 increases. If the increase is made to besuch that the reduction in re due to it cancels the increase in re due to VT, then re and gain remain constant. The TC(re) is calculated as follows, bydifferentiating re with respect to T: I dre = d VT = 1 × dVT - VT × dI E = VT × 1 × dVT - VT × 1 × dI E = r × [TC%(V ) - TC%(I )] E dT dT I E I E dT 2 dT I E VT dT I E I E dT e T E where, TC% is the fractional change in TC. The TC%(I0) = TC%(IE) can be determined as follows. The only change across R0 is due to VBE. Therefore, thefractional change in I0 with T is: TC%( I0 ) = dI 0 / dT I 0 - (dV / dT) / R = BE 0 I 0 = - dVBE / dT V - VBE = + 2 mV/°C V - VBE For TC%(I0) = TC%(VT) = 1/T @0.33%/°C, and the voltage across R0, V - VBE = 0.6 V. With −V = −1.25 V, this is not an attractivecompensation scheme. Thepolarity of TC(I0) is correct for compensation, but not its magnitude, which leads to the next scheme, shown below. I0 R1 Q0 R2 R0 -V This implementation of I0 is more versatileand more common in occurrence than theprevious scheme. The base divider provides extra freedom forsetting TC%(I0) which, for ignored TC(b), is now: R dV 2 ×V - VBE - BE TC%(I ) = 1 × dI 0 = 1 × d R1 + R2 = 1 × dT 0 I 0 dT I 0 dT R0 + (R1 || R2 ) /(b + 1) I 0 R0 + ( R1 || R2 ) /(b + 1) dV dV - BE 1 × - BE = dT = VBE dT = - TC%(VBE ) R R V R V BE 2 ×V - V R1 + R2 2 × R1 + R2 - 1 VBE 2 × R1 + R2 - 1 VBE The divider ratio that gives the correct compensation can now be found. When TC%(I0) is set equal to × TC%(VT), then: R2 V = - TC%(VBE ) + 1 = + 2 mV/°C/ VBE + 1 = 0.6 V + VBE or, R1 + R2 VBE TC%(VT ) 0.33 %/°C VBE R2 R1 + R2 = 0.6 V + VBE V @ 1.25 V V for VBE = 0.65 V. This result is interesting; whatever the value of V, the unloaded divider voltage must be 1.25 V for gain compensation. This is also the voltage of bandgap references, as well it should be. Bandgap circuits use the negative TC(VBE) and scale it to cancel thepositive TC(VT). When this is done the resulting bandgap voltage always comes out to be close to 1.25 V, and varies slightly with BJT doping levels. Another current-source variation that is often used to provide rough temperature compensation is to insert a diode in series with R2, as shown below. I0 R1 Q0 D1 R2 R0 -V The usual explanation is that the TC of the diode compensates for the TC of the BJT b-ejunction, resulting in a more stable I0. A typical example is to use a 1N4152 diode to compensate a PN3904. The diode and BJT b-ejunctions are quite different, however. The diode doping levels are far less than even the BJT base, in order to achieve a higher breakdown voltage. The emitter minority carrier concentration is made intentionally large for good emitter injection efficiency into thebase, at the expense of VBE reverse breakdown, which is typically around 7 V, much below the 60 V of the diode. The point is that although both junctions are silicon, they are rather unmatched. Suppose, however, that a similar BJT is used as a diode, with base connected to collector. Then the junction matching is better (though not as good as adjacent integrated BJTs), and allowing for a @ 1, then applying KVL around the BJT input loop: I × R - V × ln(I / I ) V × ln I 0 = I T I D × R2 - I 0 × R0 , or, I = D 2 T 0 D 0 R D 0 where, ID is the diode current. If the junction currents are equal, the TC dueto VT is removed and the TC%(I0) @ 0%/°C. This is useful for applications where a stable current source is needed, but it does not compensate re of the diff-amp. The currents must deliberately be set unequal to achieve the desired TC, and for a compensating polarity of TC, it must be positive. Consequently, we must have ID > I0. The TC%(I0) is found through implicit differentiation of I0 in the above equation: 1 dI 1 d I I dV TC%( I ) = × 0 = × - V × ln 0 - ln 0 × T 0 T I 0 dT V I 0 × R0 dT 1 I I D I D dT = - T × TC%( I ) + × ln 0 0 I 0 × R0 T I D With additional algebraic manipulation: (V / I × R ) × ln(I / I ) TC%( I 0 ) = - T 0 0 0 D T × (1 + VT / I 0 × R0 ) Then to compensate, set TC%(I0) = TC%(VT) = 1/Tand solve: I I × R D = exp1 + 0 0 I 0 VT Because of the exponentialfunction, practical current ratios require that the voltage across R0 be not much larger than VT. For I0 = 2 mA, R0 = 22 W, and VT = 26 mV, then the voltage across R0 is 44 mV, or 1.69×VT, andID = 14.77×I0 = 29.5 mA, larger than is desired in most designs. Such small values of R0 are required to keep R0 from dominating the emitter circuit so that the TC of VBE can be expressed.Yet in many designs, R0 is relatively large and its voltage drop far exceeds VT. As a consequence, the TC%(VT) of re is not correctly compensated and a TC drift in gain exists. The previous scheme, which omitted the base diode, was only slightly better in allowing for larger R0 voltage. Perhaps we should go in the oppositedirection and add a diodeor two in the emitter.The TC of the combined junctions would bemultiplied by the number of them, and that would allow RE to be made proportionally larger. It is usually not desirable to add a large number of series diodes because the static (dc) stability of I0 is not benefited. Consequently, use of the diff-amp current source to temperature-compensatere results in a circuit requiring careful static design. I0 is then made sensitive to junction parameters, and these parameters, such as IS, have a somewhatwide tolerance amongdiscrete transistors,even of the same part number.Expect as much as 50 mV of variation among PN3904 BJTs at the same current and temperature. This compensation method is best suited for monolithic integration. Closure The widespread belief that a BJT current source can temperature-compensate a BJT diff-amp is true, but it often does not. Temperature compensation of I0 for a constant re results in low voltage across the current- source external emitter resistance, R0 -- so low that it can make accurate setting of I0 infeasible. Consequently, except for more elaborate schemes which amplify VT, the dominant-RE approach to diff-amp gain stability appears to still be the best.
Sour cream is produced by fermentation and the "sour" applies to its taste. It can "go sour" (spoil or go bad), which is a different situation.
No. The jurors had to swear an oath to be neutral, impartial, and judge the case on the facts presented alone.
When your umpiring or referee your own team you can be bias to them.
biased :) <3 Apex Answered by the Jarizzle :)
That's because food containing acids are usually sour, and others with bases are bitter.