# The voltage reference: Last stop before voltage regulators - Part 3 of An introduction to analog circuit design*

**Positive TC of ΔV _{be} **

From equation [6]:

ΔV_{be} = VT*ln(x) = (k*T/q)ln(x) [18]

Taking the derivative with respect to temperature we have:

d/dT(ΔV_{be})= k*q*ln(x) =[(k*T/q)/T]ln(x)= ΔV_{be}/T [19]

Normalizing to the amplitude of ΔV_{be} we have the expression for the incremental temperature variation of ΔV_{be}:

(1/ΔV_{be})*d/dT(ΔV_{be}) = 1/T = 1/300 °C^{-1} [20]

namely, the ΔV_{be} variation in temperature, normalized to its amplitude is 0.33%/°C positive. For example if we apply [20] re-written as d/dT(ΔV_{be}) = ΔV_{be}/T, a ΔV_{be} of 18mV will have a temp variation of 18m/300=+ 0.6mV/°C and a ΔV_{be} of 600mV will have a temp variation of 600m/300= +2mV/°C.

**Negative TC of V _{be}**

The V_{be} as a well known negative temperature coefficient (TC):

d/dT(V_{be}) = (V_{be}- V_{bg})/T + 3VT/T = "2mV/°C [21]

or in relative terms for V_{be} = 0.6V:

(1/ V_{be})*d/dT(V_{be}) = "2mV/0.6V= "1/300 [22]

Comparing [20] with [22] we have:

(1/ V_{be})*d/dT(V_{be}) = (1/ ΔV_{be})*d/dT(ΔV_{be}) [23]

namely the relative variations of V_{be} and ΔV_{be} are identical in value and opposite in sign. This property is at the basis of the design of temperature in dependent circuits.

**Table 1. V**

_{be}and ΔV_{be}temperature dependency.
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Table 1 (a) and (b) formulas describe the equal in amplitude and opposite in sign temp behavior of the V_{be} and ΔV_{be}. ! (c) combines the two formulas into one.

In conclusion since ΔV_{be} and the V_{be} have opposing behavior in temperature, equal amplitudes of each summed up will always lead to a resulting voltage with null temp coefficient.

**Build a ΔV _{be} **

Now let's see how we can build practical circuits that can mix up ΔV_{be} and V_{be}. As a first step we will build a circuit that behaves like a ΔV_{be}. To this end let's repeat for convenience the expression of the V_{be}:

V_{be} = VT*ln(I/Io) [24]

Io is proportional to the emitter area such that:

Io=kA [25]

Hence two transistors of different areas, carrying different currents, will have different V_{be}s as follows:

V_{be} = VT*ln(I/kA) [26]

V_{be}' = VT*ln(I'/kA') [27]

And differentiating:

ΔV_{be} = V_{be}'-V_{be}= VT*ln[(I/I')(A'/A)] [28]

Setting I/I'=x and substituting into [28] we have:

ΔV_{be} = VT*ln(x*A'/A) [29]

For example is A'/A=10 and the two transistors carry the same current (x=1) then
ΔV_{be} = 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 V_{be} coupling of the two transistors in conjunction with the area ratio: I2 = ΔV_{be}/R2 = 60mV/600Ω = 100 µA. The voltage across R3 will be (R3/R2)* ΔV_{be} and since R2/R3 is 6Kï-/600Ω = 10, the drop across R3 is 10*60mV=600mV. This ΔV_{be}' voltage is actually an amplified ΔV_{be} and thus has all the properties of the ΔV_{be} including is positive TC.

**Figure 13. ΔV**

_{be}' circuit.
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In conclusion, Figure 13 shows a circuit that produces a 600mV voltage with positive temperature coefficient of ΔV_{be}.

**Building a voltage reference**

Adding the ΔV_{be}' voltage in figure 13 to a proper V_{be} 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 V_{be}1. Since the collector of T2 is connected to the base of T3 we will need to make the base-emitter voltage of T3, V_{be}3, identical to V_{be}1. By physically constructing T3 identical to T1 and biasing it to the same current value (100μA) its V_{be} will indeed be virtually identical to V_{be}1 fulfilling the above matching criteria.

**Figure 14. Voltage reference.**

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In figure 14 the V_{be} (600mV) of T3 is summed up to the ΔV_{be}' (600mV) of resistor R3 to add up to a temperature invariant voltage of 1.2V at the Vout node, namely Vref = V_{be} + ΔV_{be}' = 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 =V_{be} + ΔV_{be} = 1.2V [30]

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 = V_{be}/k + ΔV_{be}/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 V_{be} drop of 150mV (V_{be}3*R2/R4) and a ΔV_{be} drop of 150mV (ΔV_{be}T1-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 V_{cc} can be as low as 1V, leaving room for the 300mV reference, 660mV V_{be}4 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.

**Figure 14a. Fractional band-gap voltage reference.**

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**Next:**

**Voltage regulators**
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