Bandgap Circuit

The bandgap circuit shown in figure 1 produces a voltage, Vbg, to a first order, temperature and supply independent and approximately equal to the silicon bandgap voltage of 1.2 Volts. The voltage divider formed by R4 and R5 multiplies Vbg to produce higher voltages at Vo.


Figure 1   Band gap voltage reference outout is independent of temperature and Vcc.

The current mirror, P1, P2, acts to hold I1=I2.

I1 = n Is exp[Vbe1/VT] = I2 = Is exp[Vbe2/VT]

n exp[Vbe1/VT] = exp[Vbe2/VT]

VTln[n] = Vbe2 - Vbe1 = R2I1

I = I1 = VT ln[n]/R2

(1)

Vbg = R12 I + Vbe2

Vbg = R12 VTln[n]/R2 + Vbe2

(2)
where VT = KT/q = 0.026V at T = 300 oK

VT = 8.62 X 10-5T

VT is proportional to temperature. Vbe decreases with temperature at about -2mV/oC. The first term in equation 2 increases with temperature and the second term decreases with temperature. These changes compensate each other when changes with temperature are minimized. Taking the derivative of equation 2 with respect to temperature and setting it equal to zero and rearranging terms,
2 (R1/R2) ln[n] = .002/8.62 X 10-5

R1/R2 = 11.6/ln[n]

If n = 4

R1/R2 = 8.4

Example
If n = 4 the voltage across R2 is about 36mV. If R2 equals 450 ohms, I will equal 80 uA and R1 should equal about 3.7K. The drop across R1 is 2 I R1 = 2*80X10-6*3.7X103 = 0.6V.

The bandgap voltage, Vbg, is 0.6 + Vbe2 = 0.6 + 0.65 = 1.25V