KCC Quizzes AQQ302 about Bang Gap Reference Generation

1. I₁ must be larger than I₂ so that the ln(I₁/I₂) is positive which keeps PTAT>0 thus compensating the CTAT.

2. 86μ[ln(I₁/I₂)] = 1.8m => ln(I₁/I₂) = 20.93

            I₁/I₂ = e^20.93 = 1.23x10⁹

3. 86μ[ln(3)] x G = 1.8m => G = 19.05

We need the slope wrt to T of 86E-6*T*ln(I1/I2)=.0018.   Thus

ln(I1/I2) = 0.0018/86E-6 = 20.93.  Then I1/I2 = e^(20.93) = 1.23E9.   While doable in these days of Amps and nAmps on the same chip, this is a ridiculous ratio.  

So G*ln(I1/I2) to the rescue to reduce that factor of 20.93. 

G = 20.93/ln(3)  = 19.05.  Easy to do with an opamp and some well matched resistors. 

Q1 .

If  I1 > I2, ln(I1/I2) is positive and VBE1-VBE2 is positive to compensate the fall of Vbe with T.

Q2.

For an increase of temperature of 1K, Delta Vbe =k/q Ln(I1/I2).1= 86 10^-6.Ln(I1/I2) witch must compensate -1,8 mV. So, we have:

Ln(I1/I2)= 0.0018/0.000086 ≈ 20.93 so I1/I2=e^20.93 ≈ 1229941640.1 

Q3.

If I1/I2 =3, the gain G of the amplifier is 20.93/Ln(3) ≈ 19.0515

Q1.  I1 > I2 because Ln(I1/I2)  must be +ve.  Real band gap references don't vary the current - they vary the current density by increasing the junction area.

Q2.  Take the differential dVbe/dT and solve for equivalence with  1.8 mV (noting that the Boltzmann constant is m^2 kg S^-2 K-1).  1.23E9

Q3.   Solve for Ln(3) * Gain = 20.89.  Gain ~19.

The theory is simple but I assure you that if you have ever had to make either a discrete or integrated band gap reference then this is not so simple. 

I1 must be larger,  it must be 21, the gain would be 7

Yes! Congrats  sparky4347@gmail.com !

Bingo kevin.soch@analog.com ! Congratulations!

Right answers and right reasoning lhmgma@gmail.com, big applause!

Well done  ob1@aquacoustics.biz ! Thanks

Thanks david.taylor@electradx.com for your feedback! But not sure I understand your reasoning: how did you come from 21 to the gain of 7?