The voltage reference: Last stop before voltage regulators - Part 3 of An introduction to analog circuit design*
Positive TC of ΔVbe
From equation :
ΔVbe = VT*ln(x) = (k*T/q)ln(x) 
Taking the derivative with respect to temperature we have:
d/dT(ΔVbe)= k*q*ln(x) =[(k*T/q)/T]ln(x)= ΔVbe/T 
Normalizing to the amplitude of ΔVbe we have the expression for the incremental temperature variation of ΔVbe:
(1/ΔVbe)*d/dT(ΔVbe) = 1/T = 1/300 °C-1 
namely, the ΔVbe variation in temperature, normalized to its amplitude is 0.33%/°C positive. For example if we apply  re-written as d/dT(ΔVbe) = ΔVbe/T, a ΔVbe of 18mV will have a temp variation of 18m/300=+ 0.6mV/°C and a ΔVbe of 600mV will have a temp variation of 600m/300= +2mV/°C.
Negative TC of Vbe
The Vbe as a well known negative temperature coefficient (TC):
d/dT(Vbe) = (Vbe- Vbg)/T + 3VT/T = "2mV/°C 
or in relative terms for Vbe = 0.6V:
(1/ Vbe)*d/dT(Vbe) = "2mV/0.6V= "1/300 
Comparing  with  we have:
(1/ Vbe)*d/dT(Vbe) = (1/ ΔVbe)*d/dT(ΔVbe) 
namely the relative variations of Vbe and ΔVbe are identical in value and opposite in sign. This property is at the basis of the design of temperature in dependent circuits.
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Table 1 (a) and (b) formulas describe the equal in amplitude and opposite in sign temp behavior of the Vbe and ΔVbe. ! (c) combines the two formulas into one.
In conclusion since ΔVbe and the Vbe have opposing behavior in temperature, equal amplitudes of each summed up will always lead to a resulting voltage with null temp coefficient.
Build a ΔVbe
Now let's see how we can build practical circuits that can mix up ΔVbe and Vbe. As a first step we will build a circuit that behaves like a ΔVbe. To this end let's repeat for convenience the expression of the Vbe:
Vbe = VT*ln(I/Io) 
Io is proportional to the emitter area such that:
Hence two transistors of different areas, carrying different currents, will have different Vbes as follows:
Vbe = VT*ln(I/kA) 
Vbe' = VT*ln(I'/kA') 
ΔVbe = Vbe'-Vbe= VT*ln[(I/I')(A'/A)] 
Setting I/I'=x and substituting into  we have:
ΔVbe = VT*ln(x*A'/A) 
For example is A'/A=10 and the two transistors carry the same current (x=1) then ΔVbe = 26mV*ln10=60mV.
In Figure 13 the two transistors T1 and T2 are set the same current I1 = I2 = 100 µA, where I1 in T1 is set by the current source I and I2 is set by the Vbe coupling of the two transistors in conjunction with the area ratio: I2 = ΔVbe/R2 = 60mV/600Ω = 100 µA. The voltage across R3 will be (R3/R2)* ΔVbe and since R2/R3 is 6Kï-/600Ω = 10, the drop across R3 is 10*60mV=600mV. This ΔVbe' voltage is actually an amplified ΔVbe and thus has all the properties of the ΔVbe including is positive TC.
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In conclusion, Figure 13 shows a circuit that produces a 600mV voltage with positive temperature coefficient of ΔVbe.
Building a voltage reference
Adding the ΔVbe' voltage in figure 13 to a proper Vbe value " as described below in more detail- should do the trick, namely produce a voltage that is invariant to temperature, a so called 'reference' voltage, fundamental to any servo control mechanism. The result is the circuit in figure 14. It should be intuitive that matching of T1 and T2 is critical and best obtained if the two transistors see (are biased to) the same collector voltage. Since the collector voltage of T1 is equal to its base voltage (base and collector of T1 are shorted) it follows that the best voltage for collector of T2 is an analogous of Vbe1. Since the collector of T2 is connected to the base of T3 we will need to make the base-emitter voltage of T3, Vbe3, identical to Vbe1. By physically constructing T3 identical to T1 and biasing it to the same current value (100μA) its Vbe will indeed be virtually identical to Vbe1 fulfilling the above matching criteria.
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In figure 14 the Vbe (600mV) of T3 is summed up to the ΔVbe' (600mV) of resistor R3 to add up to a temperature invariant voltage of 1.2V at the Vout node, namely Vref = Vbe + ΔVbe' = 1.2V This Vout is temperature invariant and its value is equal to the band-gap of the silicon. We can then write that:
Vref = Vbg =Vbe + ΔVbe = 1.2V 
The analysis above is correct and a good lead to design voltage references. However it is a bit of an oversimplification. In reality any voltage reference circuit will have some slight dependence on temperature. A plot of Vref over temp is slightly curved and that curve will generally exhibit a true dVout/dT=0 at only one temperature point, typically at ambient temperature for a well done design. The circuit above, yielding a voltage equal to the silicon band-gap is referred to as a band-gap voltage reference. This particular implementation is also called a Widlar voltage reference, after its inventor. A band-gap voltage reference can yield TC flatness in the order of 50ppm/°C easily.
Fractional band-gap voltage reference
Naturally all the terms in equation 30 can be divided by any number higher or lower than one, leading to voltage references that are correspondently lower than (fractional) and higher than (multiple) the Vbg. If k is the dividing factor we can write:
V'ref = Vbg /k = Vbe/k + ΔVbe/k
For example, for k=4 we have:
V'ref = Vbg/4 = 300mV = 150mV + 150mV
The circuit in Figure 14 can be easily modified like the one in Figure 14a to produce a Vbe drop of 150mV (Vbe3*R2/R4) and a ΔVbe drop of 150mV (ΔVbeT1-T2*R2/R3). Sum the two drops (300mV drop across R2) and we first come up with a 300mV fractional band-gap voltage drop floating across R2. Then this drop is shifted down to the output by T4 (the drop across R5 is identical to the drop across R2 if T4 and T3 are identically biased). Notice also that here Vcc can be as low as 1V, leaving room for the 300mV reference, 660mV Vbe4 and some minimum headroom for the current source.
For more implementation details see for example patent number 4,628,247 (for example go to www.uspto.gov and search by the patent number or author name) by this author.
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